NEET Physics Class 12 Chapter 3 Current Electricity Notes

Current Electricity

1. Electric Current

The time rate of flow of charge through a cross-sectional area is called Current.

if Δq charge flows in time interval Δt then the average current is given by

⇒ \(\mathrm{I}_{\mathrm{av}}=\frac{\Delta \mathrm{q}}{\Delta \mathrm{t}} \text { and }\)

Instantaneous current

⇒ \(\mathrm{i}=\lim _{\Delta \mathrm{t} \rightarrow 0} \frac{\Delta \mathrm{q}}{\Delta \mathrm{t}}=\frac{\mathrm{dq}}{\mathrm{dt}} .\)

The direction of current is along the direction of the flow of positive charge or opposite to the direction of flow of negative charge. But the current is a scalar quantity.

SI unit of current is ampere and

1 Ampere = 1 column/sec

1 coulomb/sec = 1A

It is a scalar quantity because it does not obey the law of vectors.

2. Conductor

In some materials, the outer electrons of each atom or molecule are only weakly bound to it. These electrons are almost free to move throughout the body of the material and are called free electrons. They are also known as conduction electrons. When such a material is placed in an electric field, the free electrons drift in a direction opposite to the field. Such materials are called conductors.

3. Insulator

Another class of materials is called insulators in which all the electrons are tightly bound to their respective atoms or molecules. Effectively, there are no free electrons. When such a material is placed in an electric field, the electrons may slightly shift opposite to the field but they can’t leave their parent atoms or molecules and hence can’t move through long distances. Such materials are also called dielectrics.

4. Semiconductor

In semiconductors, the behavior is like an insulator at low-temperature levels. But at higher temperatures, a small number of electrons can free themselves and they respond to the applied electric field. As the number of free electrons in a semiconductor is much smaller than that in a conductor, its behavior is in between a conductor and an insulator and hence, the name semiconductor. A free electron in a semiconductor leaves a vacancy in its normal bound position.

These vacancies also help in conduction.

Current, velocity, and current density

n → no. of free charge particles per unit volume

q → charge of each free particle

i → charge flow per unit time

i = nqvA

Current density, a vector, at a point, has a magnitude equal to current per unit normal area at that point and direction is along the direction of the current at that point.

⇒ \(\vec{J}=\frac{\mathrm{di}}{\mathrm{ds}} \overrightarrow{\mathrm{n}}\)

⇒ \(\mathrm{di}=\overrightarrow{\mathrm{J}} . \mathrm{d} \overrightarrow{\mathrm{s}}\)

Current is the flux of current density.

Due to the principle of conservation of charge:

Charge entering at one end of a conductor = charge leaving at the other end, so current does not change with change in cross section and the conductor remains uncharged when current flows through it.

Solved Examples

Example 1. Find free electrons per unit volume in a metallic wire of density 10⁴ kg/m³, atomic mass number 100, and number of free electrons per atom is one.
Solution :

Number of free charge particles per unit volume

\((n)=\frac{\text { total free charge particle }}{\text { total volume }}\)

The number of free electrons per atom means total free electrons = a total number of atoms.

⇒ \(=\frac{N_A}{M_W} \times M\)

⇒ \(n=\frac{\frac{N_A}{M_W} \times M}{V}=\frac{N_A}{M_W} \times d=\frac{6.023 \times 10^{23} \times 10^4}{100 \times 10^{-3}}\)

n = 6.023 × 10²⁸ m^-3

Example 2. What will be the number of electrons passing through a heater wire in one minute, if it carries a current of 8 A?
Solution :

\(I=\frac{n e}{t} \quad \mathrm{n}=\frac{I \mathrm{t}}{\mathrm{e}}=\frac{8 \times 60}{1.6 \times 10^{-19}}=3 \times 10^{21}\) Electrons

Example 3. An electron moves in a circle of radius 10 cm. with a constant speed of 4 × 10⁶ m/sec. Find the electric current at a point on the circle.
Solution :

Consider a point A on the circle. The electron crosses this point once in every revolution. The number of revolutions made by an electron in one second is

⇒ \(\mathrm{n}=\frac{\mathrm{v}}{2 \pi \mathrm{r}}=\frac{4 \times 10^6}{2 \pi \times 10 \times 10^{-2}}=\frac{2}{\pi} \times 10^7 \text { rot. } / \mathrm{s}\)

⇒ \(\text { Current } I=\frac{n e}{t}=\frac{2}{\pi} \times 10^7 \times 1.6 \times 10^{-19}\)

(∴ t = 1 s)

⇒ \(=\frac{3.2}{\pi} \times 10^{-12} \cong 1 \times 10^{-12} \mathrm{~A}\)

Example 4. The current through a wire depends on time as i =(2 + 3t)A. Calculate the charge crossed through a cross-section of the wire in 10 s.
Solution :

⇒ \(\mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}} \quad \mathrm{r} \quad \mathrm{dq}=(2+3 \mathrm{t}) \mathrm{dt}\)

⇒ \(\int_0^{10}(2+3 t) d t \quad \mathrm{r} \quad q=\left(2 t+\frac{3 t^2}{2}\right)_0^{10}\)

⇒ \(q=2 \times 10+\frac{3}{2} \times 100=20+150=170 \mathrm{C}\)

Example 5. Current through a wire decreases uniformly from 4 A to zero in 10 s. Calculate the charge flown through the wire during this interval of time.
Solution:

Charge flown = average current × time

⇒ \(\left[\frac{4+0}{2}\right] \times 10=20 C\)

Current Electricity Multiple Choice Question And Answers

Question 1. The expression for Ohm’s law in terms of electric field E and current density J is:-

1. E = (σJ)1/2
2. J = σ/E
3. J = σE
4. σ= (J/E)1/2

Answer: 3. J = σE

Question 2. A potential difference V is applied across a copper wire of diameter d and length l. When d is doubled, the drift velocity:-

1. Increases two times
2. decreases \(\frac{1}{2} \) times
3. Does not change
4. Decreases \(\frac{1}{4} \) times

Answer: 3. Does not change

Question 3. Through a tube of radius R, 10, 000 l-particles pass per minute. The value of electric current through the tube is:-

1. 0.5 x 10^-12A
2. 2 x 10^-12A
3. 0.5 x 10^-16A
4. 2 x 10^-16A

Answer: 3. 0.5 x 10^-16A

Question 4. Two rods A and B made up of the same metal have the same length. The ratio of their resistance is 1:2 if these wires are immersed in water then loss in weight will be:-

1. More in A
2. More in B
3. Same in A and B
4. In the ratio 1: 2

Answer: 1. More in A

Question 5. For two wires A and B of the same material and the same mass, the radius of A doubles that of B. If the resistance of wire A is 34 ohm then that of B will be:-

1. 544 ohm
2. 272 ohm
3. 68 ohm
4. 17 ohm

Answer: 1. 544 ohm

Question 6. A square rod of aluminum of length 1 m and length of the side of cross-sectional surface 5 x 10^-3 m will have a resistance (Resistivity of aluminum 2.8 x 10^-8 ohm-meter):-

1. 1.24 x 10^-4Ω
2. 2.42 x 10^-3Ω
3. 1.12 x 10^-3Ω
4. 11.2 x 10^-3Ω

Answer: 3. 1.12 x 10^-3Ω

Question 7. The following graph shows the relation between the voltage and the current for the temperatures T₁ and T₂ in a metal wire. Then the relation between T₁ and T₂ is:-

1. T₁ = T₂
2. T₁ > T₂
3. T₁ < T₂
4. data insufficient

Answer: 3. T₁ < T₂

Question 8. If the temperatures of iron and silicon wires are increased from 300C to 500C, the correct statement is:-

1. Resistance of both wires increase
2. Resistance of both wires decrease
3. The resistance of iron wire increases and the resistance of silicon wire decreases
4. The resistance of iron wire decreases and the resistance of silicon wire increases

Answer: 3. Resistance of iron wire increases and the resistance of silicon wire decreases

Question 9. For a cell terminal P.D. is 2.2V when the circuit is open and reduces to 1.8V when the cell is connected to a resistance of R = 5Ω. Determine internal resistance of cell (r):-

1. \(\frac{10}{9} \Omega\)
2. \(\frac{9}{10} \Omega[\)
3. \(\frac{11}{9} \Omega\)
4. \(\frac{5}{9} \Omega\)

Answer : 1. \(\frac{10}{9} \Omega\)

Question 10. Krichoff;s Ist law based on:-

1. Energy conservation
2. Charge conservation
3. Current conservation
4. None

Answer: 2. Charge conservation

Question 11. A car battery of emf 12 V and internal resistance 5 × I 0–2 Ω, receives a current of 60 amp. from an external source, then terminal p.d of the battery is:-

1. 12 V
2. 9 V
3. 15 V
4. 20 V

Answer : 3. 15 V

Question 12. Two bulbs of (40 W, 200 V), and (100 W, 200 V). The correct relation for their resistance:-

1. R40 < R100
2. R40 > R100
3. R40 = R100
4. No relation can be predicted

Answer: 2. R40 > R100

Question 13. For the circuit shown the value of current I is:-

1. 1.3 A
2. 1.7 A
3. 3.7 A
4. 1 A

Answer: 2. 1.7 A

Question 14. The terminal voltage across a cell is more than its emf if another cell of:-

1. Higher emf is connected in parallel to it
2. Less emf is connected in parallel to it
3. Less emf is connected in series to it
4. Higher emf is connected in series to it

Answer: 1. Higher emf is connected in parallel to it

Question 15. In the following circuit if VAB = 4V, then the value of resistance X in ohm’s will be:-

1. 5
2. 10
3. 15
4. 20

Answer: 4. 20

Question 16. The resistance of each arm of the wheat stone bridge is 10 ohms. A resistance of 10 ohm is connected in series with a galvanometer then the equivalent resistance across the battery will be:-

1. 10 ohm
2. 15 ohm
3. 20 ohm
4. 40 ohm

Answer: 1. 10 ohm

Question 17. In the circuit shown the 5Ω resistor develops 20 W due to current flowing through it. Then power dissipated in 4W resistor is:-

1. 4 W
2. 6W
3. 10 w
4. 20 W

Answer: 1. 4 W

Question 18. A resistor R is connected to a cell as shown in the figure. The value of R for which in its maximum is:-

1. 3Ω
2. 6Ω
3. 9Ω
4. 12Ω

Answer: 2. 6Ω

Question 19. The bridge shown in the figure will be balanced when:-

1. R₁/R₂ = R₃/R₄
2. R₁/R₂ = R₄/R₃
3. R₁R₃ = R₂R₄
4. R₅ = 0

Answer: 1. R₁/R₂ = R₃/R₄

Question 20. In a Wheatstone bridge P = Q = 10 ohm and R = S = 15 l and G = 20 l and a cell of e.m.f. 1.5 V is connected in the circuit. The current drawn from the cell is:-

1. 0.125 A
2. 0.060 A
3. 0.025 A
4. 0.021 A

Answer: 1. 0.125 A

Question 21. For the circuit shown rate of heat produced in the 5 ohm resistance is 10 cal/sec. then the rate of heat produced in 4-ohm resistance will be:

1. 10 cal/sec.
2. 2 cal/sec.
3. 3 cal/sec.
4. 4 cal/sec.

Answer: 2. 2 cal/sec.

Question 22. Two e.m.f. source of e.m.f. E1 and E2 and internal resistance r1 and r2 are connected in parallel. The e.m.f. of this combination is:-

1. \(\frac{E_1+E_2}{2}\)
2. \(\frac{E_1 r_1+E_2 r_2}{r_1+r_2}\)
3. \(\frac{E_2 r_1+E_1 r_2}{r_1+r_2}\)
4. \(\frac{E_1+E_2}{E_1+E_2}\)

Answer: 3. \(\frac{E_2 r_1+E_1 r_2}{r_1+r_2}\)

Question 23. In a torch, there are two cells each of 1.45 volts and internal resistance of 0.15Ω. Each cell gives a current to the filament of a lamp length of resistance 1.5Ω, then the value of current in ampere is:-

1. 16.11
2. 1.611
3. 0.1611
4. 2.6

Answer: 2. 1.611

Question 24. Two cells of the same emf E and internal resistance r are connected in parallel with a resistance of R. To get maximum power in the external circuit, the value of R is:-

1. \(R=\frac{r}{2}\)
2. R = r
3. R = 2r
4. R = 4r

Answer: 1. \(R=\frac{r}{2}\)

Question 25. The post office box works on the principle of :

1. Potentiometer
2. Wheatstone bridge
3. Matter waves
4. Ampere’s law

Answer: 2. Wheatstone bridge

Question 26. While using a post office box the keys should be switched on in the following order :

1. First cell key the and then galvanometer key.
2. First the galvanometer key and then the cell key.
3. Both the keys simultaneously.
4. Any key first and then the other key.

Answer: 1. First cell key the and then galvanometer key.

Question 27. In a post office box if the position of the cell and the galvanometer are interchanged, then the :

1. The null point will not change
2. The null point will change
3. The post office box will not work
4. Nothing can be said.

Answer: 1. Null point will not change

Question 28. What is the value of R for zero deflection in a galvanometer:-

1. 20 Ω
2. 80 Ω
3. 10 Ω
4. 40 Ω

Answer: 1. 20 Ω

Question 29. A cell, an ammeter, and a variable resistance R are connected in series and a voltmeter is connected across R. For a certain value of R ammeter and voltmeter readings are 0.3 amp and 0.9 V respectively, and for some other values of R, these readings are 0.25 amp. and 1.0 V. The internal resistance of the cell is:-

1. 3.4 Ω
2. 4.3 Ω
3. 2.0 Ω
4. 4.6 Ω

Answer : 3. 2.0 Ω

Question 30. In the measurement of resistance by the Wheatstone bridge, the known and the unknown resistance are interchanged to eliminate:-

1. Minor error
2. Observational error
3. Error due to thermoelectric effect
4. Connection error

Answer: 1. Minor error

Question 31. Which of the statements is wrong:-

1. When all the resistances are equal, then the sensitivity of the Wheatstone bridge is maximum.
2. When the galvanometer and the cell are interchanged, then the balancing of the Wheatstone bridge will be affected.
3. Kirchhoff’s first law for the current meeting at the junctions in an electric circuit shows the conservation of charge.
4. Rheostat can be used as a potential divider

Answer : 3. Kirchhoff’s first law for the current meeting at the electric circuit junctions shows the charge conservation.

5. Movement Of Electrons Inside The Conductor

All the free electrons are in random motion due to the thermal energy and relationship given by

⇒ \(\frac{3}{2} \mathrm{KT}=\frac{1}{2} \mathrm{mv}^2\)

At room temperature, its speed is around 106 m/sec or 103 km/sec

but the average velocity is zero so current in any direction is zero. When a conductor is placed in an electric field. Then for a small duration electrons, do have an average velocity but its average velocity becomes zero within a short interval of time.

Thermal Speed

Conductors contain a large number of free electrons, which are in continuous random motion.

Due to random motion, the free electrons collide with positive metal ions with high frequency and a change in direction at each collision. So, the thermal velocities are randomly distributed in all possible directions.

⇒ \(\overrightarrow{u_1}, \overrightarrow{u_2}, \ldots \overrightarrow{u_N}\) is the individual thermal velocities of the free electrons at any given time.

the total number of free electrons in the conductor = N

Average velocity\(\vec{u}_{\text {ave }}=\left[\frac{\overrightarrow{u_1}+\overrightarrow{u_2}+\ldots \overrightarrow{u_N}}{N}\right]=0\)

The average velocity is zero but the average speed is non-zero.

Drift Velocity(V→ D )

Drift velocity is defined as the velocity with which the free electrons drift toward the positive terminal under the effect of the applied electric field. When the ends of a conductor are connected to a source of emf, an electric field E is established in the E = V conductor, such that,\(E=\frac{V}{\ell},\)

where V = the potential difference across the conductor and ρ = the length of the conductor.

The electric field exerts an electrostatic force \(-e \vec{E}\)on each electron in the conductor.

The acceleration of each electron\(\vec{a}=\frac{-e \vec{E}}{m}\)

m = mass of electron

e = charge of the electron

In addition to its thermal velocity, due to this acceleration, the electron acquires, a velocity component in a direction opposite to the direction of the electric field.

The gain in velocity due to the applied field is very small and is lost in the next collision.

At any given time, an electron has a velocity\(\overrightarrow{v_1}=\overrightarrow{u_1}+\vec{a} \tau_1\)

Where \(\vec{u}_1\) = the thermal velocity

⇒ \(\vec{a} \tau_1\)the velocity acquired by the electron under the influence of the applied electric field.

τ1 = the time that has elapsed since the last collision. Similarly, the velocities of the other electrons are

⇒ \(\vec{v}_2=\vec{u}_2+\vec{a} \tau_2, \quad \vec{v}_3=\vec{u}_3+\vec{a} \tau_3, \ldots, \quad \vec{v}_N=\vec{u}_N+\vec{a} \tau_N\)

The average velocity of all the free electrons in the conductor is equal to the drift velocity of the free electrons.

⇒ \(\vec{v}_d=\frac{\vec{v}_1+\vec{v}_2+\vec{v}_3+\ldots \vec{v}_N}{N}=\frac{\left(\vec{u}_1+\vec{a} \tau_1\right)+\left(\overrightarrow{u_2}+\vec{a} \tau_2\right)+\ldots+\left(\vec{u}_N+\vec{a} \tau_N\right)}{N}\)

⇒ \(\vec{v}_d=\frac{\left(\vec{u}_1+\vec{u}_2+\ldots+\vec{u}_{\mathrm{N}}\right)}{\mathrm{N}}+\overrightarrow{\mathrm{a}} \frac{\left(\tau_1+\tau_2+\ldots+\tau_{\mathrm{N}}\right)}{\mathrm{N}}\) order of drift velocity is 10–4 m/s

⇒ \(\frac{\vec{u}_1+\vec{u}_2+\ldots+\vec{u}_N}{N}=0\)

⇒ \(\overrightarrow{v_d}=\vec{a} \frac{\tau_1+\tau_2+\ldots+\tau_N}{N} \mathrm{r} \overrightarrow{v_d}=\vec{a} \tau\)

⇒ \(\vec{v}_d=-\frac{e \vec{E}}{m} \tau\)

Relaxation Time (τ) :

The average time elapsed between two successive collisions.

It is of the order of 10^-14 s

It is a temperature-dependent characteristic of the material of the conductor.

It decreases with temperature increases.

Mean Free Path (λ):

The distance traveled by a conduction electron during relaxation time is known as the mean free path λ

Mean free path of conduction electron = Thermal velocity × Relaxation time

Solved Examples

Example 6. Find the approximate total distance traveled by an electron in the time interval in which its displacement is one meter along the wire.
Solution:

⇒ \(\text { time }=\frac{\text { displacement }}{\text { drift velocity }}=\frac{\mathrm{S}}{\mathrm{V}_{\mathrm{d}}}\)

Vd = 1 mm/s = 10–3 m/s (normally the value of drift velocity is 1 mm/s)

S = 1 m

time = \(\frac{1}{10^{-3}}=10^3 \mathrm{~s}\)

Distance travelled = speed × time

∴ speed = 106 m/s

So required distance = 106 × 10³ m = 10⁹ m

6. Relation Between I And V In A Conductor

Let the number of free electrons per unit volume in a conductor = n

Total number of electrons in dx distance = n (Adx)

Total charge dQ = n (Adx)e

Cross-sectional area = A

Current \(I=\frac{d Q}{d t}=n A e \frac{d x}{d t} \quad \Rightarrow \quad \mathrm{I}=n e A v_{\mathrm{d}}\)

Current density \(J=\frac{I}{A}={nev}_{\mathrm{d}} \quad \Rightarrow \quad J=n e\left(\frac{e E}{m}\right) \tau \quad v_d=\left(\frac{e E}{m}\right) \tau\)

⇒ \(J=\left(\frac{n e^2 \tau}{m}\right) E \quad \Rightarrow \quad J=\sigma \mathrm{E} \quad \text { conductivity } \quad \sigma=\frac{n e^2 \tau}{m}\)

In vector form j = σE

σ depends only on the material of the conductor and its temperature.

As temperature (T) ↑, τ↓

Example 7. A current of 1.34 A exists in a copper wire of cross-section 1.0 mm². Assuming each copper atom contributes one free electron. Calculate the drift speed of the free electrons in the wire. The density of copper is 8990 kg/m³ and atomic mass = 63.50.
Solution :

Mass of 1m3 volume of the copper is = 8990 kg = 8990 × 10³ g

Number of moles in 1m3 \(=\frac{8990 \times 10^3}{63.5}=1.4 \times 10^5\)

Since each mole contains 6 × 10²³ atoms therefore number of atoms in 1m³

n = (1.4 × 105) × (6 × 10²³)

= 8.4 × 10²⁸ = electron density

i = need

⇒ \(\mathrm{v}_{\mathrm{d}}=\frac{i}{n e A}=\frac{1.34}{8.4 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-6}}\)

(1 mm2 = 10–6 m2 ) = 10^-4 m/s

7. Electrical Resistance

The property of a substance by which it opposes the flow of electric current through it is termed electrical resistance. Electrical resistance depends on the size, geometry, temperature, and internal structure of the conductor.

We have \(\mathrm{i}=\frac{\mathrm{nA} \mathrm{e}^2 \tau}{2 \mathrm{~m} \ell} \mathrm{V}\)

Here i ∞ V

it is known as Ohm’s law

⇒ \(i=\frac{V}{R}\)

\(\mathrm{R}=\frac{2 \mathrm{~m} \ell}{\mathrm{nAe^{2 } \tau}}\)

⇒  V = IR

Hence \(\mathrm{R}=\frac{2 \mathrm{~m}}{\mathrm{ne}^2 \tau} \cdot \frac{\ell}{\mathrm{A}}\)

⇒ \(R=\frac{\rho \ell}{A} \quad \Rightarrow \quad V=I \times \frac{\rho \ell}{A}\)

⇒ \(\frac{V}{\ell}=\frac{\mathrm{I}}{\mathrm{A}} \rho\)

⇒ \(E=J \rho \Rightarrow \quad J=\frac{I}{A}=\) current density

ρ is called resistively (it is also called specific resistance), and ρ \(=\frac{2 m}{n e^2 \tau}=\frac{1}{\sigma}, \sigma\) is called conductivity.

Therefore current in conductors is proportional to the potential difference applied across its ends. This is Ohm’s Law. Units:  R→ohm(Ω ), ρ → ohm – meter( Ω-m) also called Siemens,σ→Ω-1m-1

Important Points

• 1 ampere of current means the flow of 6.25 × 10¹⁸ electrons per second through any cross-section of the conductor.
• The electric field outside a current-carrying conductor is zero but inside a conductor is.
• Current is a scalar quantity but current density is a vector quantity.
• If A is not normal to I but makes an angle θ with the normal to current then.

I = J_A cos θ

⇒ \(J=\frac{I}{A \cos \theta}\)

Order of free e– density in conductors = 10²⁸ electrons/m³, while in semiconductors = 10¹⁶ e^–/m

If a steady current flows in a metallic conductor of the nonuniform cross-section.

Along the wire, I am the same.

Current density and drift velocity depends on area inversely so J₁ > J₂, E₁ > E₂, I ₁ = I 2, A₁ < A₂

If the temperature of the conductor increases, the amplitude of the vibrations of the positive ions in the conductor also increases. Due to this, the free electrons collide more frequently with the vibrating ions and as a result, the average relaxation time decreases.

Solved Example

Example 8. The dimensions of a conductor of specific resistance ρ are shown below. Find the resistance of the conductor across AB, CD, and EF.

Solution:

⇒ \(R_{A B}=\frac{\rho c}{a b}, \quad R_{C D}=\frac{\rho b}{a c}, \quad R_{E F}=\frac{\rho a}{b c}\)

For a condition

⇒ \(\mathrm{R}=\frac{\rho \ell}{\mathrm{A}}=\frac{\text { Re resistivity } \times \text { length }}{\text { Area of cross section }}\)

⇒ \(R_{A B}=\frac{\rho c}{a b}, R_{C D}=\frac{\rho b}{a c}, R_{E F}=\frac{\rho a}{b c}\)

7.1 Dependence Of Resistance On Various Factors

⇒ \(\mathrm{R}=\rho \frac{\ell}{\mathrm{A}}=\frac{2 \mathrm{~m}}{\mathrm{ne}^2 \tau} \cdot \frac{\ell}{\mathrm{A}}\)

Therefore R depends on as

1. ∞l
2. \(\propto \frac{1}{\mathrm{~A}}\)
3. \(\propto \frac{1}{n} \propto \frac{1}{\tau}\)
4. And in metals τ decreases as T increases ⇒ R also increases.

Results

On stretching a wire (volume constant)

If the length of the wire is taken into account then \(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\ell_1^2}{\ell_2^2}\)

If the radius of the cross-section is taken into account then \(\frac{R_1}{R_2}=\frac{r_2^4}{r_1^4}\) where R₁ and R₂ are initial and final resistances and l₁, l₂, are initial and final lengths and r₁ and r₂ initial and final radii respectively. (if the elasticity of the material is taken into consideration, the variation of the area of the cross-section is calculated with the help of Young’s modulus and Poison’s ratio)

Effect of percentage change in length of wire

⇒ \(\frac{\mathrm{R}_2}{\mathrm{R}_1}=\frac{\ell^2\left[1+\frac{\mathrm{x}}{100}\right]^2}{\ell^2}\) where l – original length and x- % increment

if x is quite small (say < 5%) then % change in R is

⇒ \(\frac{R_2-R_1}{R_1} \times 100=\left(\frac{\left(1+\frac{x}{100}\right)^2-1}{1}\right) \times 100 \cong 2 x \%\)

Solved Examples

Example 9. If a wire is stretched to double its length, find the new resistance if the original resistance of the wire was R.
Solution:

As we know that R = \(\frac{\rho \ell}{A}\)

in case R’=\(\frac{\rho \ell^{\prime}}{A^{\prime}}\)

l’=2l

A’l’= Al (volume of the wire remains constant)

⇒ \(A^{\prime}=\frac{A}{2}\)

⇒ \(\mathrm{R}^{\prime}=\frac{\rho \times 2 \ell}{\mathrm{A} / 2}=4 \frac{\rho \ell}{\mathrm{A}}=4 \mathrm{R}\)

Example 10. The wire is stretched to increase the length by 1% to find the percentage change in the Resistance.
Solution :

As we know that

⇒ \(\mathrm{R}=\frac{\rho \ell}{\mathrm{A}}\)

⇒ \(\frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{\Delta \rho}{\rho}+\frac{\Delta \ell}{\ell}-\frac{\Delta \mathrm{A}}{\mathrm{A}} \text { and } \frac{\Delta \ell}{\ell}=-\frac{\Delta \mathrm{A}}{\mathrm{A}}\)

⇒ \(\frac{\Delta R}{R}=O+1+1=2\)

Hence percentage increase in the Resistance = 2%

Note: The above method is applicable when the % change is very small.

Example 11. The figure shows a conductor of length l carrying current I and having a circular cross-section. The radius of the cross-section varies linearly from a to b. Assuming that (b – a) << l calculate current density at distance x from the left end.

Solution:

Since the radius at the left end is a and that of the right end is b, therefore increase in radius over length l is (b – a).

Hence rate of increase of radius per unit length = \(\left(\frac{b-a}{\ell}\right)\)

Increase in radius over length x =\(\left(\frac{b-a}{\ell}\right) x\)

Since the radius at the left end is a radius at distance \(\mathrm{x}=\mathrm{r}=\mathrm{a}+\left(\frac{b-a}{\ell}\right) x\)

Area at this particular section \(\mathrm{A}=\pi \mathrm{r}^2=\pi\left[a+\left(\frac{b-a}{\ell}\right) x\right]^2\)

Hence current density \(\mathrm{J}=\frac{i}{A}=\frac{i}{\pi r^2}=\frac{i}{\pi\left[a+\frac{x(b-a)}{\ell}\right]^2}\)

Note: Resistance of different shaped conductors.

⇒ \(R=\frac{\rho L}{\pi b^2}\)

⇒ \(R=\frac{\rho L}{\pi\left(b^2-a^2\right)}\)

⇒ \(R_1=\frac{\rho_1 L}{\pi\left(b^2-a^2\right)}\)

⇒ \(R=\frac{\rho L}{\pi a b}\)

The resistance between square faces

⇒ \(\mathrm{R}_{\mathrm{AB}}=\rho \frac{\text { distance between faces }}{\text { area of square }}=\rho \frac{a}{b^2}\)

The resistance between rectangular faces

⇒ \(R_{x y}=\rho \frac{b}{a \cdot b}=\frac{\rho}{a}\) (does not depends on b)

Temperature Dependence Of Resistivity And Resistance :

The resistivity of a metallic conductor nearly increases with increasing temperature. This is because, with the increase in temperature the ions of the conductor vibrate with greater amplitude, and the collision between electrons and ions becomes more frequent. Over a small temperature range (up to 100ºC), the resistivity of a metal can be represented approximately by the equation,

⇒ \(\rho(T)=\rho_0\left[1+\alpha\left(T-T_0\right)\right]……(1)\)

where ρ₀ is the resistivity at a reference temperature T₀ (often taken as 0ºC or 20ºC) and ρ(T) is the resistivity at temperature T, which may be higher or lower than T₀. The factor α is called the temperature coefficient of resistivity.

The resistance of a given conductor depends on its length and area of cross-section besides the resistivity. As temperature changes, the length and area also change. But these changes are quite small and the factor l/A may be treated as constant.

Then, R αρ and hence, R(T) = R₀ [1 + α(T – T₀)] …(ii)

In this equation, R(T) is the resistance at temperature T and R₀ is the resistance at temperature T₀, often taken to be 0ºC or 20ºC. The temperature coefficient of resistance α is the same constant that appears.

Note :

The ρ-T equation written above can be derived from the relation,

α = fractional change in resistivity per unit change in temperature

∴ \(=\frac{d \rho}{\rho \mathrm{dT}}=\alpha\)

∴ \(\frac{d \rho}{d T}=\alpha \rho\)

∴ \(\frac{d \rho}{\rho}=\alpha d T\) (ρ can be assumed constant for small temperature variation)

∴ \(\int_{\rho_0}^\rho \frac{d \rho}{\rho}=\alpha \int_{T_0}^T d T\)

∴ \(\left(\frac{\rho}{\rho_0}\right)=\alpha\left(T-T_0\right)\)

∴ \(\rho=\rho_0 e^{\alpha\left(T-T_0\right)}\)

if α (T – T₀) << 1 then

eα^(T-T₀) − can approximately be written as 1 + α(T – T₀). Hence,

In the above discussion, we have assumed α to be constant. If it is a function of temperature it will come inside the integration in Eq.

Important Points

• If a wire is stretched to n times its original length, its new resistance will be n 2 times.
• If a wire is stretched such that its radius is reduced to its original values, then resistance will increase n4 times similarly resistance will decrease n 4 times if the radius is increased n times by contraction.
• The equivalent resistance of a parallel combination is lower than the value of the lowest resistance in the combination.
• In general :
  • The resistivity of alloys is greater than their metals.
  • The temperature coefficient of alloys is lower than pure metals.
  • Resistance of most nonmetals decreases with an increase in temperature. (example.carbon)
  • The resistivity of an insulator (for example amber) is greater than the metal by a factor of 10²²
• The temperature coefficient (α) of semiconductors including carbon (graphite), insulators, and electrolytes is negative.

Solved Examples

Example 12. The resistance of a thin silver wire is 1.0 Ω at 20ºC. The wire is placed in a liquid bath and its resistance rises to 1.2 Ω. What is the temperature of the bath? (Here α = 10^-2 /ºC)
Solution :

Here change in resistance is small so we can apply

R = R 0(1 + αΔθ)

⇒ 1.2 = 1 × (1 + 10–2Δθ)

⇒  Δθ = 20ºC

⇒ θ – 20 = 20 ⇒ θ = 40º C

Electric Current In Resistance

In a resistor, current flows from high potential to low potential

High potential is represented by a positive (+) sign and low potential is represented by a negative (–) sign.

V_A – V_B = iR

If V₁ > V₂

Then current will flow from A to B

⇒ \(i=\frac{V_1-V_2}{R}\)

If V₁ < V₂

then current will go from B to A and \(i=\frac{V_2-V_1}{R}\)

Example 13. Calculate current flowing in part of the circuit shown in the figure.

Solution: V_A – V_B = i×R

⇒  \(i=\frac{6}{2}=3 A\)

8. Electrical Power :

The energy liberated per second in a device is called its power. The electrical power P delivered or consumed by an electrical device is given by P = VI, where V = Potential difference across the device and I = Current.

If the current enters the higher potential point of the device then electric power is consumed by it (i.e. acts as load). If the current enters the lower potential point then the device supplies power (i.e. acts as a source).

⇒ \(\text { Power }=\frac{\text { V.dq }}{d t}\)

= V I

P = V I

If power is constant then energy = P t

If power is variable then Energy =\(\int \mathrm{pdt}\)

Power consumed by a resistor\(P=I^2 R=V I=\frac{V^2}{R} .\)

When a current is passed through a resistor energy is wasted in overcoming the resistance of the wire.

This energy is converted into heat.

⇒ \(\mathrm{W}=\mathrm{VIt}=\mathrm{I}^2 R \mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{R}} \mathrm{t}\)

The heat generated (in joules) when a current of I ampere flows through a resistance of R ohm for t second is given by:

H = I2 Rt Joule = \(\frac{I^2 R t}{4.2} \text { Calorie }\)

1 unit of electrical energy = 1 Kilowatt hour = 1 KWh = 3.6 x 10⁶ Joule.

Solved Examples

Example 14. If the bulb rating is 100 watts and 220 V then determine

1. Resistance of filament
2. Current through filament
3. If the bulb operates at a 110 volt power supply then find the power consumed by the bulb.

Solution :

The bulb rating is 100 W and a 220 V bulb means when 220 V potential difference is applied between the two ends then the power consumed is 100 W

Here V = 220 Volt

P = 100 W

⇒ \(\frac{V^2}{R}=100 \quad \text { So } \quad R=484 \Omega\)

Since Resistance depends only on material hence it is constant for bulb

⇒ \(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{220}{22 \times 22}=\frac{5}{11} \mathrm{Amp} .\)

power consumed at 110 V

power consumed = \(\frac{110 \times 110}{484}=25 \mathrm{~W}\)

9. Battery (Cell)

A battery is a device that maintains a potential difference across its two terminals A and B. Dry cells, secondary cells, generators, and thermocouples are the devices used for producing potential differences in an electric circuit. The arrangement of the cell or battery is shown in the figure.

Electrolyte provides continuity for current.

It is often prepared by putting two rods or plates of different metals in a chemical solution. Some internal mechanism exerts force (F_n ) on the ions (positive and negative) of the solution. This force drives positive ions toward the positive terminal and negative ions toward the negative terminal.

• As a positive charge accumulates on the anode and a negative charge on the cathode a potential difference and hence an E electric field is developed from the anode to the cathode.
• This electric field exerts an electrostatic force F = qE
• F_n on the ions. This force is opposite to that of. In equilibrium (steady state) F_n = F_e and no further accumulation of charge takes place.
• When the terminals of the battery are connected by a conducting wire, an electric field is developed in the wire. The free electrons in the wire move in the opposite direction and enter the battery at the positive terminal.
• Some electrons are withdrawn from the negative terminal. Thus, potential difference and hence, F_e decreases in magnitude while F_n remains the same.
• Thus, there is a net force on the positive charge towards the positive terminal. With this, the positive charge rushes toward the positive terminal and the negative charge rushes toward the negative terminal.
• Thus, the potential difference between positive and negative terminal is maintained.

Internal Resistance (R) :

The potential difference across a real source in a circuit is not equal to the EMF of the cell. The reason is that the charge moving through the electrolyte of the cell encounters resistance. We call this the internal resistance of the source.

The internal resistance of a cell depends on the distance between electrodes (r α d), area of 1 electrode \(\left(r \propto \frac{1}{s}\right)\) and nature, concentration (r α c) and temperature of electrolyte \(\left(r \propto \frac{1}{\text { Temp. }}\right)\)

Example 15. What is the meaning of 10 Amp? hr?
Solution:

It means if the 10 A current is withdrawn then the battery will work for 1 hour. 10 Amp⎯⎯→1 hr

10 Amp →1 hr

1 Amp → 10 hr

½ Amp → 20 hr

10. Electromotive Force : (E.M.F.)

Electromotive Force Definition: Electromotive force is the capability of the system to make the charge flow. Definition II: It is the work done by the battery for the flow of 1-coulomb charge from the lower potential terminal to the higher potential terminal inside the battery.

10.1 Representation for battery :

Ideal cell: A cell in which there is no heating effect.

Nonideal cell: A cell in which there is a heating effect inside due to opposition to the current flow internally

Case 1: Battery acting as a source (or battery is discharging)

V_A–V_B = ε – ir

V_A–V_B

⇒ it is also called terminal voltage.

The rate at which the chemical energy of the cell is consumed = εi

The rate at which heat is generated inside the battery or cell = i2r electric power output = εi – i2r

= (ε – ir) i

Case 2: Battery acting as a load (or battery charging) :

V_A–V_B= ε + ir

the rate at which chemical energy is stored in the cell = εi

thermal power inside the cell = i2r electric power input = εi + i2r = (ε+ir) i = (V_A–V_B) i

Representation For Battery Definition:

The electromotive force of a cell is equal to the potential difference between its terminals when no current is passing through the circuit.

Case 3 :

When a cell is in an open circuit

i = 0 as the resistance of the open circuit is infinite (∞).

So V = ε, so the open circuit terminal voltage difference is equal to the emf of the cell.

Case 4:

Short-circuiting: Two points in an electric circuit directly connected by a conducting wire are called short-circuited, under such conditions both points are at the same potential.

When a cell is short-circuited

i = ε and V = 0, the short circuit current of a cell is maximum. r

Note: The potential at all points of a wire of zero resistance will be the same.

Earthing: If some point of the circuit is earthed then its potential is assumed to be zero.

Important Points

• At the time of charging a cell. When current is supplied to the cell, the terminal voltage is greater than the e.m.f. E V = E + Ir
• A series combination is useful when internal resistance is less than the external resistance of the cell.
• Parallel combination is useful when internal resistance is greater than external resistance of the cell.
• Power in R (given resistance) is maximum, if its value is equal to the net resistance of the remaining circuit.
• The internal resistance of the ideal cell = 0
• if external resistance is zero then the current given by the circuit is maximum.

11 Relative Potential

While solving an electric circuit it is convenient to choose a reference point and assign its voltage as zero, then all other potentials are measured concerning this point. This point is also called the common point.

Example 16. In the given electric circuit find

1. The current
2. Power output
3. Relation between r and R so that the electric power output (that means power given to R) is maximum.
4. Value of maximum power output.
5. Plot graph between the power and resistance of the load
6. From the graph we see that for a given power output there exists two values of external resistance, proving that the product of these resistances equals r2.
7. what is the efficiency of the cell when it is used to supply maximum power?

Solution :

In the circuit shown if we assume that the potential at A is zero then the potential at B is ε – ir. Now since the connecting wires are of zero resistance

∴ V_D= V_A= 0 ⇒ V_C= V_B= ε – ir

Now current through CD is also i

(It’s in series with the cell).

⇒ \(\mathrm{i}=\frac{\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{D}}}{\mathrm{R}}=\frac{(\varepsilon-\mathrm{ir})-0}{\mathrm{R}} \text { Current } \mathrm{i}=\frac{\varepsilon}{\mathrm{r}+\mathrm{R}}\)

Note: After learning the concept of series combination we will be able to calculate the current directly

Power output P = i2 R = \(\mathrm{P}=\mathrm{i}^2 \mathrm{R}=\frac{\varepsilon^2}{(\mathrm{r}+\mathrm{R})^2} \cdot \mathrm{R}\)

⇒ \(\frac{\mathrm{dP}}{\mathrm{dR}}=\frac{\varepsilon^2}{(\mathrm{r}+\mathrm{R})^2}-\frac{2 \varepsilon^2 \mathrm{R}}{(\mathrm{r}+\mathrm{R})^3}=\frac{\varepsilon^2}{(\mathrm{R}+\mathrm{r})^3}[\mathrm{R}+\mathrm{r}-2 \mathrm{R}]\) for maximum power supply

⇒ \(\frac{d P}{d R}=0 \Rightarrow\) dR= 0⇒ r + R – 2R = 0 ⇒ r = R

Here for maximum power output outer resistance should be equal to internal resistance

⇒ \(P_{\max }=\frac{\varepsilon^2}{4 r}\)

The graph between ‘P’ and R

Maximum power output at R = r

⇒ \(\mathrm{P}_{\max }=\frac{\varepsilon^2}{4 \mathrm{r}} \quad \Rightarrow \quad i=\frac{\varepsilon}{\mathrm{r}+\mathrm{R}}\)

Power output

⇒ \(P=\frac{\varepsilon^2 R}{(r+R)^2}\)

P (r² + 2rR + R²)= ε²R²

⇒ \(\mathrm{R}^2+\left(2 \mathrm{r}-\frac{\varepsilon^2}{\mathrm{P}}\right) \mathrm{R}+\mathrm{r}^2=0\)

above quadratic equation in R has two roots and for given values of ε, P, and r such that

∴ R₁R₂=r²(product of roots)

r² = R₁R₂

Power of battery spent = \(\frac{\varepsilon^2}{(\mathrm{r}+\mathrm{r})^2} \cdot 2 \mathrm{r}=\frac{\varepsilon^2}{2 \mathrm{r}} \text { power (output) }=\left(\frac{\varepsilon}{\mathrm{r}+\mathrm{r}}\right)^2 \times \mathrm{r}=\frac{\varepsilon^2}{4 \mathrm{r}}\)

Efficiency = \(\frac{\text { power output }}{\text { total power spent by cell }}=\frac{\frac{\varepsilon^2}{4 \mathrm{r}} \times 100}{\frac{\varepsilon^2}{2 \mathrm{r}}}=\frac{1}{2} \times 100=50 \%\)

Example 17. In the figure given beside find out the current in the wire BD

Solution :

Let at point D potential = 0 and write the potential of other points then current in wire AD = \(\frac{10}{2}\)

= 5 A from A to D current in wire CB = \(\frac{20}{5}\)

∴ current in wire BD = 1 A from D to B

Example 18. Find the current in each wire

Solution :

Let potential at point A is 0 volts then the potential of other points is shown in the figure.

Current in BG = \(\frac{40-0}{1}=40 \mathrm{~A} \text { from } \mathrm{G} \text { to } \mathrm{B}\)

Current in FC =\(\frac{0-(-30)}{2}=15 \mathrm{~A} \text { from } \mathrm{C} \text { to } \mathrm{F}\)

Current in DE = \(=\frac{0-(-40)}{2}=20 \mathrm{~A} \text { from } \mathrm{D} \text { to } E\)

Current in wire AH = 40 – 35 = 5 A from A to H

12. Kirchhoff’s Laws

12 . 1- Kirchhoff’s Current Law (Junction law)

This law is based on the law of conservation of charge. It states that ” The algebraic sum of the currents meeting at a point of the circuit is zero ” or total currents entering a junction equals the total current leaving the junction.

Σ Ι_In = Σ Ι_Out

It is also known as KCL (Kirchhoff’s current law).

Solved Examples

Example 19. Find relation in between current i₁, i₂, i₃, i₄, i₅ and i₆.

Solution : i₁+ i₂– i₃– i₄+ i₅+ i₆= 0

Example 20. Find the current in each wire

Solution :

Let potential at point B = 0. The potential at other points is mentioned.

∴ Potential at E is not known numerically.

Let potential at E = x

Now applying Kirchhoff’s current law at junction E. (This can be applied at any other junction also).

\(\frac{x-10}{1}+\frac{x-30}{2}+\frac{x+14}{2}=0\)

4x = 36 ⇒   x = 9

Current in EF = \(\frac{10-9}{1}=1 \mathrm{~A} \text { from } \mathrm{F} \text { to } \mathrm{E}\)

Current in BE = \(\frac{30-9}{2}=10.5 \mathrm{~A} \text { from } \mathrm{B} \text { to } \mathrm{E}\)

Current in DE = \(\frac{9-(-14)}{2}=11.5 \mathrm{~A} \text { from } E \text { to } D\)

Example 21. Find the potential at point A

Solution :

Let potential at A = x, applying Kirchhoff’s current law at junction A

⇒ \(\frac{x-20-10}{1}+\frac{x-15-20}{2}+\frac{x+45}{2}+\frac{x+30}{1}=0\)

⇒ \(\frac{2 x-60+x-35+x+45+2 x+60}{2}=0\)

⇒ \(6 x+10=0 \quad \Rightarrow \quad x=-5 / 3\)

Potential \(A=\frac{-5}{3} V\)

12.2 Kirchhoff’s Voltage Law (Loop Law) :

“The algebraic sum of all the potential differences along a closed loop is zero.

So ΙR + Σ EMF = 0”.

The closed loop can be traversed in any direction. While traversing a loop if potential increases, put a positive sign in expression, and if potential decreases put a negative sign. (Assume sign convention)

−V₁− V₂+ V₃− V₄= 0.

Boxes may contain a resistor battery or any other element (linear or nonlinear).

It is also known as KVL

Example 22. Find the current in the circuit

Solution :

All the elements are connected in a series current is all of them will be the same

let current = i

Applying Kirchhoff voltage law in the ABCDA loop

10 + 4i – 20 + i + 15 + 2i – 30 + 3i = 0

10 i = 25

i = 2.5 A

Example 23. Find the current in each wire applying only Kirchhoff voltage law

Solution: Applying Kirchhoff voltage law in loop ABEFA

i₁+ 30 + 2 (i₁+ i₂) – 10 = 0

3l₁+ 2l₂+ 20 = 0 ————– (1)

Applying Kirchoff voltage law in BEDCB

+ 30 + 2(i₁ + i₂) + 50 + 2i₂= 0

4i₂+ 2i₁+ 80 = 0

2i₂+ i₁+ 40 = 0 ————– (2)

Solving (1) and (2)

3 [−40 -2i₂] + 2i₂+ 20 = 0

– 120 – 4i₂+ 20 = 0

i2= –25 A

and i₁= 10 A

∴ i₁+ i₂= – 15 A

Current in wire AF = 10 A from A to F

Current in wire EB = 15 A from B to E

Current in wire DE = 25 A from E to D.

13. Combination Of Resistances :

Several resistances can be connected and all the complicated combinations can be reduced to two different types, namely series and parallel.

The equivalent resistance of a combination is defined as \(R_{e q}=\frac{V}{i}\)

13.1 Resistances in Series:

When the resistances (or any type of elements) are connected end to end then they are said to be in series. The current through each element is the same.

Resistances in series carry equal current but the reverse may not be true.

Example 24. Which electrical elements are connected in series?

Solution : Here S₁, S₂, R₁, R₃ connected in one series and R₄, S₃ connected in different series

The equivalent of Resistors :

The effective resistance appearing across the battery (or between the terminals A and B) is

R = R₁+ R₂+ R₃+……………. + R_n(this means Reqis greater then any resistor) and V = V₁+ V₂+ V₃+……………. + V_n.

The potential difference across a resistor is proportional to the resistance. Power in each resistor is also proportional to the resistance

V = ΙR and P = Ι2R

where Ι is the same through any of the resistors.

⇒ \(V_1=\frac{R_1}{R_1+R_2+\ldots \ldots . .+R_n} V ; V_2=\frac{R_2}{R_1+R_2+\ldots \ldots . .+R_n} V \text {; etc }\)

Solved Examples

Example 25. Find the current in the circuit

Solution :

Req= 1 + 2 + 3 = 6 Ω the given circuit is equivalent to v 30

current i = \(\frac{v}{R_{e q}}=\frac{30}{6}=5 \mathrm{~A}\)

Example 26. In the figure shown B1, B2and B3are three bulbs rated as (200V, 50 W), (200V, 100W) and (200 V, 25W) respectively. Find the current through each bulb and which bulb will give more light.

Solution :

⇒ \(R_1=\frac{(200)^2}{50} ; \quad R_2=\frac{(200)^2}{100} ; \quad R_3=\frac{(200)^2}{25}\)

the current following through each bulb is

⇒ \(=\frac{200}{R_1+R_2+R_3}=\frac{200}{(200)^2\left[\frac{2+1+4}{100}\right]}\)

⇒ \(=\frac{100}{200 \times 7}=\frac{1}{14} \mathrm{~A}\)

Since R₃> R₁> R₂

∴ Power consumed by bulb = i2R

∴ if the resistance is of higher value then it will give more light.

∴ Here Bulb B3 will give more light.

13.2 Resistances in Parallel :

A parallel circuit of resistors is one in which the same voltage is applied across all the components in a parallel grouping of resistors R₁, R₂, R₃,…….., Rn.

In figure (a) and (b) all the resistors are connected between points A and B so they are in parallel.

Equivalent Resistance :

Applying Kirchhoff’s junction law at point P

i₀= i₁+ i₂+ i₃

Therefore, \(\frac{V}{R_{e q}}=\frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3}\)

⇒ \(\frac{1}{R_{\text {eq }}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)

in general,

⇒ \(\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\ldots \ldots \ldots .+\frac{1}{R_n}\)

Conclusions: (About Parallel Combination)

The potential difference across each resistor is the same.

I = I₁ + I₂ + I₃ +………. I_n.

Effective resistance (R) then. \(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\ldots \ldots . .+\frac{1}{R_n}\)(R is less than each resistor).

Current in different resistors is inversely proportional to the resistance.

⇒ \(I_1: I_2: \ldots \ldots \ldots \ldots . I_n=\frac{1}{R_1}: \frac{1}{R_2}: \frac{1}{R_3}: \ldots \ldots \ldots: \frac{1}{R_n} \text {. }\)

⇒ \(I_1=\frac{G_1}{G_1+G_2+\ldots \ldots . .+G_n} I I, I_2=\frac{G_2}{G_1+G_2+\ldots \ldots \ldots+G_n} I \text { I, etc. }\)

where G =1R= Conductance of a resistor. [Its unit is −1 Ωor (mho)]

Solved Examples

Example 27. When two resistors are in parallel combination then determine i₁ and i₂, if the combination carries current i.
Solution :

∴ i₁R₁= i₂R₂

⇒ \(\frac{i_1}{i_2}=\frac{R_2}{R_1}\)

⇒ \(i_1=\frac{R_2 i}{R_1+R_2} \quad \Rightarrow \quad i_2=\frac{R_1 i}{R_1+R_2} \text {, }\)

Note: Remember this law of \( I \propto \frac{1}{R}\) in the resistors connected in parallel. It can be used in problems.

Example 28. Find the current passing through the battery and each resistor.

Solution:

Method (1) :

It is easy to see that the potential difference across each resistor is 30 V.

∴ current is each resistors are \(\)

∴ Current through battery is = 15 + 10 + 5 =30 A.

Method (2) :

By ohm’s law i = \(i=\frac{V}{R_{\text {eq }}}\)

⇒ \(\frac{1}{R_{e q}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1 \Omega\)

R_eq= 1 Ω ⇒ \(i=\frac{30}{1}=30 \mathrm{~A}\)

Now distribute this current in the resistors in their inverse ratio.

The current total in 3 Ω and 6 Ω is 15 A it will be divided as 10 A and 5 A.

Note: The method (Ι) is better. But you will not find such an easy case everywhere.

Exercise 29. Find the current that is passing through the battery.
Solution :

Here potential difference across each resistor is not 30 V

The battery has internal resistance. Here the concept of a combination of resistors is useful.

R_eq= 1 + 1 = 2 Ω

⇒ \(i=\frac{30}{2}=15 \mathrm{~A} \text {. }\)

Example 30. Find equivalent Resistance

Solution :

Here all the Resistance is connected between terminals A and B Modified circuit is

So\(R_{e q}=\frac{R}{3}\)

Example 31. Find the current in Resistance P if the voltage supply between A and B is V volts

Solution :

Req= \(R_{e q}=\frac{3 R}{5}\)

⇒ \(I=\frac{5 V}{3 R}\) Modified circuit

Current in P = \(=\frac{R \times \frac{5 V}{3 R}}{1.5 R+R}\)

\(=\frac{2 V}{3 R}\)

14. Wheatstone Network : (4 Terminalnetwork)

The arrangement as shown in the figure, is known as a Wheat stone bridge Here there are four terminals in which except two all are connected through resistive elements.

In this circuit if R₁ R₃= R₂ R₄ then VC= VDand current in R₅= 0 this is called balance point or null point

When the current through the galvanometer is zero (null point or balance point) Q= \(\frac{P}{Q}=\frac{R}{S}\), then PS = QR ⇒

Here in this case products of opposite arms are equal. The potential difference between C and D at a null point is zero. The null point is not affected by resistance R₅, E, and R. It is not affected even if the positions of the Galvanometer and battery (E) are interchanged.

Hence, here the circuit can be assumed to be following,

Solved Examples

Example 32. Find the equivalent resistance of the circuit between the terminals A and B.

Solution :

Since the given circuit is a wheat stone bridge it is in balance condition.

10×3 = 30 = 6×5

hence this is equivalent to

⇒ \(R_{e q}=\frac{16 \times 8}{16+8}=\frac{16}{3} \Omega\)

Example 33. Find (a) Equivalent resistance (b) and current in each resistance

Solution :

R_eq= \(R_{e q}=\left(\frac{1}{16}+\frac{1}{8}+\frac{1}{16}\right)^{-1}+1=5 \Omega\)

i = \(i=\frac{60}{4+1}=12 \mathrm{~A}\)

Hence 12 A will flow through the cell.

By using current distribution law.

Current in resistance 10Ω and 6Ω = 3A

Current in resistance 5Ω and 3Ω = 6A

Current in resistance 20Ω = 0

Current in resistance 16Ω = 3A

Example 34. Find the equivalent resistance between A and B

Solution :

This arrangement can be modified as shown in the figure

since it is a balanced wheat stone bridge

⇒ \(R_{e q}=\frac{2 R \times 2 R}{2 R+2 R}=R\)

Example 35. Determine the value of R in the circuit shown in the figure, when the current is zero in the branch CD.

Solution :

The current in the branch CD is zero, if the potential difference across CD is zero. That means, voltage at point C = voltage at point D.

Since no current is flowing, the branch CD is open-circuited. So the same voltage is applied across ACB and ADB

\(V_{10}=V \times \frac{10}{15}\) \(V_R=V \times \frac{R}{20+R}\)

∴ V10 = VRand

\(V \times \frac{10}{15}=V \times \frac{R}{20+R}\)

∴ R = 40 Ω Ans.

15. Grouping Of Cells

15.1 Cells in Series :

Equivalent EMF

E_eq= E₁ + E₂ + ……. +E_n[write EMF’s with polarity] Equivalent internal resistance

r_eq= r₁ + r₂ + r₃ + r₄ + ……. + r_n

If n cells each of emf E, arranged in series and if r is the internal resistance of each cell, then total emf = nE so current in the circuit

⇒ \(I=\frac{n E}{R+n r}\)

If nr << R then Ι =\(\frac{\mathrm{nE}}{\mathrm{R}}\) → Series combination is advantageous.

If no>> R then Ι = \(\frac{E}{r}\) → Series combination is not advantageous.

Note: If the polarity of m cells is reversed, then equivalent emf = (n-2m)E while the equivalent resistance is still nr+R, so current in R will be

⇒ \(i=\frac{(n-2 m) E}{n r+R}\)

Solved Examples

Example 36. Find the current in the loop.

Solution: The given circuit can be simplified as

⇒ \(i=\frac{35}{10+5} \quad=\frac{35}{15}\)

⇒ \(=\frac{7}{3} \mathrm{~A}\)

⇒ \(I=\frac{7}{3} A\)

15.2 Cells in Parallel :

\(E_{e q}=\frac{\varepsilon_1 / r_1+\varepsilon_2 / r_2+\ldots .+\varepsilon_n / r_n}{1 / r_1+1 / r_2+\ldots . .+1 / r_n}\)[Use emf’s with polarity]

\(\frac{1}{r_{e q}}=\frac{1}{r_1}+\frac{1}{r_2}+\ldots+\frac{1}{r_n}\)

If m cells each of emf E and internal resistance r be connected in parallel and if this combination is connected to an external resistance then equivalent emf of the circuit = E.

Internal resistance of the circuit = \(\frac{\mathrm{r}}{\mathrm{m}} \text {. }\)

\(I=\frac{E}{R+\frac{r}{m}}=\frac{m E}{m R+r}\)

If mR << r ; I = \(\frac{\mathrm{mE}}{\mathrm{r}}\) → Parallel combination is advantageous.

If mR >> r ; I = \(\frac{E}{R}\)→ Parallel combination is not advantageous.

15.3 Cells in Multiple Arc :

mn = number of identical cells.

n = number of rows

m = number of cells in each row.

The combination of cells is equivalent to a single cell of emf = mE

and internal resistance = \(\frac{m r}{n}\)

Current I =\(\frac{m E}{R+\frac{m r}{n}}\)

For maximum current nR = mr

or R =\(\frac{\mathrm{mr}}{\mathrm{n}}=\) internal resistance of the equivalent battery. nE mE

⇒ \(\mathrm{I}_{\max }=\frac{\mathrm{nE}}{2 \mathrm{r}}=\frac{\mathrm{mE}}{2 \mathrm{R}} .\)

Solved Examples

Example 37. Find the EMF and internal resistance of a single battery which is equivalent to a combination of three batteries as shown in the figure.

Solution :

Battery (B) and (C) are in parallel combination with opposite polarity. So, their equivalent

\(\varepsilon_{\mathrm{BC}}=\frac{\frac{10}{2}+\frac{-4}{2}}{\frac{1}{2}+\frac{1}{2}}=\frac{5-2}{1}=3 \mathrm{~V}\)

rBC = 1Ω

Now,

εABC = 6 – 3 = 3V

rABC = 2Ω.

16. Galvanometer

The galvanometer is represented as follows:

It consists of a pivoted coil placed in the magnetic field of a permanent magnet. Attached to the coil is a spring. In the equilibrium position, with no current in the coil, the pointer is at zero and the spring is relaxed. When there is a current in the coil, the magnetic field exerts a torque on the coil that is proportional to the current. As the coil turns, the spring exerts a restoring torque that is proportional to the angular displacement. Thus, the angular deflection of the coil and pointer is directly proportional to the coil current and the device can be calibrated to measure current.

When the coil rotates the spring is twisted and it exerts an opposing torque on the coil.

There is a resistive torque also against motion to dampen the motion. Finally in equilibrium

τ magnetic = τspring ⇒ BINA sin θ = Cφ

But by making the magnetic field radial θ = 90º.

∴ BINA = C φ

Ι ∝ φ

here B = magnetic field A = Area of the coil

Ι = Current C = torsional constant

N = Number of turns φ = angle rotated by coil.

Current sensitivity

The ratio of deflection to the current i.e. deflection per unit current is called current sensitivity (C.S.) of the galvanometer CS = \(\frac{\phi}{I}=\frac{B N A}{C}\)

Note :

Shunting a galvanometer decreases its current sensitivity.

A linear scale is obtained. The markings on the galvanometer are proportionate.

The galvanometer coil has some resistance represented by Rg. It is of the order of a few ohms. It also has a maximum capacity to carry a current known as Ιg. Is also the current required for full-scale deflection. This galvanometer is called a moving coil galvanometer.

17. Ammeter

A shunt (small resistance) is connected in parallel with a galvanometer to convert it into an ammeter; An ideal ammeter has zero resistance.

Ammeter is represented as follows –

If the maximum value of current to be measured by the ammeter is Ι then ΙG. RG= (Ι – ΙG)S

⇒ \(\frac{\phi}{I}=\frac{B N A}{C}\)

⇒ \(S=\frac{I_G \times R_G}{I} \text { when } \quad I \gg I_G \text {. }\)

where Ι = Maximum current that can be measured using the given ammeter.

For measuring the current the ammeter is connected in series.

In calculation, it is simply a resistance

Resistance of ammeter

⇒ \(R_A=\frac{R_G \cdot S}{R_G+S}\)

for S << RG ⇒ RA= S G

Solved Examples

Example 38. What is the value of a shunt that passes 10% of the main current through a galvanometer of 99 ohms?
Solution :

As in figure RgΙg= (Ι – Ιg)S

⇒ \(99 \times \frac{I}{10}=\left(I-\frac{I}{10}\right) \times S\)⇒ S = 11 Ω. 10

Example 39. Find the current in circuits (a) and (b) and also determine the percentage error in measuring the current through an ammeter.

Solution :

ln A Ι = \(\frac{10}{2}\) = 5A

ln B Ι = \(\frac{10}{2}\) = 4A

Percentage error is = \(\frac{i-i i^{\prime}}{i} \times 100=20 \%\)

Here we see that due to the ammeter, the current has reduced. A good ammeter has very low resistance as compared with other resistors so due to its presence in the circuit the current is not affected.

Example 40. Find the reading of the ammeter. Is this the current through 6 Ω?

Solution :

⇒ \(R_{e q}=\frac{3 \times 6}{3+6}+1=3 \Omega\)

Current through battery Ι = \(\frac{18}{3}=6 \mathrm{~A}\)

So, current through ammeter

⇒ \(=6 \times \frac{6}{9}=4 \mathrm{~A}\)

No, it is not the current through the 6 Ω resistor.

Note: Ideal ammeter is equivalent to zero resistance wire for calculation potential difference across it is zero.

18. Voltmeter

A high resistance is put in series with a galvanometer. It is used to measure potential differences across a resistor in a circuit.

For the maximum potential difference

⇒ \(\mathrm{V}=\mathrm{I}_{\mathrm{G}} \cdot \mathrm{R}+\mathrm{I}_{\mathrm{G}} \mathrm{R}_{\mathrm{G}} \quad \mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{G}}}-\mathrm{R}_G\)

If RG << R \(\mathrm{R}_{\mathrm{s}} \approx \frac{\mathrm{V}}{\mathrm{I}_{\mathrm{G}}}\)

For measuring the potential difference a voltmeter is connected across that element. (parallel to that element it measures the potential difference that appears between terminals ‘A’ and ‘B’.)

For calculation, it is simply a resistance

Resistance of voltmeter RV= RG+ R ≈ R

⇒ \(I_g=\frac{V_0}{R_g+R} .\) R → ∞ ⇒ Ideal voltmeter. g

A good voltmeter has a high value of resistance.

Ideal voltmeter → which has a high value of resistance.

Note :

For calculation purposes, the current through the ideal voltmeter is zero.

Percentage error in measuring the potential difference by a voltmeter is = \(\frac{V-V^{\prime}}{V} \times 100\)

Example 41. A galvanometer has a resistance of G ohm and a range of V volt. Calculate the resistance to be used in series with it to extend its range to nV volt.
Solution :

Full scale current ig= \(\frac{V}{G}\)

to change its range

V1= (G + R_s)i_g ⇒ nV = (G + R_s)\(\frac{V}{G}\) VG⇒ R_s= G(n – 1) Ans.

Example 42. Find potential differences across the resistance 300 Ω in A and B.

Solution :

In (A) : Potential difference = \(\frac{100}{200+300} \times 300=60 \text { volt }\)

In (B) : Potential difference = \(=\frac{100}{200+\frac{300 \times 600}{300+600}} \times \frac{300 \times 600}{300+600}=50 \text { volt }\)

We see that by connecting the voltmeter the voltage which was to be measured has changed. Such voltmeters are not good. If its resistance had been very large than 300 Ω then it would not have affected the voltage by much amount.

Current sensitivity

The ratio of deflection to the current i.e. deflection per unit current is called current sensitivity (C.S.) of the galvanometer CS =

Note :\(\frac{\theta}{\mathrm{I}}\)

Shunting a galvanometer decreases its current sensitivity.

Example 43. A galvanometer with a scale divided into 100 equal divisions, has a current sensitivity of 10 division per mA and voltage sensitivity of 2 division per mV. What adoptions are required to use it?

1. To read 5A full scale and
2. 1 division per volt

Solution :

Full scale deflection current ig=\(=\frac{\theta}{\mathrm{cs}}=\frac{100}{10} \mathrm{~mA}=10 \mathrm{~mA}\)

Full scale deflection voltage V_g= \(\frac{\theta}{\mathrm{vs}}=\frac{100}{2} \mathrm{mv}=50 \mathrm{mv}\)

So galvanometer resistance G = \(\frac{V_g}{i_g}=\frac{50 \mathrm{mV}}{10 \mathrm{~mA}} \quad=5 \Omega\)

To convert the galvanometer into an ammeter of range 5A, a resistance of value SΩ is connected in parallel with it such that

(Ι – i_g) S = in G

(5 – 0.01) S = 0.01 × 5

⇒ \(S=\frac{5}{499} \cong\) 0.01 Ω Ans.

To convert the galvanometer into a voltmeter which reads 1 division per volt, i.e. of range 100 V, V = ig(R + G)

100 = 10 × 10–3(R + 5)

R = 10000 – 5

R = 9995 Ω ≅ 9.995 kΩ Ans.

19. Potentiometer

Necessity of potentiometer

Practically voltameter has a finite resistance. (ideally, it should be ) in other words, it draws some current from the circuit. To overcome this problem potentiometer is used because, at the instant of measurement, it draws no current from the circuit.

Working principle of potentiometer

Any unknown potential difference is balanced on a known potential difference which is uniformly distributed over the entire length of the potentiometer wire.

This process is named as zero deflection or null deflection method.

Note :

• Potentiometer wire: Made up of alloys of magnin, constantan. Eureka.
• The specific properties of these alloys are high specific resistance and negligible temperature coefficient of resistance (α). The invariability of resistance of potentiometer wire over a long period.

Circuits of potentiometer

The primary circuit contains a constant source of voltage rheostat or Resistance Box

Secondary, Unknown, or galvanometer circuit

Let ρ = Resistance per unit length of potentiometer wire

Potential gradient (x)

The potential gradient corresponding to the unit length of the potentiometer is also called the potential gradient.

The rate of growth of potential per unit length of potentiometer wire is equal to the potential gradient. Let r = 0 and R1= 0 then V_AB = E (max. ideal) then x = E/L(V/m ; MLT-3A-1)

Always V_AB< E ;

⇒ \(\frac{V_{A B}}{L}\)

Now V_AB = I _RP(RP = resistance of potentiometer wire) L

⇒ \(\mathrm{x}=\frac{I R_p}{L}=1 \rho \quad \rho=\frac{R_P}{L}\)

current in primary circuit I = \(\frac{E}{R_1+r+R_P} ; \quad \mathrm{x}=\frac{E}{R_1+R_p+r}\left(\frac{R_p}{L}\right)\)

• If the radius is uniform = x is uniform over the entire length of the potentiometer wire. 1
• If I constant x ∝\(\frac{1}{(\text { radius })^2}\)
• ‘x’ directly depends on → ρ, r, σ, etc.

Factor affecting ‘x’

• If V_AB = const. and L = const. then for any change → x remains unchanged. 1
• If there is no information about VAB then Always take VAB as constant so (x ∝L)
• If V_AB and L are constant :
• For any change like the radius of wire, or substance of wire (σ) there is no change in x.
• Any change in the secondary circuit causes no change in x because x is an element of the primary circuit.

Note : x =\(\mathrm{x}=\frac{E}{R_p+r+R_1}\left(\frac{R_p}{L}\right)\)

x_max or x_min based on a range of rheostat or resistance box (R.B.)

If R1= 0 ⇒ xmax= \(\mathrm{x}_{\max }=\frac{E}{R_p} \times \frac{R_p}{L} \quad(\mathrm{r} \simeq 0)\)

If R1= R ⇒ \(\mathrm{x}_{\min }=\frac{E}{R_p+R}\left(\frac{R_p}{L}\right)\)

then\(\frac{x_{\max }}{x_{\min }}=\frac{R_p+R}{R_p}\)

Standardization And Sensitivity Of Potentiometer

Standardization process of evaluating x experimentally

If balanced length for standard cell (emf E) is = l0 then potential gradient \(x=\frac{E}{\ell_0}\)

Sensitivity :

x also indicates about sensitivity of the potentiometer.

If x ↓ ⇒ sensitivity ↑

To increase sensitivity → Rh ↑ (current in primary ckt should be reduced), L ↑

change in secondary ckt, no effect on sensitivity.

Balanced length for unknown potential difference ↑ ⇒ sensitivity ↑

Applications Of Potentiometer

To measure the potential difference across a resistance.

To find out the emf of a cell.

Comparison of two emfs 1

To find out the internal resistance of a primary cell.

Comparison of two resistances.

To find out an unknown resistance that is connected in series with the given resistance.

To find out the current in a given circuit.

Calibration of an ammeter or to have a check on the reading of (A)

Calibration of a voltmeter or to have a check on the reading of (V)

To find out thermocouple emf (et) (mV or μV)

Note :

For applications 3-6 no need for a standard cell and no need for a value of x.

For 7, 8, 9, and 10 – Always require a standard cell (E0= x0)

For 1 – 9 order of voltage drop (0.1 to 1v)

Comparison of emf of two cells

plug only in (1– 2) plug only in (2 – 3)

Jocky is at position J

balance length AJ = l1

E₁ = xl

Jocky is at position J’ balance length AJ’ = l₂

\(\mathrm{E}_2=\mathrm{x} \ell_2 \quad \Rightarrow \frac{E_1}{E_2}=\frac{\ell_1}{\ell_2}\)

Internal resistance of a given primary cell

E = V + I r ⇒ r = Rh

\(\mathrm{r}=\frac{E-V}{I} \text { or } \quad r=\left(\frac{E-V}{V}\right) R\)

Key K open E = xl1(AJ = 1) Key K closed T.P.D. V = xl2(AJ’ = l₂)

⇒ \(r=\left(\frac{\ell_1-\ell_2}{\ell_2}\right) R\)

Comparison Of Two Resistances

Plug only in (1–2)

The potential difference across R₁ is balanced

R1= xl₁

Plug only in (2-3)

The potential difference across (R1+R2) is balanced

⇒ \(\mathrm{I}\left(\mathrm{R}_1+\mathrm{R}_2\right)=\mathrm{x} \ell_2 \quad \frac{R_1+R_2}{R_1}=\frac{\ell_2}{\ell_1} \quad \Rightarrow \quad \frac{R_2}{R_1}=\frac{\ell_2-\ell_1}{\ell_1}\)

Measurement Of Current

Plug Only In (1–2)

E0= xl0

Plug only in (2–3)

V = I R = xl1

⇒ \(
\mathrm{I}=\frac{\ell_1}{R} \times \frac{E_0}{\ell_0}\)

Solved Examples

Example 44. The primary circuit of the potentiometer is shown in the figure determine :

1. Current in the primary circuit
2. The potential drop across potentiometer wire AB
3. Potential gradient (means potential drop per unit length of potentiometer wire)
4. Maximum potential which we can measure above potentiometer ε = 2V r = 1Ω R₁ = 20 Ω

Solution :

1. \(\mathrm{i}=\frac{\varepsilon}{\mathrm{r}+\mathrm{R}_1+\mathrm{R}}=\frac{2}{1+20+10} \Rightarrow \mathrm{i}=\frac{2}{31} \mathrm{~A}\)
2. \(V_{A B}=i R=\frac{2}{31} \times 10 \quad \Rightarrow \quad v_{A B}=\frac{20}{31} \text { volt }\)
3. \(x=\frac{V_{A B}}{L}=\frac{2}{31} \mathrm{volt} / \mathrm{m}\)
4. Maximum potential which we can measure by it = potential drop across wire AB

∴ \(\frac{20}{31} \text { volt }\)

Example 45. How to measure an unknown voltage using a potentiometer.
Solution :

The unknown voltage V is connected across the potentiometer wire as shown in the figure. The positive terminal of the unknown voltage is kept on the same side as the source of the topmost battery. When the reading of the galvanometer is zero then we say that the meter is balanced. In that condition V = xl.

Example 46. In an experiment to determine the emƒ of an unknown cell, its emf is compared with a standard cell of known emf ε1= 1.12 V. The balance point is obtained at 56cm with the standard cell and 80 cm with the unknown cell. Determine the emf of the unknown cell.

Solution:

Here, ε₁= 1.12 V; l₁= 56 cm; l₂= 80 cm

Using equation

ε₁= xl₁….(1)

ε₂= xl₂….(2)

we get \(\frac{\varepsilon_1}{\varepsilon_2}=\frac{\ell_1}{\ell_2} \quad \Rightarrow \quad \varepsilon_2=\varepsilon_1\left(\frac{\ell_2}{\ell_1}\right)\)

⇒ \(\varepsilon_2=1.12\left(\frac{80}{56}\right)=1.6 \mathrm{~V}\)

Example 47. A standard cell of emf ε0= 1.11 V is balanced against the 72 cm length of a potentiometer. The same potentiometer is used to measure the potential difference across the standard resistance R = 120 Ω. When the ammeter shows a current of 7.8 mA, a balanced length of 60 cm is obtained on the potentiometer.

1. Determine the current flowing through the resistor.
2. Estimate the error in measurement of the ammeter.

Solution : Here, l₀= 72 cm ; l= 60 cm; R = 120 Ω and ε₀= 1.11 V

By using equation ε₀= x l₀…….(1)

V = IR = xl …….(2)

From equation (1) and (2)

⇒ \(\mathrm{I}=\frac{\varepsilon_0}{\mathrm{R}}\left(\frac{\ell}{\ell_0}\right)\)

⇒ \(I=\frac{1.11}{120}\left(\frac{60}{72}\right)=7.7 \mathrm{~mA}\)

Since the measured reading is 7.8 mA (> 7.7 mA) therefore, the instrument has a positive error.

ΔI = 7.8 – 7.7 = 0.1 mA,0.1

⇒ \(\frac{\Delta \mathrm{I}}{\mathrm{I}}=\frac{0.1}{7.7} \times 100=1.3 \%\)

Example 48. The internal resistance of a cell is determined by using a potentiometer. ln an experiment, an external resistance of 60Ω is used across the given cell. When the key is closed, the balance length on the potentiometer decreases from 72 cm to 60 cm. Calculate the internal resistance of the cell.

Solution :

According to equation ε₀ = xl₀….(1)

V = IR = xl ….(2)

I =\(I=\frac{\varepsilon_0}{R+r}\) …..(3)

From equation (1), (2) and (3) we get

⇒ \(\mathrm{r}=\mathrm{R}\left(\frac{\ell_0}{\ell}-1\right)\)

Here l0= 72 cm; l = 60 cm; R = 60 Ω 72 –1

∴ r = (60) \(\left(\frac{72}{60}-1\right)\) r = 12 Ω. 60

Example 49. Comprehension A 6-volt battery of negligible internal resistance is connected across a uniform wire AB of length 100 cm. The positive terminal of another battery of emf 4V and internal resistance 1Ω is joined to point A as shown in the figure. Take the potential at B to be zero.

Question 1. The potentials at the points A and C

1. V_A= 6 V, V_C= 2V
2. V_A= 3V, V_C= 2V
3. V_A= 2V, V_C= 3V
4. None of these

Solution:

1. V_A= 6 V

V_C= 2V

Question 2. Which point D of the wire AB, the potential is equal to the potential at C.

1. AD = 200
2. AD = \(\frac{200}{3}\)
3. AD = \(\frac{100}{3}\)
4. None of these

Solution:

E = x l⇒ \(\frac{6}{100}\) ⇒ \(\frac{200}{3}\)

Question 3. If the 4V battery is replaced by 7.5 V battery, what would be the potentials at the points A and C

1. V_A= 6 V, V_C= 2V
2. V_A= 6 V, V_C= 1.5V
3. V_A= –6 V, V_C= 1.5V
4. V_A= 6 V, V_C= –1.5V

Solution:

6 V, 6 – 7.5 = – 1.5 V, no such point D exists Ans.

6 V, 6 – 7.5 = – 1.5 V

20. Metre Bridge (Use To Measure Unknown Resistance)

If AB = l cm, then BC = (100 – l) cm.

Resistance of the wire between A and B R ∝ l

[ Specific resistance ρ and cross-sectional area A are the same for the whole of the wire ]

or R = σl  …(1)

where σ is the resistance per cm of wire.

Similarly, if Q is the resistance of the wire between B and C, then

Q ∝ 100 – l

∴ Q = σ(100 – l) ….(2)

Dividing (1) by (2),

⇒ \(\frac{P}{Q}=\frac{\ell}{100-\ell}\)

Applying the condition for a balanced Wheatstone bridge, we get

R Q = P X ∴ x = R \(\frac{\mathrm{Q}}{\mathrm{P}} \quad \text { or } \quad \mathrm{X}=\frac{100-\ell}{\ell} \mathrm{R}\)

Since R and l are known, therefore, the value of X can be calculated.

Note: For better accuracy, R is so adjusted that Ω lies between 40 cm and 60 cm.

Example 50. In a meter bridge experiment, the value of unknown resistance is 2Ω. To get the balancing point at a 40cm distance from the same end, the resistance in the resistance box will be :

1. 0.5 Ω
2. 3 Ω
3. 20 Ω
4. 80 Ω

Solution :

Apply condition for balanced wheat stone bridge,

⇒ \(\frac{P}{Q}=\frac{\ell}{100-\ell}=\frac{P}{2}=\frac{100-40}{40}\)

Answer: P = 3Ω.

21. Post-Office Box (Experiment In Cbse)

Post-Office Box Introduction: It is so named because its shape is like a box and it was originally designed to determine the resistances of electric cables and telegraph wires. It was used in post offices to determine the resistance of transmission lines.

Construction: A post office box is a compact form of a Wheatstone bridge with the help of which we can measure the value of the unknown resistance correctly up to the 2nd decimal place, i.e., up to 1/100th of an ohm. Two types of post office boxes are available – plug type and dial type.

• In the plug-type instrument shown in Figure (a), each of the arms AB and BC contains three resistances of 10, 100, and 1000 ohms. These arms are called ratio arms. While resistance P can be introduced in arm AB, resistance Q can be introduced in arm BC.
• The third arm AD called the resistance arm, is a complete resistance box containing resistances from 1 Ω to 5,000 Ω. In this arm, the resistance R is introduced by taking out plugs of suitable values.
• The unknown resistance X constitutes the fourth arm CD. Thus, the four arms AB, BC, CD, and AD infect the four arms of the Wheatstone bridge (figure (b)).
• Two tap keys K1 and are also provided. While K1 is connected internally to terminal A, K2 is connected internally to B. These internal connections are shown by dotted lines in Figure (a).
• A battery is connected between C and key K1(battery key). A galvanometer is connected between D and key K2(galvanometer key).
• Thus, the circuit is the same as that shown in Figure (b). It is always the battery key which is pressed first and then the galvanometer key.
• This is because a self-induced current is always set up in the circuit whenever the battery key is pressed or released. If we first press the galvanometer key, the balance point will be disturbed on account of the induced current.
• If the battery key is pressed first, then the induced current becomes zero by the time the galvanometer key is pressed. So, the balance point is not affected.

Working: The working of the post office box involves broadly the following four steps :

Keeping R zero, each of the resistances P and Q are made equal to 10 ohms by taking out suitable plugs from the arms AB and BC respectively. After pressing the battery key first and then the galvanometer key, the direction of deflection of the galvanometer coil is noted. Now, making R infinity, the direction of deflection is again noted. If the direction is opposite to that in the first case, then the connections are correct.

Keeping both P and Q equal to 10Ω, the value of R is adjusted, beginning from 1Ω, till 1 Ω increase reverses the direction of deflection. The ‘unknown’ resistance lies somewhere between the two final values of R.
148

⇒ \(\left[X=R \frac{Q}{P}=R \frac{10}{10}=R\right]\)

As an illustration, suppose with 3Ω resistance in the arm AD, the deflection is towards left, and with 4Ω, it is towards right. The unknown resistance lies between 3Ω and 4Ω.

Making P 100 Ω and keeping Q 10 Ω, we again find those values of R between which the direction of deflection is reversed. The resistance in the arm AD will be 10 times the resistance X of the wire.

⇒ \(\left[X=R \frac{Q}{P}=R \frac{10}{100}=\frac{R}{10}\right]\)

In the illustration considered in step II, the resistance in the arm AD will now lie between 30 Ω and 40 Ω. So, in this step, we have to start adjusting R from 30 Ω onwards. If 32 Ω and 33 Ω are the two values of R which give opposite deflections, then the unknown resistance lies between 3.2 Ω and 3.3Ω.

Now, P is made 1000 Ω and Q is kept at 10 Ω. The resistance in the arm AD will now be 100 times the ‘unknown’ resistance.

⇒ \(\left[X=R \frac{10}{1000}=\frac{R}{100}\right]\)

In the illustration under consideration, the resistance in the arm AD will lie between 320 Ω and 330Ω. Suppose the deflection is to the right for 326 ohms, towards the left for 324 ohms, and zero deflection for 325Ω Then, the unknown resistance is 3.25 Ω.

The post office box method is a less accurate method for the determination of unknown resistance as compared to a meter bridge. This is because it is not always possible to arrange resistance in the four arms to be of the same order. When the arms ratio is large, large resistance is required to be introduced in the arm R.

Solved Miscellaneous Problems

Problem 1. Current is flowing from a conductor of non-uniform cross-section area if A1> then find the relation between

1. i₁and i₂
2. j₁and j₂
3. v₁and v₂ (drift velocity)
4. where i is current, j is current density and V is drift velocity.

Answer :   i₁= i₂, V₁< V₂, J₁< J₂
Solution :

i = charge flowing through a cross-section per unit time.

∴ i₁= i₂

⇒ \(j=\frac{i}{A}\)

as A₁> A2then j₁< j₂

j = need

⇒ \(v_d=\frac{j}{n e}\)

as j₁< j₂ then, v₁< v₂

Problem 2. Find the equivalent resistance between A and B

Solution :

Putting A out of the structure in the same plane

⇒\(R_{e q}=\frac{2 R \times 2 R}{2 R+2 R}=R\)

Answer: R_eq= R

Problem 3. What shunt resistance is required to convert the 1.0 mA, 20Ω galvanometer into an ammeter with a range of 0 to 50 mA?

Answer : S = \(\frac{20}{49}=0.408 \Omega\)

Solution: in Rg= (i – i_g)S

i_g= 1.0 × 10^-3 A , G = 20Ω

i = 50 × 10^-3 A

S = \(S=\frac{i_g R_g}{i-v_g}=\frac{1 \times 10^{-3} \times 20}{49 \times 10^{-3}}=0.408 \Omega\)

Problem 4. How can we convert a galvanometer with Rg= 20 Ω and i_g= 1.0 mA into a voltmeter with a maximum range of 10 V?
Answer :

A resistance of 9980 Ω is to be connected in series with the galvanometer. Solution :

v = i_g RS+ i_g R_g

10 = 1 × 10^-3 × Rs+ 1 × 10^-3 × 20

⇒ \(\mathrm{R}_{\mathrm{s}}=\frac{10-0.02}{1 \times 10^{-3}}=\frac{9.98}{10^{-3}}=9980 \Omega\)

Problem 5. A Potentiometer wire of 10 m in length and having 10-ohm resistance, emf 2 volts, and a rheostat. If the potential gradient is 1 microvolt/mm, the value of resistance in rheostat in ohms will be :

1. 1.99
2. 19.9
3. 199
4. 1990

Solution : d = 10 m , R = 10Ω ,

E = 2volts , \(\frac{\mathrm{dv}}{\mathrm{d} \ell}\)= 1µ v/mm

⇒ \(\frac{\mathrm{dv}}{\mathrm{d} \ell}=\frac{1 \times 10^{-6}}{1 \times 10^{-3}} \mathrm{v} / \mathrm{m}=1 \times 10^{-3} \mathrm{v} / \mathrm{m}\)

Across wire potential drop,

⇒ \(\frac{\mathrm{dv}}{\mathrm{d} \ell} \times \ell=1 \times 10^{-3} \times 10=0.01 \text { volts }\)

⇒ \(i=\frac{0.01}{10}=0.001=\frac{E}{R+R^{\prime}}\) R+(R’ = resistance of rheostat) ‘

⇒ \(R^{\prime}=\frac{E}{0.001}-R=\frac{2}{0.001}-10=2000-10=1990 \Omega\)

Summary Current Electricity

⇒ \(\mathrm{I}_{\mathrm{av}}=\frac{\Delta \mathrm{q}}{\Delta \mathrm{t}} \text { and } \mathrm{i}_{\text {mst. }}=\frac{\mathrm{dq}}{\mathrm{dt}} \Rightarrow \mathrm{q}=\)∫idt = area between current – time graph on time axis.

Current i = ne A Vdn = no. of free electron per unit volume, A = cross-section area of conductor, Vd= drift velocity, e = charge on electron = 1.6 × 10^-19 C

Ohm’s law V = ΙR

R = \(\mathrm{R}=\frac{\rho \ell}{\mathrm{A}}\) ρ = resistivity = \(\frac{1}{\sigma}\), σ = conductivity

Power P = VΙ ⇒ P =\(\mathrm{I}^2 \mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{R}}\)

Energy = power × time (if power is constant.) otherwise energy, E =

The rate at which the cell’s chemical energy is consumed = EiP.dt ∫where P is power.

The rate at which heat is generated inside the battery = i2r

Electric power output = (ε – ir) i

Maximum power output when net internal resistance = net external resistance, R = r Maximum power output = \(\frac{\varepsilon^2}{4 r}\)

In series combination R = R₁+ R₂+ R₃+ ……….

In parallel combination \(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\ldots . . .\)

Cell in series combination

Eeq= ε₁+ ε₂+ ε₃+ ……. + εn(write Emf’s with polarity) req= r₁+ r₂+ r₃+ ………

Cells in parallel combination

⇒ \(E_{e q}=\frac{\frac{\varepsilon_1}{r_1}+\frac{\varepsilon_2}{r_2}+\ldots . .+\frac{\varepsilon_n}{r_n}}{\frac{1}{r_1}+\frac{1}{r_2}+\ldots \ldots+\frac{1}{r_n}}\) (Use proper sign before the EMFs for polarity)

and \(\frac{1}{r_{e q}}=\frac{1}{r_1}+\frac{1}{r_2}+\ldots . .+\frac{1}{r_n}\)

• In ammeter shunt (S) = \(\frac{\mathrm{I}_{\mathrm{G}} \times \mathrm{R}_{\mathrm{G}}}{\mathrm{I}-\mathrm{I}_{\mathrm{G}}}\)
• In voltmeter V = ΙG RS+ ΙG RG
• Potential gradient in potentiometer :\(x=\frac{\varepsilon}{R+r} \times \frac{R}{L}\)

⇒ \(\frac{\theta_{\mathrm{i}}+\theta_{\mathrm{c}}}{2}=\theta_{\mathrm{n}}\) where, θi= inversion temperature

θ_C= Temperature of cold junction

θ_n= Neutral temperature

In a balanced wheat stone bridge

⇒ \(\frac{P}{R}=\frac{Q}{S}\)

Current Electricity Exercise – 1

Section (1): Definition Of Current, Current Densities, Drift

Question 1. The drift velocity of electrons in a conducting wire is of the order of 1mm/s, yet the bulb glows very quickly after the switch is put on because

1. The random speed of electrons is very high, of the order of 106 m/s
2. The electrons transfer their energy very quickly through collisions
3. The electric field is set up in the wire very quickly, producing a current through each cross-section, almost instantaneously
4. All of above

Answer: 3. The Electric field is set up in the wire very quickly, producing a current through each cross-section, almost instantaneously

Question 2. In the presence of an applied electric field (E) in a metallic conductor.

The electrons move in the direction of E

The electrons move in a direction opposite toE

The electrons may move in any direction randomly but slowly drift in the direction of E.

The electrons move randomly but slowly drift in a direction opposite to E.

Answer: 4. The electrons move randomly but slowly drift in a direction opposite to E.

Question 3. A current of 4.8 A is flowing in a conductor. The number of electrons passing through any cross-section per second is

1. 3 × 10¹⁹
2. 76.8 × 10²⁰
3. 7.68 × 10¹²
4. 3 × 10¹⁰

Answer: 1. 3 × 10¹⁹

Section (2): Resistance

Question 1. The specific resistance of a wire depends on its

1. Mass
2. Length
3. Area of cross-section
4. None of the above

Answer: 4. None of the above

Question 2. There are two wires of the same length and of the same material and radii r and 2r. The ratio of their specific resistance is

1. 1: 2
2. 1: 1
3. 1: 4
4. 4: 1

Answer: 2. 1: 1

Question 3. If the length and cross-section of a wire is doubled, then the resistance will

1. Become half
2. Increase two times
3. Remain unchanged
4. Increase four times

Answer: 3. Remain unchanged

Question 4. V-i graph for an ohmic resistance is

1. Straight line
2. Hyperbola
3. Parabola
4. Circle

Answer: 1. Straight line

Question 5. When a resistance wire is passed through a die the cross–section area decreases by 1%, and the change in resistance of the wire is

1. 1% decrease
2. 1% increase
3. 2% decrease
4. 2% increase

Answer: 4. 2% increase

Question 6. When the resistance of a copper wire is 0.1 Ω and the radius is 1 mm, then the length of the wire is (specific resistance of copper is 3.14 × 10–8 ohm x m)

1. 10 cm
2. 10 m
3. 100 m
4. 100 cm

Answer: 2. 10 m

Question 7. Three copper wires of length and cross-sectional area (L, A), (2L, A/2), and (L/2, 2A). Resistance is minimal in

1. wire of cross-sectional area A
2. wire of area A/2
3. wire of cross-sectional area 2A
4. same in all three cases

Answer: 3. wire of cross-sectional area 2A

Question 8. The resistance of a wire of cross-section ‘a’ and length ‘ Ω ’ is R ohm. The resistance of another wire of the same material and the same length but cross-section ‘4a’ will be

1. 4R
2. \(\frac{R}{4}\)
3. \(\frac{R}{16}\)
4. 16R

Answer: 2. \(\frac{R}{4}\)

Question 9. The resistance of a wire is R ohm. The wire is stretched to half of its diameter. The resistance of the wire will now be

1. 4R
2. 64 R
3. R/4
4. 16 R

Answer: 4. 16 R

Question 10. If a wire of resistance R is stretched to double its length, then the new resistance will be

1. \(\frac{\mathrm{R}}{2}\)
2. 2R
3. 4R
4. 16R

Answer: 3. 4R

Question 11. All block edges in cuboidal shape with parallel faces are unequal. Its longest edge is twice its shortest edge. The ratio of the maximum to minimum resistance between parallel faces is (a > b > c)

1. 2
2. 4
3. 8
4. Indeterminate unless the length of the third edge is specified.

Answer: 2. 4

Question 12. Read the following statements carefully :

Y: The resistivity of the semiconductor decreases with the increase of temperature.

Z: In a conducting solid, the rate of collisions between free electrons and ions increases with the increase in temperature. Select the correct statement (s) from the following :

1. Y is true but Z is false
2. Y is false but Z is true
3. Both Y and Z are true
4. Y is true and Z is the correct reason for Y

Answer: 3. Both Y and Z are true

Question 13. The dimensions of a block are 1 cm x 1 cm x 100 cm. If the specific resistance of its material is 2 10 ohm meter − 7 × ×, then the resistance between the opposite rectangular faces is

1. 2 × 10−^-8Ω
2. 2 × 10^-7Ω
3. 2 × 10^-5Ω
4. 2 × 10^-3Ω

Answer: 2. 2 × 10−7Ω

Question 14. A conductor with a rectangular cross-section has dimension (a × 2a × 4a) as shown in Fig. Resistance across AB is x, across CD, is y, and across EF is z. Then

1. x = y = z
2. x > y > z
3. y > z > x
4. x > z > y

Answer: 4. x > z > y

Question 15. The resistance of a wire is 20 ohm, it is stretched up, three times its length, then its new resistance will be

1. 6.67 Ω
2. 60 Ω
3. 120 Ω
4. 180 Ω

Answer: 4. 180 Ω

Question 16. Two wires of the same dimension but resistivities ρ1and ρ2are connected in series. The equivalent resistivity of the combination is

1. ρ1+ ρ2
2. \(\left(\frac{\rho_1+\rho_2}{2}\right)\)
3. \(\sqrt{\rho_1 \rho_2}\)
4. 2(ρ1+ ρ2)

Answer: 2. \(\left(\frac{\rho_1+\rho_2}{2}\right)\)

Question 17. In the given figure, the equivalent resistance between the points A and B is

1. 8 Ω
2. 6 Ω
3. 4 Ω
4. 2 Ω

Answer: 2. 6 Ω

Question 18. A bridge circuit is shown in the figure. The equivalent resistance between A and B will be :

1. 21Ω
2. 7Ω
3. 252
4. \(\frac{\mathrm{252}}{85}\)Ω
5. 143Ω

Answer: 4. 143Ω

Question 19. A cylindrical wire is stretched up to twice its length, then its resistance becomes :

1. 4 times
2. unchanged
3. 12times
4. 2 times

Answer: 1. 4 times

Question 20. Si and Cu are cooled to a temperature of 300K, then resistivity :

1. For Si increases and Cu decreases
2. For Cu increases and Si decreases
3. Decreases for both Si and Cu
4. Increases for both Si and Cu

Answer: 1. For Si increases and Cu decreases

Question 21. Two copper wires are of the same length one of them twice as thick as the other. Then the resistance of the two wires is in the ratio of :

1. 1: 16
2. 1: 8
3. 1: 4
4. 1: 2

Answer: 3. 1: 4

Question 22. If resistance of wire at 50ºC is 5RΩ and 100ºC is 6RΩ, find resistance of 0ºC :

1. 0RΩ
2. 2RΩ
3. 3RΩ
4. 4RΩ

Answer: 4. 4RΩ

Question 23. For a metallic wire, the ratio (V=Viapplied potential difference and i = current flowing) is

1. Independent of temperature
2. Increases as the temperature rises
3. Decreases as the temperature rises
4. Increases or decreases as temperature rises depending upon the metal

Answer: 2. Increases as the temperature rises

Question 24. The length of the given cylindrical wire is increased by 100%. Due to the consequent decrease in diameter, the change in the resistance of the wire will be

1. 200%
2. 100%
3. 50%
4. 300%

Answer: 4. 300%

Question 25. If on applying the potential of 20 V on a conductor its conductance becomes 8 (Ω)–1, then the current flowing through it will be

1. 120 A
2. 160 A
3. 90 A
4. 80 A

Answer: 2. 160 A

Question 26. The figure shows a graph of current (I) flowing through a cell against its terminal voltage (V). The EMF and internal resistance of this cell are

1. 1 V ; 0.25 Ω
2. 1 V ; 4 Ω
3. 1 V ; 1 Ω
4. 0.25 V ; 0.25 Ω

Answer: 1. 1 V; 0.25 Ω

Question 27. The length of a given cylindrical wire is increased by 100%. Due to the consequent decrease in diameter, the change in the resistance of the wire will be

1. 300 %
2. 200 %
3. 100 %
4. 50 %

Answer: 1. 300 %

Question 28. A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be :

1. Doubled
2. Four times
3. One-fourth
4. Half

Answer: 1. Doubled

Question 29. A material ‘B’ has twice the specific resistance of ‘A’. A circular wire made of ‘B’ has twice the diameter of a wire made of ‘A’. Then for the two wires to have the same resistance, the ratio IA/μ of their respective lengths must be

1. 2
2. 1
3. 1/2
4. 1/4

Answer: 3. 1/2

Question 30. The resistance of a wire is 5 ohms at 50º C and 6 ohms at 100ºC. The resistance of the wire at 0ºC will be

1. 2 ohm
2. 1 ohm
3. 4 ohm
4. 3 ohm

Answer: 3. 4 ohm

Question 31. Consider a thin square sheet of side L and thickness t, made of a material of resistivity ρ. The resistance between two opposite faces, shown by the shaded areas in the figure is :

1. Directly proportional to L
2. Directly proportional to t
3. Independent of L
4. Independent of t

Answer: 3. Independent of L

Question 32. The specific resistance of a conductor increases with :

1. Increase in temperature
2. Increases in cross-sectional area
3. Decreases in length
4. Decrease in cross-sectional area

Answer: 1. Increase in temperature

Question 33. A fuse wire is a wire of :

1. Low resistance and low melting point
2. Low resistance and high melting point
3. High resistance and high melting point
4. High resistance and low melting point

Answer: 4. High resistance and low melting point

Question 34. The electric resistance of a certain wire of iron is R. If its length and radius are both doubled, then :

1. The resistance will be doubled and the specific resistance will be halved
2. The resistance will be halved and the specific resistance will remain unchanged
3. The resistance will be halved and the specific resistance will be doubled.
4. The resistance and the specific resistance, will both remain unchanged

Answer: 2. The resistance will be halved and the specific resistance will remain unchanged

Question 35. A 6V battery is connected to the terminals of a three-metre-long wire of uniform thickness and resistance of 100Ω. The difference of potential between two points on the wire separated by a distance of 50 cm will be :

1. 2V
2. 3V
3. 1V
4. 1.5 V

Answer: 3. 1V

Question 36. When a wire of uniform cross-section a, length l, and resistance R is bent into a complete circle, the resistance between two diametrically opposite points will be :

1. \(\frac{\mathrm{R}}{4}\)
2. \(\frac{\mathrm{R}}{8}\)
3. 4R
4. \(\frac{\mathrm{R}}{2}\)

Answer: 1. \(\frac{\mathrm{R}}{4}\)

Question 37. A wire of a certain material is stretched slowly by ten percent. It’s new resistance and specific resistance become respectively.

1. 1.2 times, 1.1 times
2. 1.21 times, same
3. both remain the same
4. 1.1 times, 1.1 times

Answer: 2. 1.21 times, same

Question 38. An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water at a temperature of 20ºC? The temperature of boiling water is 100ºC

1. 6.3 min
2. 8.4 min
3. 12.6 min
4. 4.2 min

Answer: 1. 6.3 min

Section (3): Power, Energy, Battery, Emf, Terminal Voltage, Kcl, And Kvl

Question 1. In an electric circuit containing a battery, the positive charge inside the battery

1. It always goes from the positive terminal to the negative terminal
2. May go from the positive terminal to the negative terminal
3. It always goes from the negative terminal to the positive terminal
4. Does not move.

Answer: 2. May go from the positive terminal to the negative terminal

Question 2. In which of the above cells, does the potential difference between the terminals of a cell exceed its emf?

1. 1
2. 2
3. 3
4. 4

Answer: 2. 2

Question 3. The efficiency of a cell when connected to a resistance R is 60%. What will be its efficiency if the external resistance is increased to six times?

1. 80 %
2. 90%
3. 55%
4. 95%

Answer: 2. 90%

Question 4. Two bulbs 25W, 220V and 100W, 220V are given. Which has higher resistance?

1. 25W bulb
2. 100 W bulb
3. Both bulbs will have equal resistance
4. The resistance of bulbs cannot be compared

Answer: 1. 25W bulb

Question 5. The resistors whose ratio is 1: 2, are connected in parallel, the ratio of power dissipated is :

1. 1: 2
2. 1: 4
3. 4: 1
4. 2: 1

Answer: 4. 2: 1

Question 6. Find the current flowing through the resistance R1of the circuit shown in the figure if the resistances are equal to R₁= 10 Ω, R₂= 20 Ω, and R₃= 30 Ω, and the potentials of points 1, 2, and 3 are equal to ϕ₁= 10 V, ϕ₂= 6 V, and ϕ₃= 5 V.

1. 0.1 A
2. 0.2 A
3. 0.3 A
4. 0.4 A

Answer: 2. 0.2 A

Question 7. In the previous question potential at point 0 is

1. 15 V
2. 20 V
3. 25 V
4. 8 V

Answer: 4. 8 V

Question 8. In the figure a part of the circuit is shown :

1. current will flow from A to B
2. current may flow from A to B
3. current will flow from B to A
4. The direction of the current will depend on r.

Answer: 2. current may flow from A to B

Question 9. In the shown circuit, what is the potential difference across A and B

1. 50 V
2. 45 V
3. 30 V
4. 20 V

Answer: 4. 20 V

Question 10. A cell has an emf of 1.5V. When connected across an external resistance of 2Ω, the terminal potential difference falls to 1.0V. The internal resistance of the cell is :

1. 2Ω
2. 1.52Ω
3. 1.0Ω
4. 0.5Ω

Answer: 3. 1.0Ω

Question 11. A battery having e.m.f. 5 V and internal resistance 0.5 Ω is connected with a resistance of 4.5 Ω then the voltage at the terminals of the battery is

1. 4.5 V
2. 4 V
3. 0 V
4. 2 V

Answer: 1. 4.5 V

Question 12. Watt-hour meter measures

1. Current
2. Electric energy
3. Power
4. Voltage

Answer: 2. Electric energy

Question 13. Two resistors whose values are in a ratio of 2:1 are connected in parallel with one cell. Then ratio of power dissipated is

1. 2: 1
2. 4: 1
3. 1: 2
4. 1: 1

Answer: 3. 1: 2

Question 14. A hot electric iron has a resistance of 80Ω and is used on a 200 V source. The electrical energy spent, if it is used for 2 hours, will be :

1. 8000 Wh
2. 2000 Wh
3. 1000 Wh
4. 800 Wh

Answer: 3. 1000 Wh

Question 15. A current of 2 A flowing through a conductor produces 80 J of heat in 10 s. The resistance of the conductor is

1. 0.5 Ω
2. 2 Ω
3. 4 Ω
4. 20 Ω

Answer: 2. 2 Ω

Question 16. An electric heater of resistance 6 Ω is run for 10 min on a 120 V line. The energy liberated in this period is

1. 7.2 × 10³ J
2. 14.4 × 10⁵ J
3. 43.2 × 10⁴ J
4. 28.8 × 10⁴ J

Answer: 2. 14.4 × 10⁵ J

Question 17. A wire when connected to 220 V mains supply has power dissipation P₁. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is P₂. Then P₂: P₁is-

1. 1
2. 4
3. 2
4. 3

Answer: 2. 4

Question 18. A 220-volt, 1000-watt bulb is connected across a 110-volt mains supply. The power consumed will be

1. 750 watt
2. 500 watt
3. 250 watt
4. 1000 watt

Answer: 3. 250 watt

Question 19. The time taken by an 836 W heater to heat one liter of water from 10ºC to 40ºC is :

1. 50 s
2. 100 s
3. 150 s
4. 200 s

Answer: 3. 150 s

Question 20. The resistance of hot tungsten filament is about 10 times the cold resistance. What will be the resistance of 100 W and 200 V lamps when not in use :

1. 40 Ω
2. 20 Ω
3. 400 Ω
4. 200 Ω

Answer: 1. 40 Ω

Question 21. An electric bulb is rated from 220 volts – 100 watts. The power consumed by it when operated at 110 volts will be

1. 25 watt
2. 50 watt
3. 75 watt
4. 40 watt

Answer: 1. 25 watt

Question 22. The Kirchhoff’s first law \(\left(\sum \mathrm{i}=0\right)\) where the symbols have their usual meanings, are respectively based on

1. conservation of charge, conservation of energy
2. conservation of charge, conservation of momentum
3. Conservation of energy, conservation of charge
4. Conservation of momentum, conservation of charge

Answer: 1. conservation of charge, conservation of energy

Question 23. An electric kettle has two heating coils. When one of the coils is connected to an AC source, the water in the kettle boils in 10 min. When the other coil is used the water boils in 40 min. If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be :

1. 25 min
2. 15 min
3. 8 min
4. 4 min

Answer: 3. 8 min

Question 24. Two 220V, 100W bulbs are connected first in series and then in parallel. Each time the combination is connected to a 220V AC supply line. The power drawn by the combination in each case respectively will be :

1. 200W, 150W
2. 50W, 200W
3. 50W, 100W
4. 100W, 50W

Answer: 2. 50W, 200W

Question 25. A battery is charged at a potential of 15V for 8H when the current flowing is 10A. The battery on discharge supplies a current of 5A for 15H. The mean terminal voltage during discharge is 14V. The “watt-hour” efficiency of the battery is :

1. 82.5 %
2. 80%
3. 90%
4. 87.5 %

Answer: 4. 87.5 %

Question 26. In India, electricity is supplied for domestic use at 220V. It is supplied at 110V in the USA. If the resistance of a 60W bulb for use in India. is R, the resistance of a 60W bulb for use in the USA will be

1. R
2. 2R
3. \(\frac{\mathrm{R}}{4}\)
4. \(\frac{R}{2}\)

Answer: 3. \(\frac{\mathrm{R}}{4}\)

Question 27. A 5-A fuse wire can withstand a maximum power of 1W in the circuit. The resistance of the fuse wire is :

1. 0.2Ω
2. 5Ω
3. 0.4Ω
4. 0.04Ω

Answer: 4. 0.04Ω

Question 28. Kirchhoff’s first and second laws for electrical circuits are consequences of:-

1. Conservation Of Energy
2. Conservation Of Electric Charge And Energy Respectively
3. Conservation Of Electric Charge
4. Conservation Of Energy And Electric Charge Respectively

Answer: 2. Conservation Of Electric Charge And Energy Respectively

Question 29. When three identical bulbs of 60W, 200 V rating are connected in series to a 200V supply, the power drawn by them will be :

1. 60 W
2. 180 W
3. 10 W
4. 20 W

Answer: 4. 20 W

Section (4): Combination Of Resistance

Question 1. Two coils connected in series have resistances 600 Ω and 300 Ω at 20°C and temperature coefficient of resistivity 0.001 k–1 and 0.004 k–1respectively.

1. The resistance of the combination at a temperature of 50°C is

1. 426 Ω
2. 954 Ω
3. 1806 Ω
4. 214 Ω

Answer: 2. 954 Ω

2. The effective temperature coefficient of the combination is

1. \(\frac{1}{1000}\)degree–1
2. \(\frac{1}{250}\)
3. \(\frac{1}{500}\)degree–1
4. \(\frac{3}{1000}\)degree–1

Answer: 3. \(\frac{1}{500}\)degree–1

Question 2. A wire has a resistance of 12 ohms. If it is bent in the form of a circle. The effective resistance between the two points on any diameter is equal to

1. 6 Ω
2. 3 Ω
3. 9 Ω
4. 12 Ω

Answer: 2. 3 Ω

Question 3. A wire has a resistance of 12 ohms. if it is bent in the form of an equilateral triangle. The resistance between any two terminals is

1. 8/3
2. 3/4
3. 4
4. 3

Answer: 1. 8/3

Question 4. There are five resistances of 1 ohm each. If the initial three resistances are joined in parallel and rest two are joined in series, then the final resistance is

1. 3 ohm
2. 8 ohm
3. 7/3 ohm
4. 5 ohm

Answer: 3. 7/3 ohm

Question 5-11 For the following circuits, the equivalent resistance between X and Y in volts is 5. (Take R = 3 Ω)

Question 5.

1. R
2. 2R
3. 3R
4. R/2

Answer: 1. R

Question 6.

1. 4RΩ
2. 8R/3
3. R
4. 3R

Answer: 2. 8R/3

Question 7.

1. R
2. 4 R
3. 5 R
4. 6 R

Answer: 1. R

Question 8.

1. 10 Ω
2. 20 Ω
3. 30 Ω
4. ∞ Ω

Answer: 1. 10 Ω

Question 9.

1. 10 Ω
2. 20 Ω
3. 30 Ω
4. 40 Ω

Answer: 1. 10 Ω

Question 10.

1. 5 Ω
2. 10 Ω
3. 15 Ω
4. 60 Ω

Answer: 3. 15 Ω

Question 11. For the network of resistance shown in the figure, the equivalent resistance of the network between points A and B is 18 ohms. The value of unknown resistance R is:-

1. 8Ω
2. 0Ω
3. 16Ω
4. 24Ω

Answer: 3. 16Ω

Question 12. If 2 bulbs rated 2.5 W – 110 V and 100 W – 110 V are connected in series to a 220 V supply then

1. 2.5 W bulb will fuse
2. 100 W bulb will fuse
3. Both will fuse
4. Both will not fuse

Answer: 1. 2.5 W bulb will fuse

Question 13. A 50 W bulb is in series with a room heater and the combination is connected across the mains. To get max. heater output, the 50 W bulb should be replaced by

1. 25 W
2. 10 W
3. 100 W
4. 200 W

Answer: 4. 200 W

In the following Questions (14 to 19), find the potential difference between points X and Y.

Question 14.

1. 1
2. –1
3. 2
4. –2

Answer: 1. 1

Question 15.

1. 2
2. 3
3. 6
4. 9

Answer: 1. 2

Question 16.

1. 10
2. 20
3. 0
4. 5

Answer: 2. 20

Question 17.

1. 0.1
2. 2
3. 0.3
4. 0.4

Answer: 2. 2

Question 18.

1. 10
2. 50
3. 100
4. 0

Answer: 4. 0

Question 19.

1. 2/3
2. 4/3
3. 8/9
4. 5/8

Answer: 1. 2/3

Question 20. The resistance of the P, Q, and R S arms of a Wheatstone bridge are 5, 15, 20, and 60Ω. A cell of 4-volt emf and 4Ω internal resistance is connected with them, then the current flowing (in ampere) is

1. 0.1
2. 0.2
3. 1
4. 2

Answer: 2. 0.2

Question 21. The equivalent resistance between A and B in the given circuit will be

1. R
2. 2R
3. R2
4. 6R

Answer: 1. R

Question 22. In a showing figure find the equivalent resistance between A and B.

1. 2/3 ohm
2. 5/ ohm
3. 8/9 ohm
4. 7.5 ohm

Answer: 4. 7.5 ohm

Question 23. 40W, 100W, and 200 W bulbs are connected with a source of 200 V ratings of all bulbs are also 200V. Now, they are connected in series, then which bulbs will glow more :

1. 300 W
2. 100W
3. 40W
4. All give the same light

Answer: 3. 40W

Question 24. In a Wheatstone’s bridge resistance of each of the four sides is 10Ω. If the resistance of the galvanometer is also 10Ω, then the effective resistance of the bridge will be :

1. 10Ω
2. 5Ω
3. 20Ω
4. 40Ω

Answer: 1. 10Ω

Question 25. In the circuit shown, P ≠ R, the reading of the galvanometer is the same as switch S open or closed. Then

1. ΙR= ΙG
2. ΙP= ΙG
3. ΙQ= ΙG
4. ΙQ= ΙR

Answer: 1. ΙR= ΙG

Question 26. Find equivalent resistance between X and Y :

1. R
2. R/2
3. 2R
4. 5R

Answer: 1. R

Question 27. If the reading of ammeter A₁ in the figure is 2.4 A. Neglecting the resistances of the ammeters, the reading of ammeter A₂ will be :

1. 1.6 A
2. 1 A
3. 2 A
4. 3A

Answer: 1. 1.6 A

Question 28. In the previous question the reading of ammeter A₃ will be :

1. 1.6 A
2. 1.2 A
3. 4 A
4. 2 A

Answer: 3. 4 A

Question 29. The wire used in the arrangement shown in the figure has a resistance of r ohm per meter. The equivalent resistance between points A and B is

1. \(\left(\frac{6}{11}\right) r\)
2. \(\frac{2 \pi r}{(\pi+1)}\)
3. \(\frac{6 \pi r}{(16+3 \pi)}\)
4. \(\frac{3 \pi r}{(10+3 \pi)}\)

Answer: 3. \(\frac{6 \pi r}{(16+3 \pi)}\)

Question 30. For the circuit of the figure, the equivalent resistance between points A and B is

1. 1Ω
2. 2Ω
3. 4 Ω
4. 8 Ω

Answer: 3. 4 Ω

Question 31.

1. 2R/3
2. R/3
3. 2 R
4. 3 R

Answer: 1. 2R/3

Question 32. In the following circuits, the value of total resistance between X and Y in ohm is X

1. \((1+\sqrt{3}) R\)
2. \((\sqrt{3}-1) R\)
3. ∞
4. 50 r

Answer: 1. \((1+\sqrt{3}) R\)

Question 33.

1. R
2. 2R
3. R/2
4. 4 R

Answer: 2. 2R

Question 34. Arrange the order of power dissipated in the given circuits, if the same current is passing through the system. The resistance of each resistor is ‘ r’.

1. P₂> P₃> P₄> P₁
2. P₁> P₄> P₃> P₂
3. P₁> P₂> P₃> P₄
4. P₄> P₃> P₂> P1

Answer: 1. P₂> P₃> P₄> P₁

Question 35. A 100 W bulb B₁ and two 60 W bulbs B₂ and B₃ are connected to a 250 V source as shown in the figure. Now W₁, W₂, and W₃ are the output powers of the bulbs B₁, B₂and respectively. Then:

1. W₁> W₂= W₃
2. W₁> W₂> W₃
3. W₁< W₂= W₃
4. W₁< W₂< W₃

Answer: 4. W₁< W₂< W₃

Question 36. Two bars of equal resistivity ρ and radii ‘r’ and ‘2r’ are kept in contact as shown. An electric current Ι is passed through the bars. Which one of the following is correct?

1. The heat produced in bar (1) is 2 times the heat produced in bar (2)
2. The electric field in both halves is equal
3. The current density across AB is double that of BC.
4. The potential difference across BC is 4 times that across AB.

Answer: 4. Potential difference across BC is 4 times that across AB.

Question 37. From resistances of 100 ohm, each is connected in the form of a square. The effective resistance along the diagonal points P R is :

1. 100Ω
2. 180 Ω
3. 220 Ω
4. 440 Ω

Answer: 1. 100Ω

Question 38. The potential difference across BC in the following figure will be :

1. 1.2 v
2. 2 v
3. 0.8 v
4. 1 v

Answer: 3. 0.8 v

Question 39. For two resistance wires joined in parallel, the resultant resistance is 65Ω. When one of the resistance wires breaks the effective resistance becomes 2Ω. The resistance of the broken wire is

1. \(\frac{3}{5} \Omega\)
2. 2Ω
3. \(\frac{6}{5} \Omega\)
4. 3 Ω

Answer: 4. 3 Ω

Question 40. The equivalent resistance between A and B is

1. 10 Ω
2. 20 Ω
3. 30 Ω
4. 40 Ω

Answer: 1. 10 Ω

Question 41. Three resistances P, Q, and R each of 2Ω, and an unknown resistance S form the four arms of a Wheatstone’s bridge circuit. When a resistane of 6 Ω is connected in parallel to S the bridge gets balanced. What is the value of S?

1. 2 Ω
2. 3 Ω
3. 6 Ω
4. 1 Ω

Answer: 2. 3 Ω

Question 42. The equivalent resistance of the given circuit is

1. R
2. R/2
3. R/4
4. R/6

Answer: 1. R

Question 43. Two metal wires having conductivities σ1and σ2respectively have the same dimensions. If they are connected in series the effective conductivity of the combination is

1. \(\frac{\sigma_1+\sigma_2}{2}\)
2. \(\frac{\sigma_1-\sigma_2}{2}\)
3. \(\frac{\sigma_1+\sigma_2}{\sigma_1 \sigma_2}\)
4. \(\frac{2 \sigma_1 \sigma_2}{\sigma_1+\sigma_2}\)

Answer: 4. \(\frac{2 \sigma_1 \sigma_2}{\sigma_1+\sigma_2}\)

Question 44. If in the circuit, power dissipation is 150 W then R is

1. 2 Ω
2. 6 Ω
3. 5 Ω
4. 4 Ω

Answer: 2. 6 Ω

Question 45. A 3-volt battery with negligible internal resistance is connected in a circuit as shown in the figure. Currently, I will be :

1. 1/3 A
2. 1 A
3. 1.5 A
4. 2 A

Answer: 3. 1.5 A

Question 46. The total current supplied to the circuit by the battery is :

1. 1 A
2. 2 A
3. 4 A
4. 6 A

Answer: 3. 4 A

Question 47. The resistance of the series combination of two resistances is S. When they are joined in parallel, the total resistance is P. If S = nP, then the minimum possible value of n is :

1. 4
2. 3
3. 2
4. 1

Answer: 1. 4

Question 48. An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii of the wires are in the ratio of 4/3 and 2/3, then the ratio of the currents passing through the wire will be :

1. 3
2. 1/3
3. 8/9
4. 2

Answer: 2. 1/3

Question 49. The current I drew from the 5-volt source will be

1. 0.67 A
2. 0.17 A
3. 0.33 A
4. 0.5 A

Answer: 4. 0.5 A

Question 50. In a Wheat stone’s bridge, three resistances P, Q, and R are connected in the three arms, and the fourth arm is formed by two resistances S1 and S2 connected in parallel. The condition for the bridge to be balanced will be

1. \(\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{2 \mathrm{~S}_1 \mathrm{~S}_2}\)
2. \(\frac{P}{Q}=\frac{R}{S_1+S_2}\)
3. \(\frac{P}{Q}=\frac{2 R}{S_1+S_2}\)
4. \(\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{\mathrm{S}_1 \mathrm{~S}_2}\)

Answer: 4. \(\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{\mathrm{S}_1 \mathrm{~S}_2}\)

Question 51. For a cell the terminal potential difference is 2.2V when the circuit is open and reduces to 1.8V when the cell is connected to a resistance R = 5Ω, the internal resistance (r) of the cell is :

1. \(\frac{10}{9} \Omega\)
2. \(\frac{9}{10} \Omega\)
3. \(\frac{11}{9} \Omega\)
4. \(\frac{5}{9} \Omega\)

Answer: 1. \(\frac{10}{9} \Omega\)

Question 52. In a Wheatstone’s bridge, all four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is :

1. R
2. 2R
3. R/4
4. R/2

Answer: 1. R

Question 53. Resistances n, each of r ohm, when connected in parallel given an equivalent resistance of R ohm. If these resistances were connected in series, the combination would have a resistance in ohms, equal to:

1. n2R
2. R/n2
3. R/n
4. nR

Answer: 1. n2R

Question 54. Five equal resistances each of resistance R are connected as shown in the figure. A battery of V volts is connected between A and B. The current flowing in AFCEB will be :

1. \(\frac{3 V}{R}\)
2. \(\frac{V}{R}\)
3. \(\frac{V}{2 R}\)
4. \(\frac{2 V}{R}\)

Answer: 3. \(\frac{V}{2 R}\)

Question 55. For the network shown in the figure, the value of the current i is :

1. \(\frac{9 V}{35}\)
2. \(\frac{5 V}{18}\)
3. \(\frac{5 V}{9}\)
4. \(\frac{18 V}{5}\)

Answer: 2. \(\frac{5 V}{18}\)

Question 56. In the circuit shown, if a conducting wire is connected between points A and B, the current in this wire will:-

1. Flow from A to B
2. Flow in the direction which will be decided by the value of V
3. Be zero
4. Flow from B to A

Answer: 4. Flow from B to A

Question 57. The power dissipated across the 8Ω resistor in the circuit shown here is 2 watts. The power dissipated in watt units across the 3Ω resistor is:-

1. 2.0
2. 1.0
3. 0.5
4. 3.0

Answer: 4. 3.0

Question 58. The total power dissipated in watts in the circuit shown here is:-

1. 4W
2. 16W
3. 40W
4. 54W

Answer: 4. 54W

Question 59. Three resistances P, Q, and R each of 2Ω and an unknown resistance S form the four arms of a Wheatstone bridge circuit, When a resistance of 6Ω is connected in parallel to S the bridge gets balanced. What is the value of S?

1. 1Ω
2. 2Ω
3. 3Ω
4. 6Ω

Answer: 3. 3Ω

Question 60. A current of 3 A flows through the 2Ω resistor shown in the circuit. The power dissipated in the 5Ω resistor is

1. 4 W
2. 2 W
3. 1 W
4. 5 W

Answer: 4. 5 W

Question 61. In the circuit shown, the current through the 4Ω resistor is 1 A when the points P and M are connected to a DC voltage source. The potential difference between the points M and N is

1. 1.5 V
2. 1.0 V
3. 0.5 V
4. 3.2 V

Answer: 4. 3.2 V

Section (5): Combination Of Cells

Question 1. Two nonideal batteries are connected in parallel. Consider the following statements

1. The equivalent emf is smaller than either of the two emfs.
2. The equivalent internal resistance is smaller than either of the two internal resistance.
   1. Both 1 and 2 are correct
   2. 1 is correct but 2 is wrong
   3. 2 is correct but 1 is wrong
   4. Each of 1 and 2 is wrong.

Answer: 3. 2 is correct but 1 is wrong

Question 2. 12 cells each having the same emf are connected in series and are kept in a closed box. Some of the cells are wrongly connected. This battery is connected in series with an ammeter and two cells identical to each other and also identical to the previous cells. The current is 3 A when the external cells aid this battery and is 2 A when the cells oppose the battery. How many cells in the battery are wrongly connected?

1. One
2. Two
3. Three
4. None

Answer: 1. One

Question 3. Two batteries, one of emf 18V and internal resistance 2Ω and the other of emf 12 V and internal resistance 1Ω, are connected as shown. The voltmeter V will record a reading of :

1. 15V
2. 30 V
3. 14 V
4. 18 V

Answer: 3. 14 V

Question 4. n equal cell having e.m.f. E and internal resistance r, are connected in a circuit of a resistance R. The Same current flows in the circuit whether they are connected in series or parallel, if :

1. R = nr
2. R = r/n
3. R = n2r
4. R = r

Answer: 4. R = r

Question 5. Two cells of e.m.f. 10 V and 15 V are connected in parallel to each other between points A and B. The cell of e.m.f. 10 V is ideal but the cell of e.m.f. 15 V has an internal resistance of 1 Ω. The equivalent e.m.f. between A and B is:

1. \(\frac{25}{2} V \)
2. not defined
3. 15 V
4. 10 V

Answer: 4. 10 V

Question 6. Two ideal batteries of emf and V₂ and three resistances R₁ and R₃ are connected as shown in the figure. The current in resistance R₂ would be non-zero, if

1. V₁= V₂ and R₁= R₂ = R₃
2. V₁= V₂ and R₁= 2R₂ = R₃
3. V₁= 2V₂ and 2R₁= 2R₂ = R₃
4. 2V₁= V₂ and 2R₁= R₂ = R₃

Answer: 3. V₁= 2V₂and 2R₁= 2R₂ = R₃

Question 7. Two non-ideal batteries are connected in parallel. Consider the following statements.

1. The equivalent emf is smaller than either of the two emfs.
2. The equivalent internal resistance is smaller than either of the two internal resistances.
   1. Both 1 and 2 are correct
   2. 1 correct but 2 is wrong
   3. 2 is correct but 1 is wrong
   4. Both 1 and 2 are wrong

Answer: 3. 2 is correct but 1 is wrong

Question 8. Two sources of equal emf are connected to an external resistance R₁. The internal resistances of the two sources are R₁ and R₂(R₂> R₁). If the potential difference across internal resistance R₂ is zero, then:

1. \(R=\frac{R_2 \times\left(R_1+R_2\right)}{\left(R_2-R_1\right)}\)
2. \(R=R_2-R_1\)
3. \(R=\frac{R_1 R_2}{\left(R_2+R_1\right)}\)
4. \(R=\frac{R_1 R_2}{\left(R_2-R_1\right)}\)

Answer: 2. \(R=R_2-R_1\)

Question 9. In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be :

1. 200 Ω
2. 100 Ω
3. 500 Ω
4. 1000 Ω

Answer: 2. 100 Ω

Question 10. A 5 V battery with internal resistance 2 Ω and a 2V battery with internal resistance 1Ω are connected to a 10Ω resistor as shown in the figure.

The current in the 10 Ω resistor is –

1. 0.03 A P₁ to P₂
2. 0.03 A P₂ to P₁
3. 0.27 A P₁ to P₂
4. 0.27 A P₂ to P₁

Answer: 2. 0.03 A P₂ to P₁

Question 11. Two cells, having the same e.m. f., are connected in series through an external resistance R. Cells have internal resistances r₁ and r₂(r₁> r₂) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is:-

1. r₁– r₂
2. \(\frac{r_1+r_2}{2}\)
3. \(\frac{r_1-r_2}{2}\)
4. r₁ + r₂

Answer: 1. r₁– r₂

Section (6): Instrument

Question 1. The reading of the voltmeter is

1. 50V
2. 60 V
3. 40V
4. 80 V

Answer: 3. 40V

Question 2. The current through the ammeter shown in the figure is 1 A. If each of the 4Ω resistors is replaced by a 2Ω resistor, the current in the circuit will become nearly :

1. \(\frac{10}{9} \mathrm{~A}\)
2. \(\frac{5}{4}\)
3. \(\frac{9}{8} \mathrm{~A}\)
4. \(\frac{9}{8} \mathrm{~A}\)

Answer: 1. \(\frac{10}{9} \mathrm{~A}\)

Question 3. The meter-bridge wire AB shown in the figure is 50 cm long. When AD = 30 cm, no deflection occurs in the galvanometer. Find R.

1. 1 Ω
2. 2 Ω
3. 3 Ω
4. 4 Ω

Answer: 4. 4 Ω

Reading of ammeter in ampere for the following circuit is (Q. 4 to 6)

Question 4.

1. 4
2. 3
3. 1
4. 2

Answer: 3. 1

Question 5.

1. 2/15
2. 1/13
3. 2/11
4. 2/17

Answer: 2. 1/13

Question 6.

1. 0.4
2. 1
3. 0.6
4. 1.2

Answer: 2. 1

Question 7. The potential gradient of a potentiometer wire is defined as

1. The fall of potential per unit length
2. The fall of potential per unit area
3. The fall in potential across the ends of wires
4. None of the above

Answer: 1. The fall of potential per unit length

Question 8. The unit of potential gradient is

1. volt
2. volt/ampere
3. volt/meter
4. volt x meter

Answer: 3. volt/meter

Question 9. The length of the potentiometer wire is kept larger so that the value of the potential gradient may

1. Increase
2. Decrease
3. Remain uniform all over the length of its wire
4. None of the above

Answer: 2. Decrease

Question 10. For the same potential difference, a potentiometer wire is replaced by another one of high specific resistance. The potential gradient then ( r = Rh= 0)

1. Decreases
2. Remains same
3. Increases
4. Data is incomplete

Answer: 2. Remains same

Question 11. If the current in a potentiometer increases, the position of the null point will

1. Be obtained at a larger length than the previous one
2. Be equal to the previous length
3. Be obtained at a smaller length than the previous
4. None of the above

Answer: 3. Be obtained at a smaller length than the previous

Question 12. The sensitivity of a potentiometer is increased by

1. Increasing the emf of the cell
2. Increasing the length of the potentiometer wire
3. Decreasing the length of the potentiometer wire
4. None of the above

Answer: 2. Increasing the length of the potentiometer wire

Question 13. In a potentiometer wire, whose resistance is 0.5 ohm/m, a current of 2 amperes is passing. The value of the potential gradient in volt/m will be

1. 0.1
2. 0.5
3. 1.0
4. 4

Answer: 3. 1.0

Question 14. The potentiometer wire 10 m long and 20 ohm resistance is connected to a 3 volt emf battery and a 10 ohm resistance. The value of the potential gradient in volt/m of the wire will be

1. 1.0
2. 0.2
3. 0.1
4. 0.02

Answer: 2. 0.2

Question 15. The potential gradient of the potentiometer is 0.2 volt/m. A current of 0.1 amp is flowing through a coil of 2-ohm resistance. The balancing length in meters for the p.d. at the ends of this coil will be

1. 2
2. 1
3. 0.2
4. 0.1

Answer: 2. 1

Question 16. The emf of a standard cell is 1.5 volt and its balancing length is 7.5 m. The balancing length in meters for a 3.5-ohm resistance, through which a current of 0.2 A, flows will be

1. 3.5
2. 5.0
3. 5.7
4. 6.5

Answer: 1. 3.5

Question 17. In the following figure, the p.d. between the points M and N are balanced at 50 cm in length. The balancing length in cm, for the p.d. between points N and C, will be

1. 40
2. 100
3. 75
4. 25

Answer: 2. 100

Question 18. Potentiometer measures

1. Potential difference
2. Internal resistance
3. Current
4. External resistance

Answer: 4. External resistance

Question 19. The resistance of an ideal voltmeter is ;

1. very low
2. infinite
3. zero
4. none of these

Answer: 1. very low

Question 20. For changing an ammeter into a voltmeter, we should connect

1. low resistance in series
2. low resistance in parallel
3. high resistance in series
4. High resistance in parallel

Answer: 2. low resistance in parallel

Question 21. When a cell is balanced on a potentiometer wire, the balancing length is 125 cm. If the resistance of 2 ohms is connected across the ends of the cell, then the balancing length is 100 cm, then the internal resistance of the cell is : 

1. 0.5 Ω
2. 0.25 Ω
3. 0.05 Ω
4. 5 Ω

Answer: 1. 0.5 Ω

Question 22. A potentiometer measures the potential difference more accurately than a voltmeter because:

1. It has a wire of high resistance.
2. It has a wire of low resistance
3. It does not draw current from the external circuit
4. It draws a heavy current from the external circuit

Answer: 4. It draws a heavy current from the external circuit

Question 23. For which of the following meters, converted from identical galvanometers, the resistance of the converted meter is largest

1. voltmeter of range 0.5 V
2. ammeter of range 1 A
3. Voltmeter of range 1.0 V
4. ammeter of range 10 A

Answer: 3. Voltmeter of range 1.0 V

Question 24. The resistivity of the potentiometer wire is 10^-7 ohm-meter and its area of cross-section is 10^-6 m². When a current i = 0.1A flows through the wire, its potential gradient is :

1. 10–2 V/m
2. 10–4 V/m
3. 0.1 V/m
4. 10V/m

Answer: 1. 10–2 V/m

Question 25. In electrolysis, the mass deposited on an electrode is directly proportional to :

1. Current
2. Square of current
3. Concentration of solution
4. Inverse of current

Answer: 1. Current

Question 26. The material of the wire of the potentiometer is

1. Copper
2. Steel
3. Manganin
4. Aluminium

Answer: 3. Manganin

Question 27. An ammeter and a voltmeter are joined in series to a cell. Their readings are A and V respectively. If a resistance is now joined in parallel with the voltmeter,

1. both A and V will increase
2. both A and V will decrease
3. A will decrease, V will increase
4. A will increase, V will decrease

Answer: 4. A will increase, V will decrease

Question 28. In the given circuit, no current is passing through the galvanometer. If the cross-sectional diameter of the wire AB is doubled, then for the null point of a galvanometer, the value of AC would be:

1. 2 X
2. X
3. X²
4. None

Answer: 2. X

Question 29. A galvanometer can be changed into an ammeter by connecting :

1. High resistance in parallel
2. High resistance in series.
3. Low resistance in parallel
4. Low resistance in series

Answer: 3. low resistance in parallel

Question 30. The sensitivity of the potentiometer can be increased by

1. Increasing the emf of the cell
2. Increasing the length of the potentiometer wire
3. Decreasing the length of the potentiometer wire
4. None of the above

Answer: 2. Increasing the length of the potentiometer wire

Question 31. The material of the wire of the potentiometer is

1. copper
2. steel
3. manganin
4. aluminum

Answer: 3. manganin

Question 32. If an ammeter is to be used in place of a voltmeter then we must connect with the ammeter a

1. Low resistance in parallel
2. High resistance in parallel
3. High resistance in series
4. Low resistance in series

Answer: 3. High resistance in series

Question 33. An ammeter reads up to 1 ampere. Its internal resistance is 0.81 ohm. To increase the range to 10 A the value of the required shunt is.

1. 0.09 Ω
2. 0.03 Ω
3. 0.3 Ω
4. 0.9 Ω

Answer: 1. 0.09 Ω

Question 34. The length of a wire of a potentiometer is 100 cm, and the emf of its standard cell is E volt. It is employed to measure the emf of a battery whose internal resistance is 0.5 ohm. If the balance point is obtained at 30 cm from the positive end, the emf of the battery is

1. \(\frac{30 E}{100}\)
2. \(\frac{30 E}{100.5}\)
3. \(\frac{30 E}{(100-0.5)}\)
4. \(\frac{30(E-0.5 i)}{100}, \frac{30(E-0.5 i)}{100}\)where i is the current in the potentiometer 100

Answer: 1. \(\frac{30 E}{100}\)

Question 35. In a meter bridge experiment, the null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. If X < Y, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y?

1. 50 cm
2. 80 cm
3. 40 cm
4. 70 cm

Answer: 1. 50 cm

Question 36. In a potentiometer experiment, the balancing with a cell is at a length of 240 cm. On shunting the cell with a resistance of 2Ω, the balancing length becomes 120 cm. The internal resistance of the cell is :

1. 1 Ω
2. 0.5 Ω
3. 4 Ω
4. 2 Ω

Answer: 4. 2 Ω

Question 37. A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milli ampere and voltage sensitivity is 2 divisions per millivolt. So that each division reads 1 volt, the resistance in ohms needed to be connected in series with the coil will be :

1. 103
2. 105
3. 99995
4. 9995

Answer: 4. 9995

Question 38. Shown in the figure below is a meter bridge set up with null deflection in the galvanometer.

The value of the unknown resistor R is

1. 220 Ω
2. 110 Ω
3. 55 Ω
4. 13.75 Ω

Answer: 1. 220 Ω

Question 39. To convert a galvanometer into a voltmeter, one should connect a :

1. High resistance in series with galvanometer
2. Low resistance in series with galvanometer
3. High resistance in parallel with galvanometer
4. Low resistance in parallel with galvanometer

Answer: 1. High resistance in series with the galvanometer

Question 40. A galvanometer of 50Ω resistance has 25 divisions. A current of 4 × 10^-4 A gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25V, it should be connected with a resistance of

1. 2500Ω as a shunt
2. 245Ω as a shunt
3. 2550Ω in series
4. 2450Ω in series

Answer: 4. 2450Ω in series

Question 41. A galvanometer acting as a voltmeter will have :

1. A high resistance in parallel with its coil
2. A high resistance in series with its coil
3. A low resistance in parallel with its coil
4. A low resistance in series with its coil

Answer: 2. A high resistance in series with its coil

Question 42. The resistance of an ammeter is 13Ω and its scale is graduated for a current up to 100 amps. After an additional shunt has been connected to this ammeter it becomes possible to measure currents up to 750 amperes by this meter. The value of shunt-resistance is:-

1. 20Ω
2. 2Ω
3. 0.2 Ω
4. 2kΩ

Answer: 2. 2Ω

Question 43. A galvanometer of resistance 50Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full-scale deflection of 30 divisions is obtained in the galvanometer. To reduce this deflection to 20 divisions, the resistance in series should be

1. 5050 Ω
2. 5550 Ω
3. 6050 Ω
4. 4450 Ω

Answer: 4. 4450 Ω

Question 44. A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short-circuited through a resistance of 10 Ω. Its internal resistance is

1. 1.0 Ω
2. 0.5 Ω
3. 2.0 Ω
4. zero

Answer: 1. 1.0 Ω

Current Electricity Exercise – 2

Question 1. A quantity X is given by ε0L \(\frac{\Delta \mathrm{V}}{\Delta \mathrm{t}}\) where ε0 is the permittivity of free space, L is a length, ΔV is a potential difference and Δt is a time interval. The dimensional formula for X is the same as that of :

1. Resistance
2. Charge
3. Voltage
4. Current

Answer: 4. Current

Question 2. When a current flows through a conductor its temperature

1. May increase or decrease
2. Remains same
3. Decreases
4. Increases

Answer: 4. Increases

Question 3. Find the current through the 10 Ω resistors shown in the figure

1. zero
2. 1 A
3. 2A
4. 5 A

Answer: 1. zero

Question 4. Two batteries of e.m.f. 4 V and 8 V with internal resistances 1 Ω and 2 Ω are connected in a circuit with a resistance of 9 Ω as shown in the figure. The current and potential differences between the points P and Q are

1. 1/3 A 3 V
2. 1/6 A 4 V
3. 1/9 A 9 V
4. 1/2 A 12 V

Answer: 1. 1/3 A 3 V

Question 5. For driving a current of 2 A for 6 minutes in a circuit, 1000 J of work is to be done. The e.m.f. of the source in the circuit is

1. 1.38 V
2. 1.68 V
3. 2.04 V
4. 3.10 V

Answer: 1. 1.38 V

Question 6. The potential difference between points A and B is

1. \(\frac{20}{7} \mathrm{~V}\)
2. \(\frac{40}{7} \mathrm{~V}\)
3. \(\frac{10}{7} \mathrm{~V}\)
4. 0

Answer: 4. 0

Question 7. A wire is in the form of a tetrahedron. The resistance of each edge is r. The equivalent resistances between corners 1–2 and 1–3 are respectively

1. \(\frac{\mathrm{r}}{2}, \frac{\mathrm{r}}{2}\)
2. r, r
3. \(\frac{\mathrm{r}}{2}r\)
4. \(r\frac{\mathrm{r}}{2}\)

Answer: 1. \(\frac{\mathrm{r}}{2}, \frac{\mathrm{r}}{2}\)

Question 8. In the figure shown the current flowing through 2 R is :

1. From left to right
2. From right to left
3. No current
4. None of these

Answer: 2. From right to left

Question 9. The net resistance between points P and Q in the circuit shown in fig. is

1. R/2
2. 2R/5
3. 3R/5
4. R/3

Answer: 2. 2R/5

Question 10. In the given circuit, it is observed that the current Ι is independent of the value of the resistance R₆. Then the resistance values must satisfy:

1. R₁ R₂ R₅= R₃ R₄ R₆
2. \(\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_1+R_2}+\frac{1}{R_3+R_4}\)
3. R1 R₄= R₂ R₃
4. R₁ R₃= R₂ R₄= R₅ R₆

Answer: 3. R₁ R₄= R₂ R₃

Question 11. A battery of internal resistance of 4 ohms is connected to the network of resistance as shown. For the maximum power that can be delivered to the network, the value of R in ohm should be:

1. 4/9
2. 2
3. 8/3
4. 18

Answer: 2. 2

Question 12. The equivalent resistance between the points A and B is :

1. \(\frac{36}{7} \Omega\)
2. 10 Ω
3. \(\frac{85}{7} \Omega\)
4. None of these

Answer: 3. \(\frac{85}{7} \Omega\)

Question 13. The measurement (approx) of the ideal voltmeter in the following circuit is

1. 2.4 V
2. 3.4 V
3. 4.0 V
4. 6.0 V

Answer: 4. 6.0 V

Question 14. A cell is balanced at 100 cm of a potentiometer wire when the total length of the wire is 400 cm. If the length of the potentiometer wire is increased by 100 cm, then the new balancing length for the cell will be (Assume pd across potentiometer wire is constant)

1. 100 cm
2. 125 cm
3. 80 cm
4. 250 cm

Answer: 2. 125 cm

Question 15. The potential difference across the 100-ohm resistance in the following circuit is measured by a voltmeter of 900-ohm resistance. The percentage error made in reading the potential difference is

1. 10/9
2. 0.1
3. 1.0
4. 10.0

Answer: 3. 1.0

Question 16. The reading of the voltmeter in the circuit shown is

1. 3.6 V
2. 3.25 V
3. 4.25 V
4. 6.25 V

Answer: 1. 3.6 V

Question 17. One filament takes 10 min to heat a kettle and another takes 15 min. If connected in parallel they combined take………….. min to heat the same kettle :

1. 6
2. 12.5
3. 25
4. 7.5

Answer: 1. 6

Question 18. Five resistances of resistance RΩ are there, 3 are connected in parallel and are joined to them in series. Find resultant resistance :

1. \(\left(\frac{3}{7}\right) \mathrm{R} \Omega\)
2. \(\left(\frac{7}{3}\right) \mathrm{R} \Omega\)
3. \(\left(\frac{7}{8}\right) R \Omega\)
4. \(\left(\frac{8}{7}\right) \mathrm{R} \Omega\)

Answer: 2. \(\left(\frac{7}{3}\right) \mathrm{R} \Omega\)

Question 19. The potential difference between points A and B is :

1. \(\frac{20}{7} \mathrm{~V}\)
2. \(\frac{40}{7} \mathrm{~V}\)
3. \(\frac{10}{7} \mathrm{~V}\)
4. zero

Answer: 4. zero

Question 20. Find the equivalent resistance between the points A and B :

1. 2Ω
2. 4Ω
3. 8Ω
4. 16Ω

Answer: 2. 4Ω

Question 21. A circuit consists of five identical conductors as sh in n the figure. The two similar conductors are added as indicated by dotted lines. The ratio of resistances before and after addition will be

1. \(\frac{7}{5}\)
2. \(\frac{3}{5}\)
3. \(\frac{5}{3}\)
4. \(\frac{6}{5}\)

Answer: 3. \(\frac{5}{3}\)

Question 22. The electric current that passes through a metallic wire produces heat because of

1. Collisions of conduction electrons with each other
2. Collisions of the atoms of the metal with each other
3. The energy released in the ionization of the atoms of the metal
4. Collisions of the conduction electrons with the atoms of the metallic wire

Answer: 4. Collisions of the conduction electrons with the atoms of the metallic wire

Question 23. An electrical cable of copper has just one wire of radius 9 mm. Its resistance is 5 Ω. The single wire of the cable is replaced by 6 different well-insulated copper wires each of radius 3 mm. The total resistance of the cable will now be equal to

1. 270 Ω
2. 90 Ω
3. 45 Ω
4. 7.5 Ω

Answer: 4. 7.5 Ω

Question 24. A galvanometer has a resistance of 400 Ω and deflects full-scale for a current of 0.2 mA through it. The shunt resistance required to convert it into a 3 A ammeter is

1. 0.027 Ω
2. 0.054 Ω
3. 0.0135 Ω
4. None of these

Answer: 1. 0.027 Ω

Question 25. The resistance of an ammeter is 13 Ω and its scale is graduated for a current up to 100 A. After an additional shunt has been connected to this ammeter it becomes possible to measure currents up to 750 A by this meter. The value of shunt resistance is

1. 20 Ω
2. 2 Ω
3. 0.2 Ω
4. 2kΩ

Answer: 2. 2 Ω

Question 26. The length of a wire in a potentiometer is 100 cm, and the emf of its standard cell is E volt. It is employed to measure the emf of a battery whose internal resistance is 0.5. If the balance point is obtained at Ω = 30 cm from the positive end, the emf of the battery is

1. \(\frac{30 E}{100.5}\)
2. \(\frac{30 E}{100-0.5}\)
3. \(\frac{30(E-0.5 i)}{100}\), where i is the current in the potentiometer wire
4. \(\frac{30 E}{100}\)

Answer: 4. \(\frac{30 E}{100}\)

Question 27. An ammeter readsup too 1A. Its internal resistance is 0.81 Ω. To increase the range to 10 A the value of the required shunt is

1. 0.03 Ω
2. 0.3 Ω
3. 0.9 Ω
4. 0.09 Ω

Answer: 4. 0.09 Ω

Question 28. The value of current I in the circuit will be

1. 1.7 A
2. 2.1 A
3. 3 A
4. zero

Answer: 1. 1.7 A

Question 29. There is a voltameter in a circuit. To triple its range, the resistance of how much value should be used?

1. 2R
2. R/2
3. 3R
4. 4R

Answer: 1. 2R

Question 30. A wire is bent in the form of a triangle now the equivalent resistance R between its one end and the midpoint of the side is

1. \(\frac{5 R}{12}\)
2. \(\frac{7 \mathrm{R}}{12}\)
3. \(\frac{3 R}{12}\)
4. \(\frac{R}{12}\)

Answer: 4. \(\frac{R}{12}\)

Question 31. A ‘Wheatstone Bridge’ circuit has been set up as shown. The resistor R4 is an ideal carbon. The resistor R₄ is an ideal carbon resistance (tolerance = 0%) having bands of colors black, yellow, and brown marked on it. The galvanometer, in this circuit, would show a ‘null point’ when another ideal carbon resistance X is connected across R₄, having bands of colors

1. Black, brown, black, is put in parallel with R4
2. Black, brown, brown, is put in parallel with R4
3. Brown, black, brown, is put in parallel with R4
4. Black, brown, black, is put in parallel with R4

Answer: 2. Black, brown, brown, is put in parallel with R4

Current Electricity Exercise – 3

Question 1. A galvanometer having a coil resistance of 60 Ω shows full-scale deflection when a current of 1.0 A passes through it. It can be converted into an ammeter to read currents up to 5.0 A by

1. putting in parallel a resistance of 240 Ω
2. putting in series a resistance of 15 Ω
3. putting in series a resistance of 240 Ω
4. putting in parallel a resistance of 15 Ω

Answer: 3. putting in series a resistance of 240 Ω

Question 2. A wire of resistance 12Ωm–1 is bent to form a complete circle of radius 10 cm. The resistance between its two diametrically opposite points, A and B as shown in the figure, is

1. 0.6π Ω
2. 3 Ω
3. 6π Ω
4. 6Ω

Answer: 1. 0.6π Ω

Question 3. A student measures the terminal potential difference (V) of a cell (of emf ε and internal resistance r) as a function of the current (I) flowing through it. The slope and intercept of the graph between V and I, then respectively equal

1. ε and–r
2. –r and ε
3. r and –ε
4. –ε and r

Answer: 2. –r and ε

Question 4. See the electrical circuit shown in this figure. Which of the following equations is a correct equation for it?

1. ε₁– (i₁+ i2)R – i₁r₁= 0
2. ε₂– i₂r₂– ε₁– i₁r₁= 0
3. –ε₂– (i₁+ i₂)R + i₂r₂= 0
4. ε₁– (i₁+ i₂)R + i₁r₁= 0

Answer: 1. ε₁– (i₁+ i₂)R – i₁r₁= 0

Question 5. A thermocouple of negligible resistance produces an e.m.f. of 40 μV/°C in the linear range of temperature. A galvanometer of resistance 10 ohm whose sensitivity is 1μA/div, is employed with the thermocouple. The smallest value of temperature difference that can be detected by the system will be :

1. 0.5°C
2. 1°C
3. 0.1°C
4. 0.25°C

Answer: 4. 0.25°C

Question 6. In the circuit shown in the figure, if the potential at point A is taken to be zero the potential at point B is :

1. –1V
2. + 2V
3. – 2V
4. + 1V

Answer: 4. + 1V

Question 7. A milli voltmeter of 25 milli volt range is to be converted into an ammeter of 25 ampere range. The value (in ohm) of the necessary shunt will be :

1. 0.001
2. 0.01
3. 1
4. 0.05

Answer: 1. 0.001

Question 8. In the circuit shown cells A and B have negligible resistances. For V_A= 12V, R₁= 500, Ω,   and R = 100Ω the galvanometer (G) shows no deflection. The value of V_B is :

1. 4V
2. 2V
3. 12V
4. 6V

Answer: 2. 2V

Question 9. If the  voltage across a bulb rated 220 Volt – 100 Watt drops by 2.5 % of its rated value, the percentage of the rated value by which the power would decrease is :

1. 20 %
2. 2.5 %
3. 5 %
4. 10 %

Answer: 3. 5 %

Question 10. A ring is made of a wire having a resistance R0= 12 Ω. Find the points A and B as shown in the figure, at which the current-carrying conductor should be connected so that the resistance R of the subcircuit between these points is equal to 8/3 Ω.

1. \(\frac{\ell_1}{\ell_2}=\frac{5}{8}\)
2. \(\frac{\ell_1}{\ell_2}=\frac{1}{3}\)
3. \(\frac{\ell_1}{\ell_2}=\frac{3}{8}\)
4. \(\frac{\ell_1}{\ell_2}=\frac{1}{2}\)

Answer: 4. \(\frac{\ell_1}{\ell_2}=\frac{1}{2}\)

Question 11. The power dissipated in the circuit shown in the figure is 30 Watts. The value of R is :

1. 20 Ω
2. 15 Ω
3. 10 Ω
4. 30 Ω

Answer: 3. 10 Ω

Question 12. A cell having an emf ε and internal resistance r is connected across a variable external resistance R . As the resistance R is increased, the plot of potential difference V across R is given by :

Answer: 3.

Question 13. A wire of resistance 4 Ω is stretched to twice its original length. The resistance of a stretched wire would be :

1. 4 Ω
2. 8 Ω
3. 16 Ω
4. 2 Ω

Answer: 3. 16 Ω

Question 14. The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10Ω is :

1. 0.5 Ω
2. 0.8 Ω
3. 1.0 Ω
4. 0.2 Ω

Answer: 1. 0.5 Ω

Question 15. The resistances of the four arms P, Q, R, and S in a Wheatstone’s bridge are 10 ohm, 30 ohm, 30 ohm, and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 Volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be :

1. 0.2 A
2. 0.1 A
3. 2.0 A
4. 1.0 A

Answer: 1. 0.2 A

Question 16. The two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 volts and the average resistance per km is 0.5Ω The power loss in the wire is:

1. 19.2 W
2. 19.2 kW
3. 19.2 J
4. 12.2 kW

Answer: 2. 19.2 kW

Question 17. The resistance in the two arms of the meter bridge is 5Ω and R Ω, respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6Ω1. The resistance ‘R’ is :

1. 10Ω
2. 15 Ω
3. 20 Ω
4. 25 Ω

Answer:2. 15 Ω

Question 18. A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an EMF of 2.0 V and negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance, R, connects across the given cell, has values of.

1. Infinity
2. 9.5Ω

The ‘balancing lengths, on the potentiometer wire are found to be 3m and 2.85 m, respectively. The value of the internal resistance of the cell is :

1. 0.25 Ω
2. 0.95 Ω
3. 0.5 Ω
4. 0.75 Ω

Answer: 3. 0.5 Ω

Question 19. In an ammeter, 0.2% of the main current passes through the galvanometer. If the resistance of the galvanometer is G, the resistance of the ammeter will be :

1. \(\frac{1}{499} G\)
2. \(\frac{499}{500} \mathrm{G}\)
3. \(\frac{1}{500} G\)
4. \(\frac{500}{499} \mathrm{G}\)

Answer: 3. \(\frac{1}{500} G\)

Question 20. Across a metallic conductor of a non-uniform cross-section, a constant potential difference is applied. The quantity which remains constant along the conductor is

1. current
2. drift velocity
3. electric field
4. current density

Answer: 1. current

Question 21. A potentiometer wire has a length of 4m and a resistance of 8Ω. The resistance that must be connected in series with the wire and an accumulator of e.m.f 2V, to get a potential gradient of 1mV per cm on the wire is :

1. 40 Ω
2. 44 Ω
3. 48 Ω
4. 32 Ω

Answer: 4. 32 Ω

Question 22. A, B, and C are voltmeters of resistance R, 1.5R, and 3R respectively as shown in the figure. When some potential difference is applied between X and Y, the voltmeter readings are V_A, V_B, and V_C respectively. Then :

1. V_A≠ V_B= V_C
2. V_A= V_B≠ V_C
3. V_A≠ V_B≠ V_C
4. V_A= V_B= V_C

Answer: 4. V_A= V_B= V_C

Question 23. A circuit contains an ammeter, a batter of 30 V, and a resistance of 40.8 ohm all connected in series. If the ammeter has a coil of resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be:

1. 0.25 A
2. 2A
3. 1 A
4. 0.5 A

Answer: 3.1 A

Question 24. Two metal wires of identical dimensions are connected in series. If σ₁ and σ₂are the conductivities of the metal wires respectively, the effective conductivity of the combination is :

1. \(\frac{\sigma_1+\sigma_2}{2 \sigma_1 \sigma_2}\)
2. \(\frac{\sigma_1+\sigma_2}{\sigma_1 \sigma_2}\)
3. \(\frac{\sigma_1 \sigma_2}{\sigma_1+\sigma_2}\)
4. \(\frac{2 \sigma_1 \sigma_2}{\sigma_1+\sigma_2}\)

Answer: 4. \(\frac{2 \sigma_1 \sigma_2}{\sigma_1+\sigma_2}\)

Question 25. A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. E0 and a resistance r₁. An unknown e.m.f. E is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by :

1. \(\frac{E_0 r}{\left(r+r_1\right)} \cdot \frac{I}{L}\)
2. \(\frac{E_0 I}{L}\)
3. \(\frac{L E_0 r}{\left(r+r_1\right) l}\)
4. \(\frac{L E_0 r}{I r_1}\)

Answer: 1. \(\frac{E_0 r}{\left(r+r_1\right)} \cdot \frac{I}{L}\)

Question 26. The charge flowing through a resistance R varies with time t as Q = at – bt₂, where a and b are positive constants. The total heat produced in R is :

1. \(\frac{a^3 R}{b}\)
2. \(\frac{a^3 R}{6 b}\)
3. \(\frac{a^3 R}{3 b}\)
4. \(\frac{a^3 R}{2 b}\)

Answer: 2. \(\frac{a^3 R}{6 b}\)

Question 27. The potential difference (V_A – V_B) between the points A and B in the given figure is

1. + 9 V
2. – 3V
3. + 3 V
4. + 6 V

Answer: 1. + 9 V

Question 28. A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is :

1. 13 Ω
2. 230 Ω
3. 46 Ω
4. 26 Ω

Answer: 4. 26 Ω

Question 29. The resistance of a wire is ‘R’ ohm. If it is melted and stretched to ‘n’ times its original length, its new resistance will be

1. nR
2. \(\frac{R}{n}\)
3. \(n^2 R\)
4. \(\frac{R}{n^2}\)

Answer: 3. \(n^2 R\)

Question 30. A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F. because the method involves :

1. cells
2. Potential gradients
3. A condition of no current flow through the galvanometer
4. A combination of cells, galvanometer, and resistance

Answer: 3. A condition of no current flow through the galvanometer

Question 31. The current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is :

1. 40 Ω
2. 500 Ω
3. 250 Ω
4. 25 Ω

Answer: 3. 250 Ω

Question 32. A carbon resistor of (47±4.7) kΩ is to be marked with rings of different colors for identification. The color code sequence will be

1. Violet – Yellow – Orange – Silver
2. Green – Orange – Violet – Gold
3. Yellow – Green – Violet – Gold
4. Yellow – Violet – Orange – Silver

Answer: 4. Yellow – Violet – Orange – Silver

Question 33. A set of ‘n’ equal resistors, of value ‘R’ each, are connected in series to a battery of emf ‘E’ and internal resistance ‘R’. The current drawn is Ι. Now, the ‘n’ resistors are connected in parallel to the same battery. Then the current drawn from the battery becomes 10 Ι. The value of ‘n’ is:-

1. 10
2. 9
3. 20
4. 11

Answer: 1. 10

Question 34. A battery consists of a variable number ‘n’ of identical cells (having internal resistance ‘r’ each) which are connected in series. The terminals of the battery are short-circuited and the current Ι is measured. Which of the graphs shows the correct relationship between Ι and n?

Answer: 1.

Question 35. In the circuits shown below, the readings of the voltmeters and the ammeters will be:

1. V₂ > V₁ and i₁ > i₂
2. V₂ > V₁ and i₁ = i₂
3. V₁ = V₂ and i₁ > i₂
4. V₁ = V₂ and i₁ = i₂

Answer: 4. V₁ = V₂ and i₁ = i₂

Question 36. Which of the following acts as a circuit protection device?

1. Fuse
2. Conductor
3. Inductor
4. Switch

Answer: 1. Fuse

Question 37. As shown in the figure, six similar bulbs are connected with a DC source of emf E and zero internal resistance. The ratio of power consumption by the bulbs when

1. All are glowing and
2. In the situation when two from section A and one from section B are glowing, will be:

1. 2: 1
2. 4: 9
3. 9: 4
4. 1: 2

Answer: 3. 9: 4

Question 38. The reading of an ideal voltmeter in the circuit shown is

1. 0.6 V
2. 0 V
3. 0.5 V
4. 0.4 V

Answer: 4. 0.4 V

Question 39. The meter bridge shown is in a balanced position with \(\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\ell_1}{\ell_2}\). If we now interchange the positions of the galvanometer and cell, will the bridge work? If yes, what will be the balance condition?

1. \(\text { yes, } \frac{P}{Q}=\frac{\ell_1-\ell_1}{\ell_2+\ell_1}\)
2. no, no null point
3. \(\text { yes, } \frac{P}{Q}=\frac{\ell_2}{\ell_1}\)
4. \(\text { Yes, } \frac{P}{Q}=\frac{\ell_1}{\ell_2}\)

Answer: 4. \(\text { Yes, } \frac{P}{Q}=\frac{\ell_1}{\ell_2}\)

Question 40. For the circuit shown in the figure, the current I will be

1. 0.75 A
2. 1 A
3. 1.5 A
4. 0.5 A

Answer: 2. 1 A

Question 41. Two solid conductors are made up of the same material and have the same length and same resistance. One of them has a circular cross-section of area and the other one has a square cross-section of area A₂. The ratio A₁/A₂ is

1. 1.5
2. 1
3. 0.8
4. 2

Answer: 2. 1

Question 42. For the circuit given below, Kirchhoff’s loop rule for the loop BCDEB is given by the equation

1. \(-i_2 R_2+E_2-E_3+i_3 R_1=0\)
2. \(i_2 R_2+E_2-E_3-i_3 R_1=0\)
3. \(i_2 R_2+E_2+E_3+i_3 R_1=0\)
4. \(-i_2 R_2+E_2+E_3+i_3 R_1=0\)

Answer: 2. \(i_2 R_2+E_2-E_3-i_3 R_1=0\)

Question 43. The equivalent resistance between A and B for the mesh shown in the figure is

1. 7.2 Ω
2. 16 Ω
3. 30 Ω
4. 4.8 Ω

Answer: 2. 16 Ω

Question 44. Which of the following graphs represents the variation of resistivity (ρ ) with temperature (T) for copper?

Answer: 4.

Question 45. A resistance wire connected in the left gap of a meter bridge balances a 10 Ωresistance in the right gap at a point that divides the bridge wire in the ratio 3: 2. If the length of the resistance wire is 1.5m. Then the length of 1 Ω of the resistance wire is

1. 1.5 × 10^-2m
2. 1.0 × 10^-2m
3. 1.0 × 10^-1m
4. 1.5 × 10^-1m

Answer: 3. 1.0 × 10^-1m

Question 46. A charged particle having drift velocity of 1.5 × 10^-4ms^-1 in an electric field of 3 × 10^-10 Vm^-1 has a mobility in m² V^-1 S^-1 of:

1. 2.25×10^-15
2. 2.25×10¹⁵
3. 2.25×10⁶
4. 2.25×10^-6

Answer: 3. 2.25×10⁶

Question 47. The color code of a resistance is given below

The values of resistance and tolerance, respectively, are

1. 470, 5% Ω
2. 470k, 5% Ω
3. 47k, 10% Ω
4. 4.7k, 5% Ω

Answer: 1. 470 , 5% Ω

Question 48. The solids which have a negative temperature coefficient of resistance are

1. Insulators and semiconductors
2. Metals
3. Insulators only
4. Semiconductors only

Answer: 1. Insulators and semiconductors

Question 49. Column 1 gives certain physical terms associated with the current flow through a metallic conductor. Column – 2 gives some mathematical relations involving electrical quantities. Match Column – 1 and Column – 2 with appropriate relations.

1. 1) (1)-I, (2)-(S), (3)-(Q), (4)-(P)
2. (1)-I, (2)-(P), (3)-(S), (4)-(Q)
3. (1)-I, (2)-(P). (3)-(S), (4)-(P)
4. (1)-I, (2)-(S), (3)-(P), (4)-(Q)

Answer: 4. (1)-I, (2)-(S), (3)-(P), (4)-(Q)

Question 50. In a potentiometer circuit, a cell of EMF 1.5V gives a balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, does the balance point occur?

1. 21.6 cm
2. 64 cm
3. 62 cm
4. 60 cm

Answer: 4. 60 cm

Question 51. The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section, and same material is 0.25 Ω. What will be the effective resistance if they are connected in series?

1. 0.5Ω
2. 1Ω
3. 4Ω
4. 0.25Ω

Answer: 3. 4Ω

Question 52. Three resistors having resistances r₁, r₂, and r₃ are connected as shown in the given circuit. The ratio i₁ of currents in terms of resistances used in the circuit is

1. \(\frac{r_2}{r_2+r_3}\)
2. \(\frac{r_1}{r_1+r_2}\)
3. \(\frac{r_2}{r_1+r_3}\)
4. \(\frac{r_2}{r_2+r_3}\)

Answer: 1. \(\frac{r_2}{r_2+r_3}\)

Question 53. Two conductors have the same resistance at 0ºC but their temperature coefficients of resistance are α1 and α2. The respective temperature coefficients of their series and parallel combinations are nearly

1. \(\frac{\alpha_1+\alpha_2}{2}, \alpha_1+\alpha_2\)
2. \(\alpha_1+\alpha_2, \frac{\alpha_1+\alpha_2}{2}\)
3. \(\alpha_1+\alpha_2, \frac{\alpha_1 \alpha_2}{\alpha_1+\alpha_2}\)
4. \(\frac{\alpha_1+\alpha_2}{2}, \frac{\alpha_1+\alpha_2}{2}\)

Answer: 4. \(\frac{\alpha_1+\alpha_2}{2}, \frac{\alpha_1+\alpha_2}{2}\)

Question 54. If a wire is stretched to make it 0.1% longer, its resistance will :

1. Increase by 0.05%
2. Increase by 0.2%
3. Decrease by 0.2%
4. Decrease by 0.05%

Answer: 2. Increase by 0.2%

Question 56. The current in the primary circuit of a potentiometer is 0.2 A. The specific resistance and cross-section of the potentiometer wire are 4 × 10^-7 ohm meters and 8 × 10^-7 respectively. The potential gradient will be equal to :

1. 1 V/ m
2. 0.5 V/m
3. 0.1 V/m
4. 0.2 V/m

Answer: 3. 0.1 V/m

Question 57. Two electric bulbs marked 25W – 220V and 100W – 220 V are connected in series to a 440 V supply. Which of the bulbs will fuse?

1. both
2. 100W
3. 25W
4. neither

Answer: 3. 25W

Question 59. This question has Statement 1 and Statement 2. Of the four choices given after the Statements, choose the one that best describes the two Statements.

1. Statement 1: For a higher range, the resistance of the ammeter (RA ≈ shunt resistance) should be more.
2. Statement 2: To increase the range of the ammeter, an additional shunt needs to be used across it.
   1. Statement -2 is true, statement -2 is true, and Statement -2 is the correct explanation of Statement -1.
   2. Statement -1 is true, Statement- 2 is true, and Statement – 2 is not the correct explanation of Statement- 1.
   3. Statement -1 is true, Statement 2 is false.
   4. Statement -1 is false, and Statement – 2 is true.

Answer: 4. Statement -1 is false, Statement – 2 is true.

Question 60. The supply voltage to the room is 120 V. The resistance of the lead wires is 6 Ω. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?

1. zero Volt
2. 2.9 Volt
3. 13.3 Volt
4. 10.04 Volt

Answer: 4. 10.04 Volt

Question 61. In a large building, there are 15 bulbs for 40W, 5 bulbs for 100W, 5 fans for 80W, and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be:

1. 8 A
2. 10 A
3. 12 A
4. 14 A

Answer: 3. 12 A

Question 62. When a 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10^-4 ms^-1. If the electron density in the wire is 8 × 1028 m^-3, the resistivity of the material is close to:

1. 1.6 × 10^-8 Ωm
2. 1.6 × 10^-7 Ωm
3. 1.6 × 10^-6 Ωm
4. 1.6 × 10^-5 Ωm

Answer: 4. 1.6 × 10^-5 Ωm

Question 63. In the circuit shown, the current in the 1Ω resistor is :

1. 1.3 A, from P to Q
2. 0 A
3. 0.13 A, from Q to P
4. 0.13 A, from P to Q

Answer: 3. 0.13 A, from Q to P

Question 64. A galvanometer having a coil resistance of 100 Ω gives a full-scale deflection when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into an ammeter giving a full-scale deflection for a current of 10 A, is :

1. 2 Ω
2. 0.1 Ω
3. 3 Ω
4. 0.01 Ω

Answer: 4. 0.01 Ω

Question 65. The temperature dependence of resistances of Cu and undoped Si in the temperature range 300 – 400 K, is best described by :

1. Linear increase for Cu, exponential increase for Si
2. Linear increase for Cu, exponential decrease for Si
3. Linear decrease for Cu, linear decrease for Si
4. Linear increase for Cu, linear increase for Si

Answer: 2. Linear increase for Cu, exponential decrease for Si

Question 66.

In the above circuit, the current in each resistance is :

1. o A
2. 1 A
3. 0.25 A
4. 0.5 A

Answer: 1. o A

Question 67. Which of the following statements is false?

1. Krichhoff’s second law represents energy conservation.
2. Wheatstone bridge is the most sensitive when all four resistances are of the same order of magnitude
3. In a balanced Wheatstone bridge, if the cell and the galvanometer are exchanged, the null point is disturbed
4. A rheostat can be used as a potential divider.

Answer: 3. In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed

Question 68. When a current of 5mA is passed through a galvanometer having a coil of resistance 15Ω, it shows full-scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0 – 10 V is :

1. 4.005 × 103 Ω
2. 1.985 × 103 Ω
3. 2.045 × 103 Ω
4. 2.535 × 103 Ω

Answer: 2. 1.985 × 103 Ω

Question 69. On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1KΩ. How much was the resistance on the left slot before interchanging the resistances?

1. 550 Ω
2. 910 Ω
3. 990 Ω
4. 505Ω

Answer: 1. 550 Ω

Question 70. Two batteries with e.m.f 12V and 13V are connected in parallel across a load resistor of 10Ω. The internal resistance of the two batteries is 1Ω and 2Ω respectively. The voltage across the load lies between :

1. 11.4V and 11.5 V
2. 11.7V and 11.8V
3. 11.6V and 11.7V
4. 11.5V and 11.6V

Answer: 4. 11.5V and 11.6V

Question 71. A resistance is shown in the figure. Its value and tolerance are given respectively by :

1. 270Ω, 5%
2. 27 kΩ, 20%
3. 270 Ω, 10%
4. 27 kΩ, 10%

Answer: 4. 27 kΩ, 10%

Question 72. The drift speed of electrons, when 1.5A of current flows in a copper wire of cross section 5 mm², is v. If the electron density in copper is 9 × 1028 /m³the value of v in mm/s is close to (Take charge of an electron to be = 1.6 × 10^-19 C)

1. 0.2
2. 3
3. 2
4. 0.02

Answer: 4. 0.02

Question 73. A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance, if its volume remains unchanged, is :

1. 2.5%
2. 0.5%
3. 2.0%
4. 1.0%

Answer: 4. 1.0%

Question 74. When the switch S, in the circuit shown, is closed, then the value of current i will be :

1. 3A
2. 5A
3. 4A
4. 2A

Answer: 2. 5A

Question 75. In the given circuit the internal resistance of the 18v cell is negligible. If R₁ = 400Ω, R₃ = 100 Ω and R₄ = 500 Ω and the reading of an ideal voltmeter across R₄ is 5V, then the value of R₂ will be :

1. 450 Ω
2. 550 Ω
3. 230 Ω
4. 300 Ω

Answer: 4. 300 Ω

Question 76. A carbon resistance has the following color code. What is the value of the resistance?

1. 5.3 MΩ + 5%
2. 530 kΩ + 5%
3. 64 kΩ + 10%
4. 6.4 MΩ + 5%

Answer: 2. 530 kΩ + 5%

Question 77. A 2W carbon resistor is color-coded with green, black, red, and brown respectively. The maximum current which can be passed through this resistor is :

1. 20mA
2. 63 mA
3. 0.4 mA
4. 100 mA

Answer: 1. 20mA

Question 78. A potentiometer wire AB having length L and resistance 12 r is joined to a cell D of emf ε and internal resistance r. A cell C having emf ε/2 and internal resistance 3r is connected. The length AJ at which the galvanometer is shown in the figure. shows no deflection:

1. \(\frac{5}{12} \mathrm{~L}\)
2. \(\frac{11}{24} \mathrm{~L}\)
3. \(\frac{11}{12} L\)
4. \(\frac{13}{24} L\)

Answer: 4. \(\frac{13}{24} L\)

Question 79. In the given circuit the cells have zero internal resistance. The currents (in Amperes) passing through resistance R₁ and R₂ respectively are :

1. 2, 2
2. 0, 0
3. 1, 2
4. 0.5, 0

Answer: 4. 0.5, 0

Question 80. A uniform metallic wire has a resistance of 18 Ω and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is :

1. 4 Ω
2. 12 Ω
3. 2 Ω
4. 8 Ω

Answer: 1. 4 Ω

Question 81. A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11V is connected across it is :

1. 11 × 10^-4 W
2. 11 × 10⁵ W
3. 11 × 10^-3 W
4. 11 × 10^-5 W

Answer: 4. 11 × 10^-5 W

Question 82. The Wheatstone bridge shown in the figure here gets balanced when the carbon resistor used as R₁ has the color code (Orange, Red, Brown). The resistors R₂ and R₄ are 80Ω and 40Ω, respectively. Assuming that the color code for the carbon resistors gives their accurate values, the color code for the carbon resistor, used as R₃, would be :

1. Red, Green, Brown
2. Grey, Black, Brown
3. Brown, Blue, Brown
4. Brown, Blue, Black

Answer: 3. Brown, Blue, Brown

Question 83. The actual value of resistance R, shown in the figure is 30Ω. This is measured in an experiment as shown using the standard formula \(R=\frac{V}{I},\) V, where V and I are the readings of the voltmeter and ammeter, respectively. If the measured value of R is 5% less, then the internal resistance of the voltmeter is :

1. 35Ω
2. 600Ω
3. 570Ω
4. 350Ω

Answer: 3. 570Ω

Question 84. In a Wheatstone bridge (see Fig), Resistances P and Q are approximately equal. When R = 400 Ω, the bridge is balanced. On interchanging P and Q, the value of R, for balance, is 405 Ω. The value of X is close to :

1. 404.5 ohm
2. 402.5 ohm
3. 403.5 ohm
4. 401.5 ohm

Answer: 2. 402.5 ohm

Question 85. The resistance of the meter bridge AB in a given figure is 4Ω. With a cell of emf ε = 0.5 V and rheostat resistance Rh = 2Ω the null point is obtained at some point J. When the cell is replaced by another one of emf ε = ε₂ the same null point J is found for Rh = 6 Ω. The emf ε₂ is, :

1. 0.5 V
2. 0.3 V
3. 0.4 V
4. 0.6 V

Answer: 2. 0.3 V

Question 86. Two equal resistances when connected in series to a battery, consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be :

1. 120 W
2. 60 W
3. 30 W
4. 240 W

Answer: 4. 240 W

Question 87. In the experimental setup of the meter bridge shown in the figure, the null point is obtained at a distance of 40 cm from A. If a 10Ω resistor is connected in series with R₁, the null point shifts by 10 cm. The resistance that should be connected in parallel with (R₁ + 10) Ω such that the null point shifts back to its initial position is :

1. 60 Ω
2. 30 Ω
3. 20 Ω
4. 40 Ω

Answer: 1. 60 Ω

Question 88. In the circuit shown, the potential difference between A and B is

1. 1 V
2. 3 V
3. 6 V
4. 2V

Answer: 4. 2V

Question 89. A galvanometer having a resistance of 20Ω and 30 divisions on both sides has a figure of merit 0.005 ampere/division. The resistance that should be connected in series such that it can be used as a voltmeter up to 15 volts, is :

1. 80Ω
2. 100Ω
3. 120Ω
4. 125Ω

Answer: 1. 80Ω

Question 90. Two electric bulbs, rated at (25 W, 220 V) and (100 W, 220 V), are connected in series across a 220 V voltage source. If the 25W and 100 W bulbs draw powers P1 and P₂ respectively, then:

1. P₁ = 4 W, P₂ = 16W
2. P₁ = 16 W, P₂= 9W
3. P₁ = 16 W, P₂ = 4W
4. P₁ = 9 W, P₂ = 16W

Answer: 3. P₁ = 16 W, P₂ = 4W

Question 91. In a meter bridge, the wire of length 1m has a non–uniform cross-section such that, the variation \(\frac{\mathrm{dR}}{\mathrm{d} \ell}\)of its resistance R with length l is \(\frac{\mathrm{dR}}{\mathrm{d} \ell} \propto \frac{1}{\sqrt{\ell}}\)Two equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P. What is the length of AP?

1. 0.2 m
2. 0.3 m
3. 0.25 m
4. 0.35 m

Answer: 3. 0.25 m

Question 92. An ideal battery of 4V and resistance R are connected in series in the primary circuit of a potentiometer of length 1m and resistance 5Ω. The value of R, to give a potential difference of 5mV across 10 cm of potentiometer wire, is :

1. 490 Ω
2. 495 Ω
3. 480 Ω
4. 395 Ω

Answer: 4. 395 Ω

Question 93. The galvanometer deflection when key K₁ is closed but K₂ is open, equals θ0 (see figure). On closing K₂ also and adjusting R₂ to 5Ω, the deflection in the galvanometer becomes 05θ. The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery] :

1. 5 Ω
2. 12Ω
3. 25Ω
4. 22 Ω

Answer: 4. 22 Ω

Question 94. A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of 4 × 10–4 A passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V, It should be connected to a resistance of :

1. 250 ohm
2. 200 ohm
3. 6200 ohm
4. 6250 ohm

Answer: 2. 200 ohm

Question 95. In the given circuit diagram, the currents, Ι₁ = – 0.3A, Ι₄ = 0.8 A, and Ι₅ = 0.4 A, are flowing as shown. The currents Ι₂, Ι₃ and Ι₆ respectively, are :

1. 1.1 A, 0.4 A, 0.4 A
2. –0.4 A, 0.4 A, 1.1 A
3. 0.4 A, 1.1 A, 0.4 A
4. 1.1 A, –0.4 A, 0.4 A

Answer: 1. 1.1 A, 0.4 A, 0.4 A