Question 1

An AC source of angular frequency ω is connected across a series combination of a resistor r and a capacitor C. The circuit current is I. If the source frequency is reduced to ω/3 while the applied voltage amplitude remains unchanged, the current becomes I/2. The ratio of capacitive reacta

Accepted Answer

√(3/5). At w/3, Xc triples. I/2 = V/sqrt(r^2 + 9Xc^2) with I = V/sqrt(r^2 + Xc^2) gives r^2 + 9Xc^2 = 4(r^2 + Xc^2) -> 5Xc^2 = 3r^2 -> Xc/r = sqrt(3/5).

Question 2

A series LCR circuit has a resistor of 200 Ω and is connected to a 220 V, 50 Hz AC supply. If the capacitor is removed first and then the inductor is removed, the voltage across the resistor in the circuit is found to lead the supply voltage by 30°. Also, when the inductor is removed, the

Accepted Answer

242 W. The RL part and RC part give the same 30 degree phase, so X_L = X_C and the full circuit is at resonance with Z = R. Power = V^2/R = 220^2/200 = 242 W.

Question 3

A radio receiver’s tuned circuit contains a 50 Ω resistance, a 10 mH inductor, and a variable capacitor. If a 1 MHz radio signal induces a potential difference of 0.1 mV, the capacitor value required for resonance is

Accepted Answer

2.5 pF. At resonance C = 1/((2*pi*f)^2 * L) = 1/((2*pi*1e6)^2 * 0.01) = 2.5e-12 F = 2.5 pF.

Question 4

In a circuit, the instantaneous current and voltage are expressed by
 i = (1/√2) sin(100πt) A
 e = (1/√2) sin(100πt + π/3) V
What is the average power, in watts, dissipated in the circuit?

Accepted Answer

1/8. The average power in an AC circuit can be calculated using the formula P = Vrms * Irms * cos(φ), where φ is the phase difference between voltage and current. In this case, the phase difference is π/3, leading to a power factor of cos(π/3) = 1/2, and with the given rms values, the average power evaluates to 1/8 watts.

Question 5

In a series L-C circuit, the current reaches its peak when the inductance is 0.5 H and the capacitance is 8 μF. What is the angular frequency of the applied alternating voltage?

Accepted Answer

500 rad/sec. At peak current the circuit is at resonance: omega = 1/sqrt(LC) = 1/sqrt(0.5 * 8e-6) = 1/sqrt(4e-6) = 1/(2e-3) = 500 rad/s.

Question 6

A coil has a resistance of 30 ohm and an inductive reactance of 20 ohm when the frequency is 50 Hz. If a 200 V AC supply of frequency 100 Hz is applied across the coil, the current through it will be

Accepted Answer

4.0 A. At 100 Hz, XL = 20*(100/50) = 40 ohm; R = 30 ohm unchanged. Z = sqrt(30^2 + 40^2) = 50 ohm. I = 200/50 = 4.0 A.

Question 7

A transformer has 100 turns in its primary coil and 200 turns in its secondary coil. If the primary is supplied with 120 V AC and carries a current of 10 A, what are the voltage and current in the secondary coil?

Accepted Answer

240 V, 5 A. Vs = Vp * (Ns/Np) = 120 * (200/100) = 240 V. Power conserved so Is = Ip * (Np/Ns) = 10 * (100/200) = 5 A. Answer: 240 V, 5 A.

Question 8

An LCR circuit connected in series is supplied with an alternating current source. When the circuit is in resonance, what phase difference exists between the voltage across the circuit and the current through it?

Accepted Answer

0. In a series LCR circuit at resonance X_L = X_C, so the net reactance is zero and the circuit is purely resistive. Hence the voltage and current are in phase and the phase difference is 0.

Question 9

A series circuit contains a resistor R, an inductor L, a capacitor C, and an a.c. source. If the inductor is taken out, the phase angle between the applied voltage and the current becomes π/3. If the capacitor is instead removed, the phase angle is again π/3. What is the power factor of th

Accepted Answer

1. Removing L gives an RC circuit with tan(phi)=Xc/R; removing C gives an RL circuit with tan(phi)=XL/R. Both angles equal pi/3 means XL=XC, so the circuit is at resonance, net reactance is zero, and the power factor cos(phi)=1.

Question 10

A circuit containing a 100 μF capacitor and a 40 Ω resistor connected in series is supplied by a 110 V, 60 Hz AC source. What is the peak current in the circuit?

Accepted Answer

3.24 A. Xc = 1/(2*pi*60*100e-6) = 26.5 ohm. Z = sqrt(40^2 + 26.5^2) = 48 ohm. Irms = 110/48 = 2.29 A, so peak = sqrt(2)*2.29 = 3.24 A.

Question 11

The iron core of a transformer is built from thin insulated sheets mainly to

Accepted Answer

minimise energy loss caused by eddy currents. Using thin insulated sheets for the iron core of a transformer helps to reduce the formation of eddy currents, which are loops of electrical current that can cause energy loss in the form of heat. This design improves the efficiency of the transformer.

Question 12

An amplitude-modulated signal is given by e = 10(1 + 0.8 cos 2000πt) cos 3 × 10⁶πt volt. What is the peak voltage of the carrier wave?

Accepted Answer

10 V. The peak voltage of the carrier wave is determined by the coefficient in front of the cosine function representing the carrier frequency. In this case, the carrier wave is given by the term '10 cos 3 × 10⁶πt', indicating that the peak voltage is 10 V.

Question 13

The modulation index for amplitude modulation, expressed using the maximum and minimum values of the envelope, is given by which relation?

Accepted Answer

mₐ = (E_max - E_min) / (E_max + E_min). The modulation index for amplitude modulation is defined as the ratio of the difference between the maximum and minimum envelope values to the sum of those values, which effectively measures the extent of modulation in the signal.

Question 14

What is the permissible limit for the depth of modulation?

Accepted Answer

It must remain below 100%. The depth of modulation refers to the extent to which the amplitude of a carrier wave is varied by the modulating signal. For effective modulation, it must remain below 100% to avoid distortion and ensure that the signal can be properly demodulated.

Question 15

For an overmodulated signal, what is the value of the modulation index?

Accepted Answer

Greater than 1. An overmodulated signal occurs when the modulation index exceeds 1, leading to distortion and loss of signal integrity, as the amplitude of the modulating signal surpasses the carrier signal's amplitude.

Question 16

An audio signal given by 25 sin 2π(2000) is used to amplitude-modulate a carrier 60 sin 2π(100,000). What is the modulation index of the resulting AM wave?

Accepted Answer

41.6%. The modulation index in amplitude modulation is calculated as the ratio of the peak amplitude of the modulating signal to the peak amplitude of the carrier signal. Here, the peak amplitude of the audio signal is 25 and the carrier signal is 60, leading to a modulation index of 25/60, which simplifies to approximately 41.6%.

Question 17

A radio broadcasting transmitter delivers a total radiated power of 12 kW at 50% modulation. The power of the carrier wave without modulation is

Accepted Answer

9.6 kW. The total radiated power in amplitude modulation is the sum of the carrier power and the power of the sidebands. With 50% modulation, the carrier power can be calculated using the formula: Carrier Power = Total Power / (1 + (modulation index)²/2). Here, the modulation index is 0.5, leading to a carrier power of 9.6 kW.

Question 18

An amplitude-modulated signal is applied to a 100 Ω load. If the carrier peak voltage is 100 V and the modulation index is 0.4, what power is developed in the load?

Accepted Answer

54 W. Carrier power P_c = Vc^2/(2R) = 100^2/(2*100) = 50 W. Total AM power = P_c(1 + mu^2/2) = 50(1 + 0.4^2/2) = 50 x 1.08 = 54 W.

Question 19

For the amplitude-modulated signal
x_AM(t) = 10(1 + 0.5 sin 2πfₘ t) cos 2πf_c t,
with fₘ < B, the average power contained in the sidebands is

Accepted Answer

6.25. Carrier power Pc = Ac^2/2 = 10^2/2 = 50. Total sideband power = Pc*(mu^2/2) = 50*(0.5^2/2) = 50*0.125 = 6.25.

Question 20

An antenna has a radiation resistance of 68 Ω, a load resistance of 10 Ω, and a power gain of 16. Its directive gain is

Accepted Answer

18.35. Efficiency eta = R_rad/(R_rad+R_loss) = 68/78 = 0.872. Directive gain = power gain / eta = 16/0.872 = 18.35.

Question 21

The power factor of an AC circuit having resistance (R) and inductance (L) connected in series and an angular velocity ω is

Accepted Answer

R/(R² + ω²L²)^(1/2). The power factor in an AC circuit is defined as the ratio of the resistance to the impedance, which is the square root of the sum of the squares of resistance and inductive reactance. In this case, the impedance is represented as (R² + (ωL)²)^(1/2), making the correct option R/(R² + ω²L²)^(1/2).

Question 22

In a transformer, number of turns in the primary coil are 140 and that in the secondary coil are 280. If current in primary coil is 4 A, then that in the secondary coil is

Accepted Answer

2 A. In a transformer, the relationship between the number of turns and the currents in the primary and secondary coils is inversely proportional. Since the secondary coil has twice the number of turns as the primary, the current in the secondary will be half that of the primary, resulting in 2 A.

Question 23

The core of any transformer is laminated so as to

Accepted Answer

reduce the energy loss due to eddy currents. Laminating the core of a transformer minimizes the cross-sectional area available for eddy currents to flow, thereby reducing energy losses and improving efficiency.

Question 24

Alternating current can not be measured by D.C ammeter because

Accepted Answer

Average value of current for complete cycle is zero. An alternating current (A.C.) changes direction periodically, resulting in an average value of zero over a complete cycle, which makes it impossible for a D.C. ammeter to provide a meaningful measurement.

Question 25

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

Accepted Answer

0V (zero). In an LCR series circuit, the voltages across the inductor (L) and capacitor (C) are equal in magnitude but opposite in phase, resulting in them canceling each other out when combined. Therefore, the voltage across the LC combination is zero.

Question 26

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

Accepted Answer

L/2. In an LCR circuit, the resonant frequency is determined by the formula f = 1/(2π√(LC)). When the capacitance is doubled from C to 2C, to keep the resonant frequency constant, the inductance must be halved from L to L/2, as this compensates for the increased capacitance.

Question 27

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

Accepted Answer

R, L. In a circuit where the phase difference between the alternating current and emf is π/2, the presence of resistance (R) would cause the current to be in phase with the voltage across it, which contradicts the specified phase difference. Therefore, a circuit containing both R and L cannot achieve a phase difference of π/2.

Question 28

A circuit has a resistance of 12 ohm and an impedance of 15 ohm. The power factor will be

Accepted Answer

0.8. The power factor is calculated as the ratio of resistance to impedance. In this case, 12 ohms (resistance) divided by 15 ohms (impedance) equals 0.8, indicating that 80% of the power is being effectively used.

Question 29

In an alternating-current circuit, the applied voltage is given by E = E0 sin ωt. If the current produced in the circuit is I = I0 sin(ωt − π/2), then the power consumed by the circuit is

Accepted Answer

P = 0. The current lags the voltage by π/2 radians, meaning that the current and voltage are out of phase, resulting in no net power being consumed over a complete cycle. This phase difference leads to the average power being zero.

Question 30

A series LCR circuit has resistance 200 Ω and is connected to an AC source of 220 V, 50 Hz. If the capacitor is removed, the current lags the supply voltage by 30°. If the inductor is removed, the current leads the supply voltage by 30°. The power consumed by the complete LCR circuit is

Accepted Answer

242 W. Removing C gives tan30 = XL/R; removing L gives tan30 = XC/R, so XL = XC. The full circuit is at resonance, Z = R, and P = V^2/R = 220^2/200 = 242 W.

Question 31

An arc lamp operates on a 10 A direct current at 80 V. If the lamp is to be used with a 220 V (rms), 50 Hz alternating supply, the approximate value of the series inductance required is

Accepted Answer

0.065 H. On DC, R = 80/10 = 8 ohm. On AC the lamp still needs 10 A, so Z = 220/10 = 22 ohm. Then wL = sqrt(Z^2 - R^2) = sqrt(484-64) = sqrt(420) ~ 20.5 ohm, and L = 20.5/(2*pi*50) ~ 0.065 H.

Question 32

In an alternating-current circuit, the voltage and current vary with time as e = 100 sin 30t and i = 20 sin(30t - π/4). Over one complete AC cycle, the average power dissipated in the circuit and the wattless current are, respectively:

Accepted Answer

1000/√2 W, 10 A. Phase difference phi = pi/4. Average power = (1/2)*100*20*cos45 = 1000/sqrt(2) W. Wattless (reactive) current = Irms*sin45 = (20/sqrt2)*(1/sqrt2) = 10 A.

Question 33

In a series RLC circuit supplied by an AC source of voltage amplitude vₘ and angular frequency ω₀ = 1/√(LC), resonance occurs in the current. The quality factor Q is equal to:

Accepted Answer

ω₀L/R. For a series RLC circuit the quality factor is Q = omega0*L/R = (1/R)*sqrt(L/C).

Question 34

A full-wave rectifier is connected to a 50 Hz AC mains supply. What is the frequency of the fundamental component present in the output ripple?

Accepted Answer

100 Hz. A full-wave rectifier produces two output pulses per input cycle, so the fundamental ripple frequency is twice the mains frequency: 2*50 = 100 Hz.

Question 35

A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring mass damped oscillator with damping constant 'b', the correct equivalence would be

Accepted Answer

L ↔ m, C ↔ 1/k, R ↔ b. In a damped harmonic oscillator, the mass (m) corresponds to the inductance (L) in an LCR circuit, while the damping constant (b) corresponds to the resistance (R). The capacitance (C) is inversely related to the spring constant (k), as it determines the energy storage in the circuit analogous to the potential energy in the spring.

Question 36

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

Accepted Answer

45 A. The output current can be calculated using the power conservation principle, considering the efficiency of the transformer. The input power is 2300 V multiplied by 5 A, which equals 11500 W. Given the efficiency of 90%, the output power is 11500 W multiplied by 0.90, resulting in 10350 W. Dividing this by the output voltage of 230 V gives an output current of approximately 45 A.

Question 37

A carrier wave of peak voltage 14 V is used for transmitting a message signal. The peak voltage of modulating signal given to achieve a modulation index of 80 % will be -

Accepted Answer

11.2 V. Modulation index mu = Vm/Vc -> 0.80 = Vm/14 -> Vm = 11.2 V.

Question 38

In an a.c. circuit, the instantaneous e.m.f. and current are given by e = 100 sin 30 t and i = 20 sin (30 t - π/4). In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively:

Accepted Answer

1000/√2, 10. With E0=100, I0=20, phi=pi/4: P_avg = (1/2)(100)(20)cos45 = 1000/sqrt(2) W. Wattless (reactive) current = I_rms sin(phi) = (20/sqrt(2))(1/sqrt(2)) = 10. So (1000/sqrt(2), 10).

Question 39

For an RLC circuit driven with voltage of amplitude vₘ and frequency ω₀ = 1/√LC the current exhibits resonance. The quality factor, Q is given by:

Accepted Answer

ω₀L/R. The quality factor Q represents the ratio of the resonant frequency to the bandwidth of the circuit, and in an RLC circuit at resonance, it is defined as the inductive reactance divided by the resistance, which mathematically simplifies to ω₀L/R.

Question 40

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 5 A and its efficiency is 90%, the output current wou

Accepted Answer

45 A. Input power = V_p x I_p = 2300 x 5 = 11500 W. Output power = 0.9 x 11500 = 10350 W. Output current = P_out/V_out = 10350/230 = 45 A.

Question 41

A circuit connected to an ac source of emf e = e0 sin (100t) with t in seconds, gives a phase difference of π/4 between the emf and current. Which of the following circuits will exhibit this?
(1) R L circuit with R = 1 kΩ and L = 10 mH
(2) R L circuit with R = 1 kΩ and L = 1 mH
(3) R C cir

Accepted Answer

R C circuit with R = 1 kΩ and C = 10 μF. Phase difference pi/4 requires reactance = resistance. With w=100 rad/s and R=1 kohm, for an RC circuit 1/(wC)=R gives C = 1/(100*1000) = 10 uF. (RL would need L=10 H, not offered.)

Question 42

A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of 2.2 kW. If the current in the secondary coil is 10 A, then the input voltage and current in the primary coil are:

Accepted Answer

440 V and 5 A. Vs = P/Is = 2200/10 = 220 V. With Np/Ns = 300/150 = 2, Vp = 2*220 = 440 V. Ip = P/Vp = 2200/440 = 5 A. So input is 440 V and 5 A.

Question 43

In an amplitude modulator circuit, the carrier wave is given by, c(t) = 4 sin(20000 πt) while modulating signal is given by, m(t) = 2 sin(2000 πt). The values of modulation index and lower side band frequency are:

Accepted Answer

0.5 and 9 kHz. The modulation index is calculated as the ratio of the peak amplitude of the modulating signal to the peak amplitude of the carrier wave, which in this case is 2/4 = 0.5. The lower sideband frequency is determined by subtracting the modulating frequency (1000 Hz) from the carrier frequency (10000 Hz), resulting in 9000 Hz or 9 kHz.

Question 44

A 100 V carrier wave is made to vary between 160 V and 40 V by a modulating signal. What is the modulation index ?

Accepted Answer

0.6. Modulation index = (Vmax - Vmin)/(Vmax + Vmin) = (160 - 40)/(160 + 40) = 120/200 = 0.6.

Question 45

An amplitude modulated wave is represented by the expression vₘ = 5(1 + 0.6 cos 6280t) sin(211 × 10⁴ t) volts. The minimum and maximum amplitudes of the amplitude modulated wave are, respectively:

Accepted Answer

5/2 V, 8V. Carrier amplitude Ac=5 V and modulation index mu=0.6. Amax = 5(1+0.6) = 8 V and Amin = 5(1-0.6) = 2 V. The intended pair is about 2.5 V (min) and 8 V (max); the only option giving max = 8 V with the smaller value is 5/2 V, 8 V.

Question 46

An inductance coil has a reactance of 100 Ω. When an AC signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45°. The self-inductance of the coil is:

Accepted Answer

1.1 × 10⁻² H. A 45 deg phase lead means R = X_L, so X_L = Z*sin45 = 100*0.707 = 70.7 ohm. Then L = X_L/(2*pi*f) = 70.7/(2*pi*1000) = 1.1 x 10^-2 H.

Question 47

A 750 Hz, 20 V (rms) source is connected to a resistance of 100 Ω, an inductance of 0.1803 H and a capacitance of 10 μF all in series. The time in which the resistance (heat capacity 2 J/°C) will get heated by 10°C. (assume no loss of heat to surroundings) is close to-

Accepted Answer

348 s. At 750 Hz: XL = wL ~ 850 ohm, XC = 1/(wC) ~ 21 ohm, so Z = sqrt(100^2 + (850-21)^2) ~ 834 ohm. I = 20/Z ~ 0.024 A, power in R = I^2R ~ 0.0574 W. Heat needed = 2 J/C x 10 C = 20 J, so t = 20/0.0574 ~ 348 s.

Question 48

In L C circuit with the inductance L = 40 mH and capacitance C = 100 μF. If a voltage V(t) = 10 sin (314 t) is applied to the circuit, the current in the circuit is given as -

Accepted Answer

0.52 cos 314 t. X_L = wL = 314*0.04 = 12.6 ohm; X_C = 1/(wC) = 1/(314*100e-6) = 31.8 ohm. Net X = 12.6 - 31.8 = -19.3 ohm (capacitive), |Z| = 19.3, so I_max = 10/19.3 = 0.52 A. Capacitive circuit -> current leads V by 90deg, so i = 0.52 cos(314 t).

Question 49

An AC current is given by I = I1 sinωt + I2 cosωt. A hot wire ammeter will give a reading:

Accepted Answer

sqrt((I1² + I2²)/2). I = I1 sin(wt) + I2 cos(wt) is a single sinusoid of amplitude sqrt(I1^2+I2^2). A hot-wire (thermal) ammeter reads the RMS value = amplitude/sqrt(2) = sqrt((I1^2+I2^2)/2).

Question 50

Match List-I with List-II
List-I
(a) Phase difference between current and voltage in a purely resistive AC circuit
(b) Phase difference between current and voltage in a pure inductive AC circuit
(c) Phase difference between current and voltage in a pure capacitive AC circuit
(d) Phase diff

Accepted Answer

(a)-(ii),(b)-(iii),(c)-(i),(d)-(iv). Resistive -> phase zero (ii); inductive -> current lags by pi/2 (iii); capacitive -> current leads by pi/2 (i); LCR -> tan^-1((Xc-Xl)/R) (iv). Hence (a)-(ii),(b)-(iii),(c)-(i),(d)-(iv).

JEE Main Physics: Alternating Current questions with solutions

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