NEET Physics Class 12 Notes Chapter 1 Alternating Current

Alternating Current

Ac And DC Current:

A current that changes its direction periodically is called alternating current (AC). If a current maintains its direction constant it is called direct current (DC).

If a function’s supposed current, varies with time as i = I_msin (ωt+Φ), it is called a sinusoidally varying function. Here I_m is the peak or maximum current and i is the instantaneous current. The factor (ωt+Φ) is called phase. ω It is called the angular frequency, its unit rad/s. Also, ω =2π f where f is the frequency, its unit s-1 or Hz. Also, frequency f = 1/T where T is called the period.

Average Value

Average value of a function, from t1 to t2, is defined as <f> = \(\frac{\int_{t_1}^{t_2} f d t}{t_2-t_1}\). We can find the value of \(\int_{t_1}^{t_2} f d t\) graphically if the graph is simple. It is the area of the f-t graph from t1 to t2.

• The average value of the sin function is zero in a period or an integral multiple of the period. If i = I_m sin ωt then the average value of i in a period is zero.
•  The average value of the square of the sin function is \(\frac{1}{2}.\) in a period or an integral multiple of the period. If i=I_m sin2πt then the average value is \(\frac{1}{2}.\)

Solved Examples

Exercise 1. Find the average value of current shown graphically, from t = 0 to t = 2 sec.

Solution:

From the i – t graph, the area from t = 0 to t = 2 sec

⇒ \(=\frac{1}{2} \times 2 \times 10=10 \text { Amp. sec. }\)× 2 × 10 = 10 Amp. sec.

Average Current = \(\frac{10}{2}=5 \mathrm{Amp} .\)

Root Mean Square Value

Root Mean Square Value of a function, from t1 to t2, is defined as firms = \(f_{m s}=\sqrt{\frac{\int_{t_1}^{t_2} f^2 d t}{t_2-t_1}}\)

If the current varies as i = msin t then the root mean square value of current is \(\frac{1}{\sqrt{2}}\) times of maximum current.

⇒ \(i_{\mathrm{rms}}=\frac{\mathrm{I}_{\mathrm{m}}}{\sqrt{2}}\)

The R M S Values For One Cycle And Half Cycle (Either Positive Half Cycle Or Negative Half Cycle) Is Same.

Solved Examples

Example 2. Find the effective value of current i = 2 sin 100 π t + 2 cos (100 π t + 30º).
Solution :

The equation can be written as i = 2 sin 100  t + 2 sin (100 π t + 120º)
so phase difference Φ = 120º

⇒ \(=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \varphi}\)

⇒ \(=\sqrt{4+4+2 \times 2 \times 2\left(-\frac{1}{2}\right)}=2\) so effective value or rms value = 2 /√2 = √2 A

Ac Sinusoidal Source

The figure shows a coil rotating in a magnetic field. The flux in the coil changes as φ = NBA cos (ωt + Φ). d Emf induced in the coil, from Faraday’s law is \(\frac{-\mathrm{d} \phi}{\mathrm{dt}}\) = N B A ω sin (ωt +Φ). Thus the emf between the points A and B will vary as E = E₀ sin (ωt +Φ). The potential difference between points A and B will also vary as V = V₀ sin (ωt + Φ). The symbolic notation of the above arrangement is. We do not put any + or – sign on the AC source.

Power Consumed Or Supplied In An AC Circuit

Consider an electrical device which may be a source, a capacitor, a resistor, an inductor, or any combination of these. Let the potential difference be v = V_A–V_B = V_m sinΦt. Let the current through it be i = I sin(ωt + Φ). Instantaneous power P consumed by the device = v i =(V_m sin ωt ) (I_m sin(ωt + Φ))

Average power consumed in a cycle = \(\frac{\int_0^{\frac{2 \pi}{\omega}} P d t}{\frac{2 \pi}{\omega}}=V_m I_m \cos \phi\)

⇒ \(=\frac{V_m}{\sqrt{2}} \cdot \frac{I_m}{\sqrt{2}} \cdot \cos \phi=V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi\)

Here cos Φ is called power factor.

Power Factor

The factor cos Φ present about the average power of an AC circuit is called the power factor

So \(\cos \phi=\frac{P_{\mathrm{ac}}}{E_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}}}=\frac{P_{\mathrm{av}}}{P_{\mathrm{v}}}\)

• Thus, the ratio of average power and virtual power in the circuit is equal to the power factor.
• The power factor is also equal to the ratio of the resistance and the impedance of the AC circuit.
• Thus \(\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}\)
• The power factor depends upon the nature of the components used in the circuit.
• If a pure resistor is connected in the AC circuit then Φ = 0, cos Φ = 1

⇒ \(P_{\mathrm{av}}=\frac{\mathrm{E}_0 \mathrm{I}_0}{2}=\frac{\mathrm{E}_0^2}{2 R}=E_{\mathrm{rms}} \mathrm{I}_{\mathrm{mss}}\)

• Thus the power loss is maximum and electrical energy is converted in the form of heat.
• If a pure inductor or a capacitor is connected to the AC circuit, then

Φ = ± 90º, cos  = 0

P_av = 0 (minimum)

1. Thus there is no loss of power.
2. If a resistor and an inductor or a capacitor are connected in an AC circuit, then

Φ ≠ 0 or ± 90º

• This  is between 0 and 90º.
• If the components L, C, and R are connected in series in an AC circuit, then

⇒ \(\tan \varphi=\frac{X}{R}=\frac{(\omega L-1 / \omega C)}{R}\)

and \(\cos \varphi=\frac{R}{Z}=\frac{R}{\left[R^2+(\omega L-1 / \omega C)^2\right]^{1 / 2}}\)

• Power factor cos Φ \(\cos \phi=\frac{R}{Z}\)
• Power factor is a unit less quantity.
• If there is only an inductance coil in the circuit, there will be no loss of power and energy will be stored in the magnetic field.
• If a capacitor is only connected in the circuit, even then there will be no loss of power and energy will be stored in the electrostatic field.
• In reality, an inductor and a capacitor do have some resistance, so there is always some loss of power.
• In the state of resonance, the power factor is one.

Solved Examples

Example 3. When a voltage v_s = sin (ωt + 15º) is applied to an AC circuit the current in the circuit is found to be i = 2 sin (ω t +π/4) then the average power consumed in the circuit is

1. 200 watt
2. 400 √2 watt
3. 100 √6 watt
4. 200 √2 watt

Solution:

P_av = V_rmsI_rms cos Φ

⇒ \(=\frac{200 \sqrt{2}}{\sqrt{2}} \frac{2}{\sqrt{2}} \cdot \cos \left(30^{\circ}\right)=100 \sqrt{6} \text { watt }\)

Some Definitions

The factor cos Φ is called the Power factor.

Im in Φ is called wattless current.

Impedance Z is defined as \(Z=\frac{V_m}{I_m}=\frac{V_{r m s}}{I_{r m s}}\)

I_L is called inductive reactance and is denoted by X_L.

⇒ \(\frac{1}{\omega C}\) is called capacitive reactance and is denoted by X_C

Phasor Diagram

It is a diagram in which AC voltages and currents are represented by rotating vectors. The phasor represented by a vector of magnitude proportional to the peak value rotates counterclockwise with an angular frequency ω about the origin. The projection of the phasor on the vertical axis gives the instantaneous value of the alternating quantity involved.

E = E₀ sin ωt

I = I₀ sin (ωt – π/2)

= –I₀ cosωt

Purely Resistive Circuit

Writing KVL along the circuit

V_S – I_R = 0

⇒ \(I=\frac{V_s}{R}=\frac{V_m \sin \omega t}{R}=I_m \sin \omega t\)

We see that the phase difference between potential difference across resistance, V_R, and I_R is 0.

⇒ \(I_m=\frac{V_m}{R} \quad \Rightarrow \quad I_{r m s}=\frac{V_{r m s}}{R}\)

⇒ \(\langle P\rangle=V_{m s} I_{m s} \cos \phi=\frac{V_{r m s}{ }^2}{R}\)

Graphical and vector representations of E and I are shown below :

Purely Capacitive Circuit

Writing KVL along the circuit,

⇒ \(v_s-\frac{q}{C}=0\)

⇒ \(\mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{Cv})}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{CV}_{\mathrm{m}} \sin \omega \mathrm{t}\right)}{\mathrm{dt}}=C V_m \omega \cos \omega \mathrm{t}=\frac{V_m}{1 / \omega \mathrm{C}} \cos \omega \mathrm{t}=\frac{V_m}{\mathrm{X}_{\mathrm{C}}} \cos \omega \mathrm{t}=\mathrm{I}_{\mathrm{m}} \cos \omega \mathrm{t} .\)

⇒ \(\mathrm{X}_{\mathrm{c}}=\frac{1}{\omega \mathrm{C}}\) and is called capacitive reactance. Its unit is ohm Ω.

From the graph of current versus time and voltage versus time, it is clear that current attains its peak value at a time \(\frac{T}{4}\) before the time at which voltage attains its peak value. Corresponding to \(\frac{T}{4}\)the phase difference = \(=\omega \Delta t=\frac{2 \pi}{T} \frac{T}{4}=\frac{2 \pi}{4}=\frac{\pi}{2}\) is leads v_C by π/2 Diagrammatically (phasor diagram) it is represented as

SinΦ = 90º, <P> = V_rmsI_rmscosΦ = 00

The graphical and vector representations of E and I are shown in the following figures :

Solved Examples

Example 4. An alternating voltage E = 200 sin (100 t) V is connected to a 1 F capacitor through an AC ammeter (it reads rms value). What will be the reading of the ammeter?

Solution : Comparing E = 200 sin (100 t) with E = E0 sin t we find that,

E0 = 200 √2 and ω = 100 (rad/s)

So, \(X_C=\frac{1}{\omega C}=\frac{1}{100 \times 10^{-6}}=10^4 \Omega\)

And as ac instruments read rms value, the reading of ammeter will be, \(I_{r m s}=\frac{E_{r m s}}{X_C}=\frac{E_0}{\sqrt{2} X_C}\left[\text { as } E_{r m s}=\frac{E_0}{\sqrt{2}}\right]\)

i.e \(I_{\mathrm{rms}}=\frac{200 \sqrt{2}}{\sqrt{2} \times 10^4}=20 \mathrm{~mA}\)

Purely Inductive Circuit

Writing KVL along the circuit,

⇒ \(v_s-L \frac{d i}{d t}=0\)

⇒ \(\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}\)

⇒ \(\int \mathrm{Ldi}=\int \mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{tdt}\)

⇒ \(i=-\frac{V_m}{\omega L} \cos \omega t+C\)

< i > = 0

C = 0

⇒ \(\mathrm{i}=-\frac{\mathrm{V}_{\mathrm{m}}}{\omega \mathrm{L}} \cos \omega \mathrm{t}\)

⇒ \(I_m=\frac{V_m}{X_L}\)

From the graph of current versus time and voltage versus time \(\frac{T}{4}\), it is clear that voltage attains its peak value at a time before the time at which current attains its peak value. Corresponding to \(\frac{T}{4}\)the 2 phase difference =\(=\omega \Delta t=\frac{2 \pi}{T} \frac{T}{4}=\frac{2 \pi}{4}=\frac{\pi}{2} .\). Diagrammatically (phasor diagram) it is represented as an. i_L lags behind v_L by π/2.

SinΦ = 90º, <P> = V_rmsI_rmscosΦ = 0

Graphical and vector representations of E and I are shown in the following figures

Summary :

RC Series Circuit With An AC Source

Let i = I_m sin (t + )  ⇒ V_R=i_R= I_mR sin (ωt+Φ)

⇒ \(v_c=I_m X_c \sin \left(\omega t+\phi-\frac{\pi}{2}\right) \Rightarrow \quad v_S=v_R+v_C\)

or V_m sin (ωt+Φ)= mR sin (ωt +Φ) + I_m X_Csin (ωt + \(\phi-\frac{\pi}{2}\))

⇒ \(V_m=\sqrt{\left(I_m R\right)^2+\left(I_m X_c\right)^2+2\left(I_m R\right)\left(I_m X_c\right) \cos \frac{\pi}{2}}\)

⇒ \(\text { or } I_m=\frac{V_m}{\sqrt{R^2+X c^2}} \quad Z=\sqrt{R^2+X c^2}\)

Using the phasor diagram also we can find the above result.

⇒ \(\tan \phi=\frac{I_m X_c}{I_m R}=\frac{X_c}{R}, \quad X_c=\frac{1}{\omega c}\)

Solved Examples

Example 5. In an RC series circuit, the rms voltage of the source is 200V and its frequency is 50 Hz. If 100 R =100 Ω and C= \(\frac{100}{\pi} \mu \mathrm{F}\) μF, find

1. Impedance of the circuit
2. Power factor angle
3. Power factor
4. Current
5. Maximum current
6. Voltage across R
7. Voltage across C
8. The max voltage across R
9. The max voltage across C
10. < P >
11. < PR >
12. < PC >

Solution:

⇒ \(X_c=\frac{10^6}{\frac{100}{\pi}(2 \pi 50)}=100 \Omega\)

⇒ \(Z=\sqrt{R^2+X c^2}=\sqrt{100^2+(100)^2}=100 \sqrt{2} \Omega\)

⇒ \(\tan \phi=\frac{X c}{R}=1 \phi=45^{\circ}\)

⇒ \(\text { Power factor }=\cos \phi=\frac{1}{\sqrt{2}}\)

⇒ \(\text { Current } \mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{Z}}=\frac{200}{100 \sqrt{2}}=\sqrt{2} \mathrm{~A}\)

⇒ \(\text { Maximum current }=\mathrm{I}_{\mathrm{ms}} \sqrt{2}=2 \mathrm{~A}\)

⇒ \(\text { voltage across } R=V_{R, m s}=I_{m s} R=\sqrt{2} \times 100 \text { Volt }\)

⇒ \(\text { voltage across } \mathrm{C}=\mathrm{V}_{\mathrm{c}, \mathrm{rms}}=\mathrm{I}_{\mathrm{rms}} \mathrm{X}_{\mathrm{c}}=\sqrt{2} \times 100 \text { Volt }\)

⇒ \(\text { max voltage across } R=\sqrt{2} V_{R, r m s}=200 \text { Volt }\)

⇒ \(\text { max voltage across } C=\sqrt{2} \mathrm{~V}_{\mathrm{c}, \mathrm{rms}}=200 \text { Volt }\)

⇒ \(\langle\mathrm{P}\rangle=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{mms}} \cos \phi=200 \times \sqrt{2} \times \frac{1}{\sqrt{2}}=200 \text { Watt }\)

<P_R> = I_rms2R = 200 W

<P_C> = 0

Example 6. In the above question if vs(t) = 200 sin (2π 50 t), find (a) i (t), (b) V_R, and (c) V_C(t)
Solution :

i(t) = I_m sin (ωt + Φ) = 2 sin (2π 50 t + 45º)

V_R = i_R. R = i(t) R = 2 × 100 sin (100 πt + 45º)

V_C (t) = i_CX_C (with a phase lag of 90º) = 2 ×100 sin (100 πt + 45 – 90)

Example 7. An AC source of angular frequency ω is fed across a resistor R and a capacitor C in series. The current registered is I. If now the frequency of the source is changed to ω/3 (but maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency ω.
Solution :

According to a given problem,

⇒ \(I=\frac{V}{Z}=\frac{V}{\left[R^2+(1 / C \omega)^2\right]^{1 / 2}}\)

⇒ \(\frac{I}{2}=\frac{V}{\left[R^2+(3 / C \omega)^2\right]^{1 / 2}}\)

Substituting the value of  from Equation (1) in (2),

⇒ \(4\left(R^2+\frac{1}{C^2 \omega^2}\right)=R^2+\frac{9}{C^2 \omega^2} \text {. i.e., } \frac{1}{C^2 \omega^2}=\frac{3}{5} R^2\)

So that \(\frac{X}{R}=\frac{(1 / C \omega)}{R}=\frac{\left(\frac{3}{5} R^2\right)}{R}=\sqrt{\frac{3}{5}} \text { Ans. }\)

LR Series Circuit With An AC Source

From the phasor diagram

⇒ \(\mathrm{V}=\sqrt{(\mathrm{IR})^2+(\mathrm{IXL})^2}=\mathrm{I} \sqrt{(\mathrm{R})^2+\left(\mathrm{XL}_{\mathrm{L}}\right)^2}=\mathrm{IZ} \Rightarrow \tan \phi=\frac{\mathrm{I} \mathrm{XL}_{\mathrm{L}}}{\mathrm{IR}}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}\)

Example 8. A \(\) H inductor and a 12 ohm resistance are connected in series to a 225 V, 50 Hz AC source. Calculate the current in the circuit and the phase angle between the current and the source voltage.
Solution:

⇒ \(X_L=\omega L=2 \pi f L=2 \pi \times 50 \times \frac{9}{100 \pi}=9 \Omega\)

⇒ \(I=\frac{V}{Z}=\frac{225}{15}=15 \mathrm{~A}\)

⇒ \(I=\frac{V}{Z}=\frac{225}{15}=15 \mathrm{~A}\)

⇒ \(\phi=\tan ^{-1}\left(\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}\right)=\tan ^{-1}\left(\frac{9}{12}\right)\)

= tan–1 3/4 = 37º the current will lag the applied voltage by 37º in phase.

Example 9. When an inductor coil is connected to an ideal battery of EMF 10 V, a constant current of 2.5 A flows. When the same inductor coil is connected to an AC source of 1 0 V and 50 Hz then the current is 2A. Find out the inductance of the coil.
Solution :

When the coil is connected to the DC source, the final current is decided by the resistance of the coil.

⇒ \(r=\frac{10}{2.5}=4 \Omega\)

When the coil is connected to an AC source, the final current is decided by the impedance of the coil.

⇒ \(Z=\frac{10}{2}=5 \Omega\)

⇒ \(Z=\sqrt{(r)^2+\left(X_L\right)^2}\)

X_L² = 5²– 4² = 9

X_L= 3Ω

ωL = 2π fL = 3

2π50L = 3

∴ L = 3/100π Henry

Example 10. A bulb is rated at 100 V and 100 W, it can be treated as a resistor. Find out the inductance of an inductor (called choke coil ) that should be connected in series with the bulb to operate the bulb at its rated power with the help of an AC source of 200 V and 50 Hz.
Solution :

From the rating of the bulb, the resistance of the bulb is R= \(R=\frac{V_{\text {rms }}{ }^2}{P}=100 \Omega\)

For the bulb to be operated at its rated value the rms current through it should be 1 A

Also, \(I_{m s}=\frac{V_{r m s}}{Z}\)

⇒ \(1=\frac{200}{\sqrt{100^2+(2 \pi 50 L)^2}}\)

⇒ \(L=\frac{\sqrt{3}}{\pi} H\)

Example 11. A choke coil is needed to operate an arc lamp at 160 V (rms) and 50 Hz. The arc lamp has an effective resistance of 5 Ω when running at 10 A (rms). Calculate the inductance of the choke coil. If the same arc lamp is to be operated on 160 V (dc), what additional resistance is required? Compare the power losses in both cases.
Solution :

As for lamp V_R = IR = 10 × 5 = 50 V, so when it is connected to a 160 V AC source through a choke in series,

V² = V_R² + V_L²,

⇒ \(V_L=\sqrt{160^2-50^2}=152 \mathrm{~V}\)

and as, V_L = IX_L = IωL= 2πfLI

So \(L=\frac{V_L}{2 \pi \mathrm{fI}}=\frac{152}{2 \times \pi \times 50 \times 10}=4.84 \times 10^{-2} \mathrm{H}\)

Now the lamp is to be operated at 160 V dc; instead of choke if additional resistance r is put in series with it,

V = I(R + r), i.e., 160 = 10(5 + r)

i.e., r = 11Ω

In the case of AC, as the choke has no resistance, power loss in the choke will be zero while the bulb will consume, P = I² R = 10² × 5 = 500 W However, in the case of DC as resistance r is to be used instead of choke, the power loss in the resistance r will be.

PL = 10² × 11 = 1100 W while the bulb will still consume 500 W, i.e., when the lamp is run on resistance r instead of choke more than double the power consumed by the lamp is wasted by the resistance r.

Lc Series Circuit With An AC Source

From the phasor diagram

V = I | X_L – X_C| = IZ

Φ = 90º

RLC Series Circuit With An AC Source

From the phasor diagram

⇒ \(V=\sqrt{(\mathrm{IR})^2+(\mathrm{IX} \mathrm{L}-\mathrm{IXc})^2}=\mathrm{I} \sqrt{(\mathrm{R})^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{Xc}^2\right)^2}=\mathrm{IZ}\)

⇒ \(\mathrm{Z}=\sqrt{(R)^2+\left(\mathrm{X}_L-\mathrm{X}_C\right)^2}\)

⇒ \(\tan \phi=\frac{I\left(X_L-X_C\right)}{I R}=\frac{\left(X_L-X_C\right)}{R}\)

Resonance :

The amplitude of current (and therefore I rms also) in an RLC series circuit is maximum for a given value of Vm and R if the impedance of the circuit is minimum, which will be when X_L-X_C =0. This condition is called resonance.

So at resonance: X_L-X_C =0.

Or \(\omega L=\frac{1}{\omega C} \quad \text { or } \quad \omega=\frac{1}{\sqrt{L C}}\)

Example 12. In the circuit shown in the figure, find

1. The reactance of the circuit.
2. Impedance of the circuit
3. The current
4. Readings of the ideal AC voltmeters (these are hot wire instruments and read rms values, they act on heating effect).

Solution : 1.  XL = 2 μ f L = 2μ × 50 ×\(\frac{2}{\pi}=200 \Omega\)

⇒ \(X_c=\frac{1}{2 \pi 50 \frac{100}{\pi} \times 10^{-6}}=100 \Omega\)

The reactance of the circuit X = X_L–X_C = 200-100 = 100 Ω

Since X_L > X_C, the circuit is called inductive

2.  Impedance of the circuit Z =\(Z=\sqrt{R^2+X^2}=\sqrt{100^2+100^2}=100 \sqrt{2} \Omega\)

3. The current \(I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{Z}=\frac{200}{100 \sqrt{2}}=\sqrt{2} \mathrm{~A}\)

4. Readings of the ideal voltmeter

V₁: I_rmsX_L = 200 √2 Volt

V₂: I_rmsR = 100 √2 Volt

V₃: I_rms X_C =100 √2 Volt

V₄: \( I_{r m s} \sqrt{R^2+X_L^2}=100 \sqrt{10} \text { Volt }\)

V₅: I_rmsZ = 200 Volt, which also happens to be the voltage of the source.

Admittance, Susceptance And Conductance

• Admittance :

The reciprocal of the impedance of an AC circuit is called admittance. It is represented by Y

∴ Admittance \(=\frac{1}{\text { Impedance }} \quad \Rightarrow \quad Y=\frac{1}{Z}\)

The unit of admittance is (ohm)–1 or mho.

• Susceptance :

The reciprocal of the reactance of an AC circuit is called susceptance. It is represented by S.

∴ Susceptance \(=\frac{1}{\text { Reactance }} \quad \text { or } \quad S=\frac{1}{X}\)

The unit of susceptance is (ohm)–1 or mho.

The susceptance of a coil of inductance L is called inductive susceptance. It is equal to the reciprocal of inductive reactance.

∴ Inductive susceptance \(=\frac{1}{\text { Inductive reactance }}\)

The susceptance of a capacitor of capacitance C is called capacitive susceptance. It is equal to the reciprocal of capacitive reactance.

∴ Capacitive susceptance \(=\frac{1}{\text { Capacitive reactance }}\)

⇒ \(S_C=\frac{1}{X_C}=\frac{1}{1 / \omega C}=\omega C \text { mho }\)

• Conductance :

The reciprocal of resistance of a circuit is called conductance. It is represented by G.

∴ Conductance = \(=\frac{1}{\text { Resistance }} \quad \text { or } \quad G=\frac{1}{R}\)

The unit of conductivity is also (ohm)–1 or mho.

In a circuit in which different components are connected in parallel and the same EMF is applied to them its analysis in terms of admittance, susceptance, and conductance becomes simpler because current in a component = voltage or (Impedance or Reactance or Resistance) = Voltage × (Admittance or Susceptance or Conductance)

Half-Power Points Or Frequencies, Band Width, And Quality Factor Of A Series Resonant Circuit

Half power frequencies

• The frequencies at which the power in the circuit is half of the maximum power (the power at resonance), are called half-power frequencies. Thus at these frequencies

⇒ \(P=\frac{P_{\max }}{2}\)

• The current in the circuit at half-power frequencies is \(\frac{1}{\sqrt{2}}\)or 0.707 or 70.7% of the maximum current Imax (current at resonance).
• Thus \(I=\frac{I_{\max }}{\sqrt{2}}=0.707 I_{\max }\)
• There are two half-power frequencies f₁ and f₂ :

Lower half power frequency (f₁) :

This half-power frequency is less than the resonant frequency. At this frequency the circuit is capacitive.

Upper half power frequency (f₂) :

This half-power frequency is greater than the resonant frequency. At this frequency the circuit is inductive.

Band Width (Δf) :

• The difference between half-power frequencies f₁ and f₂ is called bandwidth (Δf)
• Bandwidth Δf = (f₁ – f₂)
• For series resonant circuit :

⇒ \(\omega_0=\frac{1}{\sqrt{L C}}\)

Quality factor (Q) :

• In an AC circuit Q is defined by the following ratio :

⇒ \(\mathrm{Q}=2 \pi \times \frac{\text { Maximum energy stored }}{\text { Energy dissipation per cycle }}=\frac{2 \pi}{\mathrm{T}} \times \frac{\text { Maximum energy stored }}{\text { Mean power dissipated }}\)

• For an L–C–R series resonant circuit :

⇒ \(Q=\frac{\omega_{\mathrm{r}} L}{R}=\frac{1}{\omega_{\mathrm{r}} C R}=\frac{\omega_0}{\Delta \omega}=\frac{2 \pi f_0}{\left(f_2-f_1\right) 2 \pi}\)

• Quality factor in terms of bandwidth :

\(Q=\frac{\omega_r}{\omega_2-\omega_1}=\frac{2 \pi f_r}{2 \pi\left(f_2-f_1\right)}=\frac{f_r}{\left(f_2-f_1\right)}=\frac{f_r}{\Delta f}\)

• Quality factor \(=\frac{\text { Resonant frequency }}{\text { Band width }}\)
• Thus the ratio of the resonant frequency and the bandwidth is equal to the quality factor of the circuit.
• In the state of resonance, the voltage across the resistor R will be equal to the applied voltage E. The magnitudes of voltage across the inductor and the capacitor will be equal and their values will be equal QE. Thus

⇒ \(V_L=I \omega L=\frac{E}{R} \omega L=E Q\)

And, \(V_c=I\left(\frac{I}{\omega C}\right)=\frac{E}{\omega C R}=E Q\)

Sharpness Of Resonance :

• For an AC circuit, Q measures the sharpness of resonance.
• When Q is large, the resonance is sharp and when Q is small, the resonance is flat.
• The sharpness of resonance is inversely proportional to the bandwidth and the resistance R.
• For resonance to be sharp the resistance of the circuit should be small.

Form Factor

• The form factor for a sinusoidal current is defined as :

⇒ Form factor = \(\frac{r m s \text { value of ac }}{\text { Average value of positive half cycle }}=\frac{\mathrm{I}_{\mathrm{rms}}}{2 \mathrm{I}_0 / \pi}=\frac{\mathrm{I}_0}{\sqrt{2}} \cdot \frac{\pi}{2 \mathrm{I}_0}=\frac{\pi}{2 \sqrt{2}}\)

• Similarly form factor for a sinusoidal voltage :

⇒ \(F=\frac{r m s \text { value of alternating voltage }}{\text { Average value of positive half cycle }}=\frac{\pi}{2 \sqrt{2}}\)

Transformer

• It is an instrument that changes the magnitude of alternating voltage or current.
• The magnitude of D.C. voltage or current cannot be changed by it.
• It works with alternating current but not with direct current.
• It converts magnetic energy into electrical energy.
• It works on the principle of electromagnetic induction.
• It consists of two coils :

Primary coil: in which input voltage is applied.

Secondary coil: from which output voltage is obtained

• The frequency of the output voltage produced by the transformer is the same as that of the input voltage, i.e., frequency remains unchanged.

• The transformer core is laminated and is made of soft iron.
• Let the number of turns in the primary coil be np and the voltage applied to it be Ep the number of turns in the secondary coil be ns and the voltage output by Es, then

⇒ \(\frac{E_s}{E_p}=\frac{n_s}{n_p}=K\)

Thus the ratio of voltage obtained in the secondary coil to the voltage applied in the primary coil is equal to the ratio of several turns of respective coils. This ratio is represented by K and it is called the transformer ratio.

• If n_s > n_p, then E_s > E_p and K > 1. The transformer is called a step-up transformer.
• If n_s < n_p, then E_s < E_p and K < 1. The transformer is called a step-down transformer.
• In ideal transformer

Input power = output power

E_p I_p= E_s I_s

where I p – current in the primary coil

Is – current in the secondary coil

Or \(\frac{I_p}{I_s}=\frac{E_s}{E_p}=\frac{n_s}{n_p}=K\)

Or \(\frac{I_s}{I_p}=\frac{E_p}{E_s}=\frac{1}{K}\)

Thus the ratio of currents in the secondary coil and the primary coil is inverse of the ratio of respective voltages.

• As the voltage changes by the transformer, the current changes in the same ratio but in the opposite sense, i.e., the current decreases with the increase of voltage, and similarly the current increases with the decrease of voltage. Due to this reason the coil in which voltage is lesser, the current will be higher and therefore this coil is thicker in comparison to the other coil so that it can bear the heat due to flow of high current.
• In step-up transformer

no > n_p , K > 1  E_S > E_p and I_S < I_p

• and in step down transformer

no < n_p , K < 1  E_S < E_p and I_S> I_p

• If Z_p and Z s are impedances of primary and secondary coils respectively, then

⇒ \(\frac{E_s}{E_p}=\frac{I_p}{I_s}=\frac{n_s}{n_p}=\sqrt{\frac{Z_s}{Z_p}}\)

• The law of conservation of energy is applicable in the transformer.
• Efficiency of transformer \(\frac{\text { Power obtained from sec ondary coil }}{\text { Power applied in primary coil }} \times 100 \%\)
• Generally, the efficiency of transformers is found in between 90% to 100%.
• Energy losses in transformers: Losses of energy are due to the following reasons :
  • Copper losses due to resistance of coils
  • Eddy’s current losses in the core.
  • Hysteresis losses in core.
  • Flux leakage due to poor linking of magnetic flux.
• Uses of transformer :
• • Step-down and step-up transformers are used in electrical power distribution.
  • Audio frequency transformers are used in radiography, television, radio, telephone, etc.
  • Ratio frequency transformers are used in radio communication.
  • Transformers are also used in impedance matching.

Example 13. A 50 Hz a.c. the current of crest value 1A flows through the primary of a transformer. If the mutual inductance between the primary and secondary is 1.5 H, the crest voltage induced in the secondary is

1. 75V
2. 150V
3. 225V
4. 300V

Solution :

The crest value is attained in T/4 time where T is the period of A.C.

Thus dI = 1A in dt = T/4 sec.

⇒ \(\mathrm{T}=\frac{1}{50} \quad \text { or } \quad \mathrm{dt}=\frac{1}{200}\)

The induced emf is |E2| = \(M \frac{d I_1}{d t}=1.5 \times \frac{1}{(1 / 200)}=1.5 \times 200=300 \mathrm{~V}\)

The correct answer is (4)

Solved Miscellaneous Problems

Problem 1. The peak voltage in a 220 V AC source is

1. 220 V
2. about 160 V
3. about 310 V
4. 440 V

Solution : V₀ = √2 V_rms = √2 × 220  310 V

Answer:  (3)

Problem 2. An AC source is rated 220 V, 50 Hz. The average voltage is calculated in a time interval of 0.01 s. It

1. Must be zero
2. May be zero
3. Is never zero
4. Is (220/√2)V

Solution: May be zero

Answer: is (2)

Problem 3. Find the effective value of current i = 2 + 4 cos 100  t.

Solution : \(I_{r m s}=\left[\int_0^T \frac{(2+4 \cos 100 \pi t)^2 d t}{T}\right]^{1 / 2}=2 \sqrt{3}\)

Problem 4. The peak value of an alternating current is 5 A and its frequency is 60 Hz. Find its rms value. How long will the current take to reach the peak value starting from zero?

Solution: \(I_{\text {RMS }}=\frac{I_0}{\sqrt{2}}=\frac{5}{\sqrt{2}} A, \quad t=\frac{T}{4}=\frac{1}{240} s\)

Problem 5. An alternating current having a peak value of 14 A is used to heat a metal wire. To produce the same heating effect, a constant current i can be used where i is

1. 14 A
2. about 20 A
3. 7 A
4. about 10 A

Solution : \(\mathrm{I}_{\text {RMS }}=\frac{\mathrm{I}_0}{\sqrt{2}}=\frac{14}{\sqrt{2}} \simeq 10\)

Answer: is (4)

Problem 6. Find the average power consumed in the circuit if a voltage vs = 200√2 sinωt is applied to an AC circuit and the current in the circuit is found to be i = 2 sin (ωt + π/4).

Solution : \(\mathrm{P}=\mathrm{V}_{\mathrm{RMS}} \mathrm{I}_{\mathrm{RMS}} \cos \phi=\frac{200 \sqrt{2}}{\sqrt{2}} \times \frac{2}{\sqrt{2}} \times \cos \frac{\pi}{4}=200 \mathrm{~W}\)

Problem 7. A capacitor acts as an infinite resistance for

1. DC
2. AC
3. DC as well as AC
4. neither AC nor DC

Solution:

xC = \(x_c=\frac{1}{\omega c} \text { for } D C \omega=0 . \text { so, } x_c=\infty\)

Answer: is (1)

Problem 8. A 10 μF capacitor is connected with an AC source E = 200 sin (100 t) V through an AC ammeter (it reads rms value). What will be the reading of the ammeter?

Solution : \(I_0=\frac{V_0}{x_C}=\frac{200 \sqrt{2}}{1 / \omega C} ; I_{R M S}=\frac{I_0}{\sqrt{2}}=200 \mathrm{~mA}\)

Problem 9. Find the reactance of a capacitor (C = 200 μF) when it is connected to

1. 10 Hz AC source,
2. a 50 Hz AC source and
3. a 500 Hz AC source.

Solution:

1. \(x_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C} \simeq 80 \Omega \text { for } f=10 \mathrm{~Hz} \text { AC source, }\)
2. \(x_c=\frac{1}{\omega C}=\frac{1}{2 \pi f C} \simeq 16 \Omega \text { for } f=50 \mathrm{~Hz} \text { and }\)
3. \(x_c=\frac{1}{\omega C}=\frac{1}{2 \pi f C} \simeq 1.6 \Omega \text { for } f=500 \mathrm{~Hz}\)

Problem 10. An inductor (L = 200 mH) is connected to an AC source of peak current. What is the instantaneous voltage of the source when the current is at its peak value?
Solution :

Because the phase difference between voltage and current is π/2 for the pure inductors. So, Ans. is zero

Problem 11. An AC source producing emf ξ = ξ₀[cos(100 π s^-1)t + cos(500 π s^-1)t]is connected in series with a capacitor and a resistor. The current in the circuit is found to be i = i₁ cos[(100 π s^-1)t + φ₁]+i₂ cos[(500 π s^-1)t+ φ₁]

1. i₁ > i₂
2. i₁ = i₂
3. i₁ < i₂
4. The information is insufficient to find the relation between i₁ and i₂

Solution : Impedance z is given by z = \(\sqrt{\left(\frac{1}{\omega C}\right)^2+R^2}\)

For higher ω, z will be lower so the current will be higher

Ans is (3)

Problem 12. An alternating voltage of 220-volt r.m.s. at a frequency of 40 cycles/sec is supplied to a circuit containing a pure inductance of 0.01 H and a pure resistance of 6 ohms in series.

1. The current,
2. Potential differences across the resistance,
3. Potential differences across the inductance,
4. The time lag,
5. Power factor

Solution:

1. \( z=\sqrt{(\omega L)^2+R^2}=\sqrt{\left(2 \pi \times 40 \times 0.01^2\right)^2+6^2}=\sqrt{(42.4)}\)
   • \(I_{R M S}=\frac{220}{z}=33.83 \mathrm{amp}\)
2. \(V_{\text {RMS }}=I_{R M S} \times R=202.98 \text { volts }\)
3. \( \omega \mathrm{L} \times \mathrm{I}_{\text {RMS }}=96.83 \text { volts }\)
4. \(\mathrm{t}=\mathrm{T} \frac{\varphi}{2 \pi}=0.01579 \mathrm{sec}\)
5. \(\cos \phi=\frac{R}{Z}=0.92\)

Problem 13. Which of the following plots may represent the reactance of a series LC combination?

Solution: D

Problem 14. A series AC circuit has a resistance of 4Ω and a reactance of 3Ω. The impedance of the circuit is

1. 5 Ω
2. 7 Ω
3. 12/7 Ω
4. 7/12 Ω

Solution: \(Z=\sqrt{4^2+3^2}=5 \Omega\)

Answer: is (1)

Problem 15. Comprehension – 1 A voltage source V = V₀ sin (100 t) is connected to a black box in which there can be either one element out of L, C, R, or any two of them connected in series.

At Steady state. the variation of current in the circuit and the source voltage are plotted together with time, using an oscilloscope, as shown

1. The element(s) present in the black box is/are :

1. only C
2. L C
3. L and R
4. R and C

Solution:

As the current is leading the source voltage, so circuit should be capacitive and as the phase difference is not \(\frac{\pi}{2}\)it must contain the resistor also.

2. Values of the parameters of the elements, present in the black box are –

1. R = 50Ω , C = 200µf
2. R = 50Ω , L = 2mµ
3. R = 400Ω , C = 50µf
4. None of these

Solution:

⇒ \(\text { Time delay }=\frac{\varphi}{\omega}=\frac{\pi}{400} \Rightarrow \varphi=\frac{\pi}{4}\)

⇒ \(\tan ^{-1}\left(\frac{1}{R \omega C}\right)=\frac{\pi}{4} \quad \Rightarrow \frac{1}{\omega C}=R\)

⇒ \(i_0=\frac{v_0}{\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}}\)

⇒ \(\sqrt{2}=\frac{100}{\sqrt{R^2+R^2}} \rightarrow R=50 \Omega \text { and } C=\frac{1}{50 \times 100}=200 \mu \mathrm{F}\)

3. If the AC source is removed, the circuit is shorted for some time so that the capacitor is fully discharged and then a battery of constant EMF is connected across the black box. At t = 0, the current in the circuit will –

1. increase exponentially with time constant = 0.02 sec.
2. decrease exponentially with time constant = 0.01 sec.
3. oscillate with angular frequency 20 rad/sec
4. first increase and then decrease

Solution: For DC circuit \(\mathrm{i}=\mathrm{i}_0 \mathrm{e}^{-\frac{\mathrm{t}}{R C}} \text { and } \mathrm{RC}=0.01 \mathrm{sec} \text {. }\)

Problem 16. Comprehension-2 An AC generator G with an adjustable frequency of oscillation is used in the circuit, as shown.

1. Current drawn from the AC source will be maximum if its angular frequency is –

1. 10⁵ rad/s
2. 10⁴ rad/s
3. 5000 rad/s
4. 500 rad/s

Solution: The current drawn is maximum at a resonant angular frequency

L_eq = 4 mH C_eq = 10 μF

⇒ \(\omega=\frac{1}{\sqrt{\text { LC }}}=5000 \mathrm{rad} / \mathrm{s}\)

2. To increase the resonant frequency of the circuit, some of the changes in the circuit are carried out. Which change(s) would certainly increase in resonant frequency?

1. R is increased.
2. L₁ is increased and C₁ is decreased.
3. L₂ is decreased and C₂ is increased.
4. C₃ is removed from the circuit.

Solution: (4) C_eq decreases thereby increasing resonant frequency.

3. If the AC source G is of 100 V rating at the resonant frequency of the circuit, then the average power supplied by the source is –

1. 50 W
2. 100 W
3. 500 W
4. 1000 W

Solution: At resonance \(i_{\mathrm{ms}}=\frac{100}{100}=1 \mathrm{~A}\)

Power supplied = V_rms I_rms cos Φ (Φ = 0 at resonance) P = 100 W

Key Concept

AC and DC Current:

A current that changes its direction periodically is called alternating current (AC). If a current maintains its direction constant it is called direct current (DC).

Average Value :

Average value of a function, from t1 to t2, is defined as < f > \(=\frac{\int_{t_1}^{t_2} f . d t}{t_2-t_1} .\)

Root Mean Square Value:

Root Mean Square Value of a function, from t₁to t₂, is defined as frms \(=\sqrt{\frac{\int_{t_1}^{t_2} f^2 d t}{t_2-t_1}}.\)

Power Consumed or Supplied in an AC Circuit:

Instantaneous power P consumed by the device = V i =(V_m sin ωt ) (I_m sin(ωt + Φ))

Average power consumed in a cycle \(=\frac{\int_0^{\frac{2 \pi}{\omega}} P d t}{\frac{2 \pi}{\omega}}=\frac{1}{2} V_m I_m \cos \phi=\frac{V_m}{\sqrt{2}} \cdot \frac{I_m}{\sqrt{2}} \cdot \cos \phi=V_{r m s} I_{r m s} \cos \phi .\)

Here cosΦ is called power factor.

Some Definitions:

The factor cosΦ is called the Power factor.

I_m in Φ is called wattless current.

Impedance Z is defined as Z

⇒ \(Z=\frac{V_m}{I_m}=\frac{V_{r m s}}{I_{\mathrm{rms}}}\)

_ωL is called inductive reactance and is denoted by X_L.

\(\frac{1}{ \mathrm{\omega} C}\) is called capacitive reactance and is denoted by X_C.

Purely Resistive Circuit:

⇒ \(I=\frac{V_s}{R}=\frac{V_m \sin \omega t}{R}=I_m \sin \omega t\)

⇒ \(I_m=\frac{V_m}{R} \Rightarrow I_{r m s}=\frac{V_{r m s}}{R}<P>=V_{m s} I_{r m s} \cos \phi=\frac{V_{r m s}{ }^2}{R}\)

Purely Capacitive Circuit:

⇒ \(I=\frac{d q}{d t}=\frac{d(C V)}{d t}=\frac{d\left(C V_m \sin \omega t\right)}{d t}=C V_m \omega \cos \omega t=\frac{V_m}{1 / \omega C} \cos \omega t=\cos \omega t=\frac{V_m}{X_C} I_m \cos \omega t .\)

⇒ \(X_c=\frac{1}{\omega C}\) and is called capacitive reactance

IC leads VC by π/2 Diagrammatically (phasor diagram) it is represented as.

SinceΦ =90º, <P> = V_rms I_rmsCosΦ = 0

RLC Series Circuit With An AC Source :

From the phasor diagram

⇒ \(\mathrm{V}=\sqrt{(\mathrm{IR})^2+\left(\mathrm{I} \mathrm{X}_{\mathrm{L}}-\mathrm{IXC}\right)^2}\)

⇒ \(=I \sqrt{(R)^2+\left(X_L-X_C\right)^2}=I Z\)

⇒ \(Z=\sqrt{(R)^2+\left(X_L-X_C\right)^2}\)

⇒ \(\tan \phi=\frac{\mathrm{I}\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)}{\mathrm{IR}}=\frac{\left(\mathrm{XLL}_{\mathrm{L}}-\mathrm{Xc}_{\mathrm{C}}\right)}{\mathrm{R}}\)

Resonance :

The amplitude of current (and therefore I rms also) in an RLC series circuit is maximum for a given value of Vm and R if the impedance of the circuit is minimum, which will be when X_L-X_C =0. This condition is called resonance.

So at resonance: X_L-X_C =0.

⇒ \(\omega L=\frac{1}{\omega C} \quad \text { or } \quad \omega=\frac{1}{\sqrt{L C}} \text {. Let us denote this } \omega \text { as } \omega_r \text {. }\)

Quality factor: \(: Q=\frac{X_L}{R}=\frac{X_C}{R}\)

\(Q=\frac{\text { Re sonance freq. }}{\text { Band width }}=\frac{\omega_R}{\Delta \omega}=\frac{f_R}{f_2-f_1}\)

where f₁ and f₂ are half-power frequencies.

Transformer :

A transformer changes an alternating potential difference from one value to another of greater or s s p smaller value using the principle of mutual induction. For an ideal transformer \(\frac{E_s}{E_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}\), where denotations have their usual meanings.

E_S N and  are the emf, number of turns, and current in the coils.

N_S > N_P  E_S > E_P → step-up transformer.

N_S < N_P  E_S < E_P → step down transformer.

Energy Losses In Transformers are due to

1. Resistance of the windings.

2. Eddy Current.

3. Hysteresis.

4. Flux Leakage.

Alternating Current Exercise-1

Section (1): Average, Peak, And Rms Values

Question 1. If the value of potential in an A.C. circuit is 10V, then the peak value of potential is

1. \(\frac{10}{\sqrt{2}}\)
2. 10√2
3. 20√2
4. \(\frac{20}{\sqrt{2}}\)

Answer: 2. 10√2

Question 2. In an A.C. circuit of capacitance, the phase of current from potential is

1. Forward
2. Backward
3. Both are in the same phase
4. None of these

Answer: 1. Forward

Question 3. A coil of 200 Ω resistance and 1.0 H inductance is connected to an AC source of frequency 200/2π Hz. The phase angle between potential and current will be –

1. 30º
2. 90º
3. 45º
4. 0º

Answer: 3. 45º

Question 4. The hot wire ammeter measures :

1. D.C. current
2. A.C. Current
3. None of above
4. Both (1) and (2)

Answer: 4. Both (1) and (2)

Question 5. A capacitor is a perfect insulator for :

1. constant direct current
2. alternating current
3. direct as well as alternating current
4. variable direct current

Answer: 1. constant direct current

Question 6. A choke coil should have :

1. High inductance and high resistance
2. Low inductance and low resistance
3. High inductance and low resistance
4. Low inductance and high resistance

Answer: 3. High inductance and low resistance

Question 7. An AC voltage source V = 200 sin 100 t is connected across a circuit containing an AC ammeter (it reads rms value) and capacitor of capacity 1 μF. The reading of the ammeter is :

1. 10 mA
2. 20 mA
3. 40 mA
4. 80 mA

Answer: 2. 20 mA

Question 8. The average value of A.C. current in a half-time period may be :

1. Positive
2. Negative
3. zero
4. All of these

Answer: 4. All of these

Question 9. An alternating current is given by I = I₁ cosωt + I₂ sinωt. The rms current is given by

1. \(\frac{1}{\sqrt{2}}\left(\mathrm{I}_1+\mathrm{I}_2\right)\)
2. \(\frac{1}{\sqrt{2}}\left(I_1+I_2\right)^2\)
3. \(\frac{1}{\sqrt{2}}\left(\mathrm{I}_1^2+\mathrm{I}_2^2\right)^{1 / 2}\)
4. \(\frac{1}{2}\left(\mathrm{I}_1^2+\mathrm{I}_2^2\right)^{1 / 2}\)

Answer: 3. \(\frac{1}{\sqrt{2}}\left(\mathrm{I}_1^2+\mathrm{I}_2^2\right)^{1 / 2}\)

Question 10. The peak value of A.C. is amp., and its apparent value is- (in the amp.)

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

Answer: 2. 2

Question 11. r.m.s. value of current i = 3 + 4 sin (ωt + π/3) is:

1. 5A
2. √17A
3. \(\frac{5}{\sqrt{2}} A\)
4. \(\frac{7}{\sqrt{2}} \mathrm{~A}\)

Answer: 2. √17A

Question 12. The peak value of an alternating e.m.f E given by E = E₀cos ω t is 10 volts and the frequency is 50 Hz. At time t = (1/600) sec, the instantaneous value of e.m.f is :

1. 10 volt
2. 5 volt
3. 5 volt
4. 1 volt

Answer: 2. 5 volt

Question 13. An alternating voltage is given by: e = e₁ sin ω t + e₂ cos ω t. Then the root mean square value of voltage is given by :

1. \(\sqrt{e_1^2+e_2^2}\)
2. \(\sqrt{e_1 e_2}\)
3. \(\sqrt{\frac{e_1 e_2}{2}}\)
4. \(\sqrt{\frac{\mathrm{e}_1^2+\mathrm{e}_2^2}{2}}\)

Answer: 4. \(\sqrt{\frac{\mathrm{e}_1^2+\mathrm{e}_2^2}{2}}\)

Question 14. An AC voltage is given by :

⇒ \(E=E_0 \sin \frac{2 \pi t}{T}\)

Then the mean value of voltage was calculated over a time interval of T/2 seconds :

1. Is always zero
2. Is never zero
3. Is (2e₀/π) always
4. May be zero

Answer: 4. May be zero

Question 15. The voltage of an AC source varies with time according to the equation, V = 100 sin 100 π t cos 100 π t. Where t is in second and V is in volt. Then :

1. The peak voltage of the source is 100 volt
2. The peak voltage of the source is (100/√2 ) volt
3. The peak voltage of the source is 50 volt
4. The frequency of the source is 50 Hz

Answer: 3. The peak voltage of the source is 50 volt

Question 16. An AC voltage of V = 220v2 sin \(\left(2 \pi 50 t+\frac{\pi}{2}\right)\) is applied across a DC voltmeter, its reading will be

1. 220 √2 V
2. √2V
3. 220 V
4. zero

Answer: 4. zero

Question 17. The r.m.s value of an A.C. of 50 Hz is 10 amp. The time taken by the alternating current to reach from zero to maximum value to find the peak value will be

1. 2 × 10–2 sec and 14.14 amp
2. 1 × 10–2 sec and 7.07 amp
3. 5 × 10–3 sec and 7.07 amp
4. 5 × 10–3 sec and 14.14 amp

Answer: 4. 5 × 10–3 sec and 14.14 amp

Question 18. If instantaneous current is given by i = 4 cos (ωt + Φ) amperes, then the r.m.s value of current is –

1. 4 amperes
2. amperes
3. amperes
4. zero amperes

Answer: 2. amperes

Question 19. A 40Ω electric heater is connected to a 200V, 50 Hz mains supply. The peak value of electric current flowing in the circuit is approximately-

1. 2.5 A
2. 5.0 A
3. 7 A
4. 10 A

Answer: 3. 7 A

Question 20. When a magnet is approached near a glowing bulb, the vibrations are produced in filament carrying current, then the current through the filament is-

1. D.C.
2. A.C.
3. A mixture of A.C. and D.C.
4. Nothing can be said

Answer: 2. A.C.

Question 21. An AC ammeter is used to measure current in a circuit. When a given direct constant current passes through the circuit, the AC ammeter reads 3 amperes. When an alternating current passes through the circuit, the AC ammeter reads 4 amperes. Then the reading of this ammeter if DC and AC flow through the circuit simultaneously, is :

1. 3 A
2. 4 A
3. 7 A
4. 5 A

Answer: 4. 5 A

Section (2): Power Consumed In An AC Circuit

Question 1. A choke coil is preferred to a rheostat in an AC circuit:

1. It consumes almost zero power
2. It increases current
3. It increases power
4. It increases voltage

Answer: 1. It consumes almost zero power

Question 2. The average power consumed in an A.C. series circuit is given by (symbols have their usual meaning) :

1. E_rms I_rms cosΦ
2. (Arms)2 R
3. \(\frac{E_{\max }{ }^2 R}{2(|z|)^2}\)
4. All of these

Answer: 4. All of these

Question 3. A circuit with e.m.f. E = 200 sin ωt, contains a capacitance and inductance, then the value of the power factor will be 

1. 0
2. 1
3. 0.6
4. 0.3

Answer: 1. 0

Question 4. If a choke coil of negligible resistance works on a 220-volt source and 5m. Amp. current is flowing through it, then the loss of power in the choke coil is-

1. 0
2. 11 watt
3. 44 × 103 watt
4. 1.1 watt

Answer: 3. 44 × 103 watt

Question 5. The value of current at half power point is-

1. Im√2
2. \(\frac{I_{\mathrm{m}}}{\sqrt{2}}\)
3. 2Im
4. \(\frac{\mathrm{I}_{\mathrm{m}}}{2}\).

Answer: 2. \(\frac{I_{\mathrm{m}}}{\sqrt{2}}\)

Question 6. Expressions for emf and current in an A.C. circuit are E = 200 sin \(\left(314+\frac{\pi}{3}\right)\) volt and amp. The power factor is-

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

Answer: 1. \(\frac{1}{2}\)

Question 7. In a series LR circuit, the voltage drop across the inductor is 8 volts, and across the resistor is 6 volts. Then voltage applied and power factor of the circuit respectively are:

1. 14 V, 0.8
2. 10 V, 0.8
3. 10 V, 0.6
4. 14 V, 0.6

Answer: 3. 10 V, 0.6

Question 8. In an a.c. circuit the e.m. f, (e) and the current (i) at any instant are given respectively by :

e = E 0 sinωt

i = l0 sin (ωt – Φ)

The average power in the circuit over one cycle of a.c. is :

1. \(\frac{E_0 1_0}{2} \cos \varphi\)
2. \(\mathrm{E}_0 \mathrm{I}_0\)
3. \(\frac{E_0 l_0}{2}\)
4. \(\frac{E_0 1_0}{2} \sin \varphi\)

Answer: 1. \(\frac{E_0 1_0}{2} \cos \varphi\)

Question 9. The power factor of an A.C. circuit having resistance R and inductance L (connected in series) and an angular velocity ω is –

1. \(\frac{\mathrm{R}}{\omega \mathrm{L}}\)
2. \(\frac{R}{\left(R^2+\omega^2 L^2\right)^{1 / 2}}\)
3. \(\frac{\omega L}{R}\)
4. \(\frac{R}{\left(R^2-\omega^2 L^2\right)^{1 / 2}}\)

Answer: 2. \(\frac{R}{\left(R^2+\omega^2 L^2\right)^{1 / 2}}\)

Question 10. The average power delivered to a series AC circuit is given by (symbols have their usual meaning) :

1. E_rms I_rms
2. E_rms I_rms cos Φ
3. E_rms I_rms sin Φ
4. zero

Answer: 2. Erms Irms cos Φ

Question 11. The potential difference between V across and the current I flowing through an instrument in an AC circuit are given by :

V = 5 cos ω t volt

I = 2 sin ω t volt

The power dissipated in the instrument is :

1. zero
2. 5 watt
3. 10 watt
4. 2.5 watt

Answer: 1. zero

Question 12. A direct current of 2 A and an alternating current having a maximum value of 2 A flow through two identical resistances. The ratio of heat produced in the two resistances in the same time interval will be:

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

Answer: 3. 2: 1

Question 13. A sinusoidal AC flows through a resistor of resistance R. If the peak current is I_P, then the average power dissipated is :

1. \(\mathrm{I}_{\mathrm{p}}^2 R \cos \theta\)
2. \(\frac{1}{2} I_P^2 R\)
3. \(\frac{4}{\pi} I_p^2 R\)
4. \(\frac{1}{\pi^2} I_p^2 R\)

Answer: 2. \(\frac{1}{2} I_P^2 R\)

Question 14. What is the rms value of an alternating current which when passed through a resistor produces heat, which is thrice that produced by a current of 2 ampere in the same resistor in the same time interval?

1. 6 ampere
2. 2 ampere
3. 2√3 ampere
4. 0.65 ampere

Answer: 3. 2√3 ampere

Question 15. If a current I given by I0 sin\(\left(\omega t-\frac{\pi}{2}\right)\) flows in an A.C. circuit across which an A.C. potential of E = E 0 sin t has been applied, then the power consumption P in the circuit will be

1. \(P=\frac{E_0 I_0}{\sqrt{2}}\)
2. \(P=\sqrt{2} E_0 I_0\)
3. \(P=\frac{E_0 I_0}{2}\)
4. P=0.

Answer: 4. P=0.

Question 16. You have two copper cables of equal length for carrying current. One of them has a single wire of area A of cross-section A, the other has ten wires of cross-section 10 each. Judge their suitability for transporting AC and DC

1. Only single strand for DC and only multiple strand for AC
2. Either for DC, only multiple strands for AC
3. Only single strand for AC, either for DC
4. Only single strand for DC, either for AC

Answer: 2. Either for DC, only multiple strands for AC

Question 17. The self-inductance of a choke coil is 10 mH. When it is connected with a 10V D.C. source, then the loss of power is 20 watts. When it is connected to 10 volt A.C. source loss of power is 10 watts. The frequency of A.C. source will be 

1. 50 Hz
2. 60 Hz
3. 80 Hz
4. 100 Hz

Answer: 3. 80 Hz

Question 18. If the frequency of the source e.m.f. in an AC circuit is n, the power varies with a frequency :

1. n
2. 2 n
3. n/2
4. zero

Answer: 2. 2 n

Question 19. A coil of inductive reactance 31 Ω has a resistance of 8Ω It is placed in series with a condenser of capacitative reactance 25Ω. The combination is connected to an a.c. source of 110 volts. The power factor of the circuit is :

1. 0.56
2. 0.64
3. 0.80
4. 0.33

Answer: 3. 0.80

Question 20. A circuit has a resistance of 12 ohms and an impedance of 15 ohms. The power factor of the circuit will be

1. 0.8
2. 0.4
3. 1.25
4. 0.125

Answer: 1. 0.8

Question 21. A coil of inductance 5.0 mH and negligible resistance is connected to an alternating voltage V = 10 sin (100 t). The peak current in the circuit will be :

1. 2 amp
2. 1 amp
3. 10 amp
4. 20 amp

Answer: 4. 20 amp

Question 22. A resistor and a capacitor are connected to an AC supply of 200 volts, 50 Hz in series. The current in the circuit is 2 ampere. If the power consumed in the circuit is 100 watts, then the resistance in the circuit is:

1. 100 Ω
2. 25 Ω
3. \(\sqrt{125 \times 75} \Omega\)
4. 400 Ω

Answer: 2. 25 Ω

Question 23. The impedance of a series circuit consists of 3-ohm resistance and 4-ohm reactance. The power factor of the circuit is :

1. 0.4
2. 0.6
3. 0.8
4. 1.0

Answer: 2. 0.6

Section (3): Ac Source With R, L, C Connected In Series

Question 1. A series LCR circular connected to a.c. source of variable frequency ‘f’. The graphical representation of the variation of impedance ‘z’ of the circuit with frequency f will be

Answer: 1.

Question 2. With the increase in the frequency of an AC supply, the inductive reactance :

1. Decreases
2. Increases directly proportional to frequency
3. Increases as the square of the frequency
4. Decreases inversely with frequency

Answer: 2. Increases directly proportional to frequency

Question 3. With the increase in the frequency of an AC supply, the capacitive reactance :

1. Varies inversely with frequency
2. Varies directly with frequency
3. Varies directly as a square of frequency
4. Remains constant

Answer: 1. Varies inversely with frequency

Question 4. In an a.c. circuit consisting of resistance R and inductance L, the voltage across R is 60 volts, and that across L is 80 volts. The total voltage across the combination is

1. 140 V
2. 20 V
3. 100 V
4. 70 V

Answer: 3. 100 V

Question 5. An e.m.f. E = 4 cos 1000t volt is applied to an LR circuit of inductance 3mH and resistance 4. The amplitude of the current in the circuit is-

1. 1.0 A
2. 0.8 A
3. \(\frac{4}{\sqrt{7}} \mathrm{~A}\)
4. \(\frac{5}{7} \mathrm{~A}\)

Answer: 2. 0.8 A

Question 6. A coil has a reactance of 100Ω when the frequency is 50 Hz. If the frequency becomes 150 Hz, then the reactance will be

1. 100Ω
2. 300Ω
3. 450Ω
4. 600Ω

Answer: 2. 300Ω

Question 7. The current in a circuit containing a capacitance C and a resistance R in series lead over the applied voltage of frequency \(\frac{\omega}{2 \pi} \text { by. }\)

1. \(\tan ^{-1}\left(\frac{1}{\omega C R}\right)\)
2. \(\tan ^{-1}(\omega \mathrm{CR})\)
3. \(\tan ^{-1}\left(\omega \frac{1}{R}\right)\)
4. \(\cos ^{-1}(\omega C R)\)

Answer: 1. \(\tan ^{-1}\left(\frac{1}{\omega C R}\right)\)

Question 8. In a circuit, an inductance of 0.1 Henry and a resistance of 1 Ω are connected in series with an AC source of voltage V = 5 sin 10 t. The phase difference between the current and applied voltage will be

1. π
2. 2π
3. π/4
4. 0

Answer: 3. π/4

Question 9. If a resistance of 30Ω, a capacitor of reactance 20Ω, and an inductor of inductive reactance 60Ω are connected in series to a 100 V, 50 Hz power source, then

1. A current of 2.0 A flows
2. A current of 3.33 A flows
3. The power factor of the circuit is zero
4. The power factor of the circuit is 2/5

Answer: 1. A current of 2.0 A flows

Question 10. The output of an AC generator is given by: E = E_msin(ωt – Φ/4) and the current is given by i = i_msin(ωt – 3Φ/4). The circuit contains a single element other than the generator. It is :

1. A capacitor.
2. A resistor.
3. An inductor.
4. not possible to decide due to lack of information.

Answer: 3. An inductor.

Question 11. If the readings of v₁ and v₃ are 100 volts each then reading of v₂ is :

1. 0 volt
2. 100 volt
3. 200 volt
4. cannot be determined by the given information.

Answer: 3. 200 volt

Question 12. In an LRC series circuit at resonance current in the circuit is A. If the frequency of the source is changed such that now current lags by 45° then applied voltage in the circuit. Which of the following is correct :

1. Frequency must be increased and current after the change is 10 A
2. Frequency must be decreased and current after the change is 10 A
3. Frequency must be decreased and the current must be the same as that of the initial value
4. The given information is insufficient to conclude anything [Made 2006, CSS, GRSTU]

Answer: 1. Frequency must be increased and current after the change is 10 A

Question 13. In a purely inductive circuit or an A.C. circuit containing inductance only, the current-

1. Leads the e.m.f. by 90°
2. Lags behind the e.m.f. by 90°
3. Sometimes leads and sometimes lags behind the e.m.f.
4. Is in phase with the e.m.f.

Answer: 2. Lags behind the e.m.f. by 90°

Question 14. A series combination of R, L, and C is connected to an a.c. source. If the resistance is 3Ω and the reactance is 4Ω, the power factor of the circuit is-

1. 0.4
2. 0.6
3. 0.8
4. 1.0

Answer: 2. 0.6

Question 15. A 12Ω resistor and a 0.21 Henry inductor are connected in series to an AC source operating at 20 volts, 50 Hz. The phase angle between the current and the source voltage is-

1. 30°
2. 40°
3. 80°
4. 90°

Answer: 3. 80°

Question 16. When 100 V DC is applied across a solenoid, a steady current of 1 flows in it. When 100 V AC is applied across the same solenoid, the current drops to 0.5 A. If the frequency of the AC source is 150 √3/π Hz, the impedance and inductance of the solenoid are :

1. 200 Ω and 1/3 H
2. 100 Ω and 1/16 H
3. 200 Ω and 1.0 H
4. 1100 Ω and 3/117 H

Answer: 1. 200 Ω and 1/3 H

Question 17. If in a series LCR AC circuit, the rms voltage across L, C, and R are V₁, V₂, and V₃ respectively, then the I_rms voltage of the source is always :

1. Equal to V₁ + V₂ + V₃
2. Equal to V₁ – V₂ + V₃
3. more than V₁ + V₂ + V₃
4. None of these is true

Answer: 4. None of these is true

Question 18. In the series LCR circuit, as shown in the figure, the voltmeter and ammeter readings are :

1. V = 100 volt, I = 2 amp
2. V = 100 volt, I = 5 amp
3. V = 1000 volt, I = 2 amp
4. V = 300 volt, I = 1 amp

Answer: 1. V = 100 volt, I = 2 amp

Question 19. An AC voltage source of variable angular frequency ω and fixed amplitude V connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When ω is increased :

1. The bulb glows dimmer
2. The bulb glows brighter
3. The total impedance of the circuit is unchanged
4. The total impedance of the circuit increases

Answer: 2. The bulb glows brighter

Question 20. In a circuit L, C, and R are connected in series with an alternating voltage source of frequency f. The current leads the voltage by 450 The value of C is :

1. \(\frac{1}{2 \pi f(2 \pi f L-R)}\)
2. \(\frac{1}{2 \pi f(2 \pi f L+R}\)
3. \(\frac{1}{\pi f(2 \pi f L-R)}\)
4. \(\frac{1}{\pi f(2 \pi f L+R)}\)

Answer: 2. \(\frac{1}{2 \pi f(2 \pi f L+R}\)

Question 21. In an LCR series a.c. circuit, the voltage across each of the components. L, C, and R is 50 V. The voltage across the LC combination will be :

1. 50 V
2. 50√3
3. 100 V
4. 0 V (zero)

Answer: 4. 0 V (zero)

Question 22. The phase difference between the alternating current and emf is π/2. Which of the following cannot be the constituent of the circuit?

1. C alone
2. R, L
3. L, C
4. L alone

Answer: 2. L, C

Question 23. An alternating voltage E= 200√2 sin (100 t) is connected to a 1 microfarad capacitor through an A.C. ammeter. The reading of the ammeter shall be –

1. 10 mA
2. 20 mA
3. 40 mA
4. 80 mA

Answer: 2. 20 mA

Question 24. A 0.21 -H inductor and an 88- resistor are connected in series to a 220-V, 50-Hz AC source. The current in the circuit and the phase angle between the current and the source voltage are respectively. Use π = 22/7.

1. 2 A, tan^-1 ¾
2. 14.4 A, tan^-1 7/8
3. 14.4 A, tan^-1 8/7
4. 3.28 A, tan^-1 2/11

Answer: 1. 2 A, tan^-1 ¾

Question 25. An LCR series circuit with 100  resistance is connected to an AC source of 200 V and an angular frequency of 300 radians per second. When only the capacitance is removed, the current lags behind the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60º. Then the current and power dissipated in the LCR circuit are respectively

1. 1A, 200 watt.
2. 1A, 400 watt.
3. 2A, 200 watt.
4. 2A, 400 watt.

Answer: 2. 2A, 400 watt.

Question 26. A 100-volt AC source of angular frequency 500 rad/s is connected to an LCR circuit with L = 0.8 H, C = 5 μF, and R = 10 Ω, all connected in series. The potential difference across the resistance is

1. \(\frac{100}{\sqrt{2}} \text { volt }\)
2. 100 volt
3. 50 volt
4. 50v3

Answer: 2. 100-volt

Question 27. A pure resistive circuit element X when connected to an AC supply of peak voltage 200 V gives a peak current of 5 A which is in phase with the voltage. A second circuit element Y, when connected same AC supply also gives the same value of peak current but the current lags behind by 90°. If the series combination of X and Y is connected to the same supply, what will be the rms value of the current?

1. \(\frac{10}{\sqrt{2}} \mathrm{amp}\)
2. \(\frac{5}{\sqrt{2}} \mathrm{amp}\)
3. \(\frac{5}{2} \mathrm{amp}\)
4. 5 amp

Answer: 3. \(\frac{5}{2} \mathrm{amp}\)

Question 28. In an L-R series circuit (L = \(\frac{175}{11}\) mH and R = 12Ω), a variable emf source (V = V0 sin ωt) of V rms 2 = 130 V and frequency 50 Hz is applied. The current amplitude in the circuit and phase of current concerning voltage are respectively (Use π = 22/7)

1. 14.14A, 30°
2. \(10 \sqrt{2} \mathrm{~A}, \tan ^{-1} \frac{5}{12}\)
3. \(10 \mathrm{~A}, \tan ^{-1} \frac{5}{12}\)
4. \(20 \mathrm{~A}, \tan ^{-1} \frac{5}{12}\)

Answer: 4. \(20 \mathrm{~A}, \tan ^{-1} \frac{5}{12}\)

Question 29. In an AC circuit, a resistance of R ohm is connected in series with an inductance L. If the phase angle between voltage and current is 45°, the value of inductive reactance will be.

1. R/4
2. R/2
3. R
4. cannot be found in the given data

Answer: 3. R

Question 30. In an AC circuit, the potential differences across an inductance and resistance joined in series are respectively 16 V and 20 V. The total potential difference across the circuit is

1. 20 V
2. 25.6 V
3. 31.9 V
4. 53.5 V

Answer: 2. 25.6 V

Question 31. An alternating current of frequency ‘f’ is flowing in a circuit containing a resistance R and a choke L in series. The impedance of this circuit is –

1. \(R+2 \pi f L\)
2. \(\sqrt{R^2+4 \pi^2 f^2 L^2}\)
3. \(\sqrt{R^2+L^2}\)
4. \(\sqrt{R^2+2 \pi f L}\)

Answer: 2. \(\sqrt{R^2+4 \pi^2 f^2 L^2}\)

Question 32. An alternating current flows through a circuit consisting of inductance L and resistance R. Periodicity of the supply is \(\frac{\omega}{2 \pi}\) which of the following is true-

1. The limiting value of impedance is L for low-frequency
2. The limiting value of impedance for high frequency is Lω
3. The limiting value of impedance for high frequency is R
4. The limiting value of impedance for low frequency is Lω

Answer: 2. The limiting value of impedance for high frequency is Lω

Question 33. In the following diagram voltage on L and C is-

1. In some phase
2. With a phase angle of 90°
3. In phase angle of 180°
4. It will depend on the value of L and C

Answer: 3. In phase angle of 180°

Question 34. An LCR series circuit is connected to a source of alternating current. At resonance, the applied voltage and the current flowing through the circuit will have a phase difference of-

1. Ω
2. \(\frac{\pi}{2}\)
3. \(\frac{\pi}{4}\)
4. 0

Answer: 4. 0

Question 35. The same current is flowing in two alternating circuits. The first circuit contains only inductance and the other contains only a capacitor. If the frequency of the e.m.f. is increased, the effect on the value of the current will be

1. Increase in the first circuit and decrease in the other
2. Increase in both circuits
3. Decrease in both circuits
4. Decrease in first circuit and increase in other

Answer: 4. Decrease in the first circuit and increase in other

Question 36. Energy dissipates in the LCR circuit in the:

1. L only
2. C only
3. R only
4. all of these

Answer: 3. R only

Question 37. A coil has an inductance of H and is joined in series with a resistance of 220 . When an alternating e.m.f. of 220 V at 50 cps is applied to it, then the wattless component of the rms current in the circuit is

1. 5 ampere
2. 0.5 ampere
3. 0.7 ampere
4. 7 ampere

Answer: 2. 0.5 ampere

Question 38. An electric bulb and a capacitor are connected in series with an AC source. On increasing the frequency of the source, the brightness of the bulb :

1. Increase
2. Decreases
3. Remains unchanged
4. Sometimes increases and sometimes decreases

Answer: 1. Increase

Question 39. By what percentage the impedance in an AC series circuit should be increased so that the power factor changes from (1/2) to (1/4) (when R is constant)?

1. 200%
2. 100%
3. 50%
4. 400%

Answer: 2. 100%

Section (4): Resonance

Question 1. The self-inductance of the motor of an electric fan is 10 H. To impart maximum power at 50 Hz, it should be connected to a capacitance of :

1. 4μF
2. 8μF
3. 1 μF
4. 2μF

Answer: 3. 1 μF

Question 2. In an LCR circuit, capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to :

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

Answer: 3. L/2

Question 3. In an oscillating LC circuit the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is :

1. Q/2
2. Q/ √3
3. Q/ √2
4. Q

Answer: 3. Q/ √2

Question 4. What is the value of inductance L for which the current is maximum in a series LCR circuit with C =10 μ F and ω= 1000 radian/s?

1. 10 mH
2. 100mH
3. 1 mH
4. Cannot be calculated unless R is Known

Answer: 2. 100mH

Question 5. A transistor–oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produces oscillations of frequency f. If L is doubled and C is changed to 4C, then the frequency will be:

1. \(\frac{f}{4}\)
2. 8f
3. \(\frac{f}{2 \sqrt{2}}\)
4. \(\frac{f}{2}\)

Answer: 3. \(\frac{f}{2 \sqrt{2}}\)

Question 6. A 2μF capacitor is initially charged to 20 Volts and then shorted across an 8μH inductor. The maximum value of the current in the circuit is :

1. 10.0 A
2. 7.5 A
3. 12.0 A
4. 8.2 A

Answer: 1. 10.0 A

Question 7. For an A.C. circuit at the condition of resonance-

1. impedance = R
2. impedance = \(=\left(\omega L-\frac{1}{\omega C}\right)\)
3. the potential difference across L and C in the same phase.
4. The current and emf have a phase difference .

Answer: 1. impedance = R

Question 8. A series A.C. circuit consists of an inductor and a capacitor. The inductance and capacitance are respectively. 1 Henry and 25 µF. If the current is maximum in the circuit then the angular frequency will be

1. 200
2. 100
3. 50
4. \(\frac{200}{2 \pi}\)

Answer: 1. 200

Question 9. The value of power factor cosΦ in a series LCR circuit at resonance is :

1. zero
2. 1
3. 1/2
4. 1/2 ohm

Answer: 2. 1/2

Question 10. A series LCR circuit containing a resistance of 120 ohm has an angular resonance frequency of 4 × 103 rad s–1. At resonance, the voltage across resistance and inductance are 60V and 40V respectively. The values of L and C are respectively :

1. 20 mH, 25/8 μF
2. 2mH, 1/35 μF
3. 20 mH, 1/40 μF
4. 2mH, 25/8 nF

Answer: 1. 20 mH, 25/8 μF

Question 11. In an LCR circuit, the capacitance is made one-fourth, when in resonance. Then what should be the change in inductance, so that the circuit remains in resonance?

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

Answer: 1. 4 times

Question 12. A resistor R, an inductor L, and a capacitor C are connected in series to an oscillator of frequency n. If the resonant frequency is n r, then the current lags behind voltage, when :

1. n = 0
2. n < nr
3. n = rr
4. n > nr

Answer: 4. n > nr

Question 13. A 10-ohm resistance 0.5 mH coil and 10µF capacitor are joined in series when a suitable frequency of alternating current source is joined to this combination, and the circuit resonates. If the resistance is halved the resonance frequency-

1. Is halved
2. Is doubled
3. Remains unchanged
4. Is quadrupled

Answer: 3. remains unchanged

Question 14. At a frequency more than the resonance frequency, the nature of an anti-resonant circuit is-

1. resistive
2. capacitive
3. Inductive
4. All of the above

Answer: 3. inductive

Question 15. If the resonance frequency is f and then the capacity is increased 4 times, then the new resonance frequency

1. \(\frac{f}{2}\)
2. 2f
3. f
4. \(\frac{f}{4}\)

Answer: 1. \(\frac{f}{2}\)

Section (5): Transformer

Question 1. The core of any transformer is laminated to: 

1. Reduce the energy loss due to eddy currents
2. Make it lightweight
3. Make it robust
4. Increase the secondary voltage

Answer: 1. Reduce the energy loss due to eddy currents

Question 2. In a transformer, the number of turns in the primary is 140, and in the secondary is 280. If the current in the primary is 4 A, then that in the secondary is :(assume that the transformer is ideal)

1. 4 A
2. 2 A
3. 6 A
4. 10 A

Answer: 2. 2 A

Question 3. A transformer is used for a 100-watt, 20-volt electric bulb at a place where the A.C. mains potential is 200 volts and the current drawn is 0.6 A. The efficiency of the transformer is nearly

1. 48%
2. 68%
3. 30 %
4. 83%

Answer: 4. 83%

Question 4. In a step-up transformer, the turns ratio is 10. If the frequency of the current in the primary coil is 50 Hz then the frequency of the current in the secondary coil will be

1. 500 Hz
2. 5 Hz
3. 60 Hz
4. 50 Hz

Answer: 4. 50 Hz

Question 5. A power (step-up) transformer with a 1: 8 turn ratio has 60 Hz and 120 V across the primary; the load in the secondary is 104 Ω. The current in the secondary is

1. 96 A
2. 0.96 A
3. 9.6 A
4. 96 mA

Answer: 4. 96 mA

Question 6. A transformer is used to light a 140-watt, 24-volt lamp from 240 V AC mains. The current in the main cable is 0.7 amp. The efficiency of the transformer is :

1. 48%
2. 63.8%
3. 83.3%
4. 90%

Answer: 3. 83.3%

Question 7. In a step-up transformer, the voltage in the primary is 220 V and the current is 5A. The secondary
voltage is found to be 22000 V. The current in the secondary (neglect losses) is

1. 5 A
2. 50 A
3. 500 A
4. 0.05 A

Answer: 4. 0.05 A

Question 8. The core of a transformer is laminated to reduce

1. Eddy current loss
2. Hysteresis loss
3. Copper loss
4. Magnetic loss

Answer: 1. Eddy current loss

Paragraphs for Questions 9 and 10

A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 20 km away from the power plant for consumers’ usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers’ end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with a power factor unity. All the currents and voltages mentioned are rms values.

Question 9. If the direct transmission method with a cable of resistance 0.4 Ω km–1 is used, the power dissipation (in %) during transmission is :

1. 20
2. 30
3. 40
4. 60

Answer: 4. 60

Question 10. In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1: 10. If the power to the consumers has to be supplied at 200V, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is :

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

Answer: 1. 200: 1

Alternating Current Exercise-2

Question 1. An AC voltage source V = V₀ sin ωt is connected across resistance R and capacitance C as shown in the figure. It is given that R =\(\frac{1}{\omega \mathrm{C}}\) . The peak current is I₀. If the angular frequency of the voltage source is changed  , keeping the voltage amplitude constant, then the new peak current in the circuit is :

1. \(\frac{\mathrm{I}_0}{2}\)
2. \(\frac{\mathrm{I}_0}{\sqrt{2}}\)
3. \(\frac{I_0}{\sqrt{3}}\)
4. \(\frac{I_0}{3}\)

Answer: 1. \(\frac{\mathrm{I}_0}{\sqrt{2}}\)

Question 2. An AC voltage source of variable angular frequency  and fixed amplitude V connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When ω is increased :

1. The bulb glows dimmer
2. The bulb glows brighter
3. The total impedance of the circuit is unchanged
4. The total impedance of the circuit increases

Answer: 2. The bulb glows brighter

Question 3. Alternating current cannot be measured by D.C. ammeter because :

1. A.C. current pass through d.C. ammeter
2. A.C. change direction
3. The average value of current for the complete cycle is zero
4. D.C. ammeter will get damaged

Answer: 3. Average value of current for the complete cycle is zero

Question 4. The overall efficiency of a transformer is 90%. The transformer is rated for an output of 9000 watts. The primary voltage is 1000 volts. The ratio of turns in the primary to the secondary coil is 5: 1. The iron losses at full load are 700 watts. The primary coil has a resistance of 1 ohm.

1. The voltage in the secondary coil is :

1. 1000 volt
2. 5000 volt
3. 200 volt
4. zero volt

Answer: 3. 200 volt

2. In the above, the current in the primary coil is :

1. 9 amp
2. 10 amp
3. 1 amp
4. 4.5 amp

Answer: 2. 10 amp

3. In the above, the copper loss in the primary coil is :

1. 100 watt
2. 700 watt
3. 200 watt
4. 1000 watt

Answer: 1. 100 watt

4 . In the above, the copper loss in the secondary coil is :

1. 100 watt
2. 700 watt
3. 200 watt
4. 1000 watt

Answer: 3. 200 watt

5 . In the above, the current in the secondary coil is :

1. 45 amp
2. 46 amp
3. 10 amp
4. 50 amp

Answer: 2. 46 amp

6. In the above, the resistance of the secondary coil is approximately :

1. 0.01 Ω
2. 0.1 Ω
3. 0.2 Ω
4. 0.4 Ω

Answer: 2. 0.1 Ω

Alternating Current Exercise – 3

Question 1. The power dissipated in an LCR series circuit connected to an a.c.source of emf is :

1. \(\varepsilon^2 \mathrm{R} /\left\{\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right\}\)
2. \(\varepsilon^2 R / \sqrt{\left[R^2+\left(L \omega-\frac{1}{C \omega}\right)^2\right]}\)
3. \(\varepsilon^2 / \sqrt{\left.R^2+\left(L \omega-\frac{1}{C}\right)^2\right]} R\)
4. \(\frac{\varepsilon^2\left[R^2+\left(L \omega-\frac{1}{C \omega}\right)^2\right]}{R}\)

Answer: 2. \(\varepsilon^2 R / \sqrt{\left[R^2+\left(L \omega-\frac{1}{C \omega}\right)^2\right]}\)

Question 2. The r.m.s. value of potential difference V shown in the figure is :

1. √0
2. √0/ √2
3. √0 /2
4. √0 /√3

Answer: 2. √0/ √2

Question 3. A coil has a resistance of 30 ohm and inductive reactance of 20 Ohm at 50 Hz frequency. If an ac source, of 200volts, 100 Hz, is connected across the coil, the current in the coil will be :

1. 4.0 A
2. 8.0 A
3. \(\frac{20}{\sqrt{13}} A\)
4. 2.0 A

Answer: 1. 4.0 A

Question 4. In an electrical circuit R, L, C, and an a.c. voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and the current in the circuit is π/3. If instead, C is removed from the circuit, the phase difference is again π/3. The power factor of the circuit is :

1. 1/2
2. 1/√2
3. 1
4. v3/2

Answer: 3. 1

Question 5. The instantaneous values of alternating current and voltages in a circuit are given as

• \(i=\frac{1}{\sqrt{2}} \sin (100 \pi t) \text { amper }\)
• \( e=\frac{1}{\sqrt{2}} \sin (100 \pi t+\pi / 3) \text { Volt }\)

The average power in Watts consumed in the circuit is :

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

Answer: 4. \(\frac{1}{8}\)

Question 6. A coil of self-inductance L is connected in series with a bulb B and an AC source. The brightness of the bulb decreases when :

1. number of turns in the coil is reduced.
2. a capacitance of reactance X_C = X_L is included in the same circuit
3. an iron rod is inserted into the coil
4. frequency of the AC source is decreased

Answer: 3. An iron rod is inserted into the coil

Question 7. A resistance ‘R’ draws power ‘P’ when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes ‘Z’, the power drawn will be:

1. \(P \sqrt{\frac{R}{Z}}\)
2. \(\mathrm{P}\left(\frac{\mathrm{R}}{\mathrm{Z}}\right)\)
3. P
4. \(P\left(\frac{\mathrm{R}}{\mathrm{Z}}\right)^2\)

Answer: 4. \(P\left(\frac{\mathrm{R}}{\mathrm{Z}}\right)^2\)

Question 8. An inductor of 20 mH, a capacitor of 50 μ F, and a resistor. 40Ω are connected in series across a source of emf V = 10 sin 340t. The power loss in the A.C. circuit is

1. 0.89 W
2. 0.51 W
3. 0.67W
4. 0.76W

Answer: 2. 0.51 W

Question 9. A small signal voltage V(t) = V₀ sinωt is applied across an ideal capacitor C :

1. Current I(t), leads voltage V(t) by 180°
2. Current I(t), lags voltage V(t) by 90°
3. Over a full cycle, the capacitor C does not consume any energy from the voltage source.
4. Current I(t) is in phase with voltage V(t)

Answer: 3. Over a full cycle the capacitor C does not consume any energy from the voltage source.

Question 10. Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication?

1. R = 25 ω, L = 1.5 H, C = 45 μF
2. R = 20 ω, L = 1.5 H, C = 35 μF
3. R = 25 ω, L = 2.5 H, C = 45 μF
4. R = 15 ω, L = 3.5 H, C = 30 μF

Answer: 4. R = 15 ω, L = 3.5 H, C = 30 μF

Question 11. The potential differences across the resistance, capacitance,e, and inductance are 80 V, 4, V, and 100 V respectively in an L-C-R circuit. The power factor of this circuit is

1. 1.0
2. 0.4
3. 0.5
4. 0.8

Answer: 4. 0.8

Question 12. A 100 Ω resistance and a capacitor of 100 Ω reactance are connected in series across a 220 V source. When the capacitor is 50% charged, the peak value of the displacement current is

1. 11√2A
2. 2.2 A
3. 11 A
4. 4.4 A

Answer: 2. 2.2 A

Question 13. An inductor of 20 mH a capacitor of 100 μF and a resistor of 50 Ω are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is :

1. 0.79 W
2. 1.13 W
3. 2.74 W
4. 0.43 W

Answer: 1. 0.79 W

Question 14. A circuit when connected to an AC source of 12 V gives a current of 0.2A. The same circuit when connected to a DC source of 12 V, gives a current of 0.4 A. The circuit is :

1. series LR
2. series RC
3. series LC
4. series LCR

Answer: 1. series RC

Question 15. A 40 μF capacitor is connected to a 200V-50 Hz AC supply. The rms value of the current in the circuit is, nearly:

1. 25.1A
2. 1.7A
3. 2.05A
4. 2.5A

Answer: 4. 2.5A

Question 16. A series LCR circuit is connected to an AC voltage source. When L is removed from the circuit, the phase difference between current and voltage is \(\frac{\pi}{3}\). If instead C is removed from the circuit, the phase difference is again \(\frac{\pi}{3}\) between current and voltage. The power factor of the circuit is

1. -10
2. Zero
3. 0.5
4. 1.0

Answer: 4. 1.0

Question 17. A capacitor of capacitance ‘C’, is connected across an AC source of voltage V, given by The displacement current between the plates of the capacitor, which would then be given by

1. \(I_d=\frac{V_0}{\omega C} \cos \omega t\)
2. \(\mathrm{I}_{\mathrm{d}}=\frac{\mathrm{V}_0}{\omega \mathrm{C}} \sin \boldsymbol{\omega}t\)
3. \(\mathrm{I}_{\mathrm{d}}=\mathrm{V}_0 \omega \mathrm{sin} \omega \mathrm{t}\)
4. \(I_d=V_0 \omega C \cos \omega t\)

Answer: 4. \(I_d=V_0 \omega C \cos \omega t\)

Question 18. An inductor of inductance L, a capacitor of capacitance C, and a resistor of resistance ‘R’ are connected in series to an AC source of potential difference ‘V’ volts as shown in the figure. The potential difference across L, C, and R is 40 V, 10 V, and 40 V, respectively. The amplitude of the current flowing through LCR series circuit 10 2 A is. The impedance of the circuit is

1. 5/√2Ω
2. 4 Ω
3. 5 Ω
4. 4√2Ω

Answer: 3. 5 Ω

Question 19. A series LCR circuit containing a 5.0 H inductor, capacitor,r, and resistor is connected to a  230 V variable frequency ACCc source. The angular frequencies of the source at which power is transferred to the circuitries half the power at the resonant angular frequency are likely to be :

1. 50 rad/s and 25 rad/s
2. 46 rad/s and 54 rad/s
3. 42 rad/s and 58 rad/s
4. 25 rad/s and 75 rad/s

Answer: 2. 46 rad/s and 54 rad/s

Question 20. A step-down transformer connected to an AC mains supply of 220 V is made to operate at 11 V, 44 W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit?

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

Answer: 4. 0.2 A

Question 21. In a series LCR circuit R = 200 Ω and the voltage and the frequency of the main supply are 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30º. On taking out the inductor from the circuit the current leads the voltage by 30º. The power dissipated in the LCR circuit is

1. 305 W
2. 210 W
3. Zero W
4. 242 W

Answer: 4. 242 W

Question 22. An arc lamp requires a direct current of 10 A at 80 V to function. if it is connected to a 220 V(rms), 50 Hz AC supply, the series inductor needed for it to work is close to :

1. 0.08 H
2. 0.044 H
3. 0.065 H
4. 80 H

Answer: 3. 0.065 H

Question 23. For an RLC circuit driven with a voltage of amplitude and frequency \(\ omega_=\frac {1}{\sqrt{L C}}\) the current exhibits resonance. The quality factor, Q is given by :

1. \(\frac{\mathrm{R}}{\left(\omega_0 \mathrm{C}\right)}\)
2. \(\frac{\mathrm{CR}}{\omega_0}\)
3. \(\frac{\omega_0 L}{R}\)
4. \(\frac{\omega_0 R}{L}\)

Answer: 3. \(\frac{\omega_0 L}{R}\)

Question 24. In an a.c circuit, the instantaneous e.m.f and current are given by e = 100 sin 30t i = \(20 \sin \left(30 t-\frac{\pi}{4}\right)\) In one cycle of a.c the average power consumed by the circuit and the wattless current are, respectively:

1. \(\frac{50}{\sqrt{2}}, 0\)
2. 50,0
3. 50,10
4. \(\frac{1000}{\sqrt{2}}, 10\)

Answer: 4. \(\frac{1000}{\sqrt{2}}, 10\)

Question 25. A power transmission line feeds input power at 2300 V to a step-down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90%, the output current would be :

1. 45 A
2. 25 A
3. 35 A
4. 50 A

Answer: 1. 45 A

Question 26. A series AC circuit containing an inductor (20 mH), a capacitor (120 μF), and a resistor (60Ω) is driven by an AC source of 50Hz 24 V. The energy dissipated in the circuit in 60 s is :

1. 3.39 × 103 J
2. 5.65 × 102 J
3. 5.17 × 102 J
4. 2.26 × 103 J

Answer: 3. 5.17 × 102 J

Question 27. In the circuit shown the switch S₁ is closed at time t = 0 and the switch S₂ is kept open. At some later time (t₀), the switch S₁ is opened and S₂ is closed. The behavior of the current I as a function of time ‘t’ given by :

Answer: (Bonus), the correct graph will be

Question 28. In the above circuit\(C=\frac{\sqrt{3}}{2} m F, R_2=20 \Omega L=\frac{\sqrt{3}}{10} H\) L = and R₁ = 10Ω. Currently in the L-R₁ path is I₁ and in the CV 200 2sin(100t)= R₂ path it is I₂. The voltage of the A.C. source is given by V= 200√2sin (100t) volts, the phase difference between I₁ and I₂ is :

1. 60º
2. 0º
3. 30º
4. 90º

Answer: 4. 90º