JEE Physics Alternating Current MCQs Set 10

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Multiple Choice Questions

Question 1: A 120 V, 60 Hz supply feeds the primary of a step-up transformer with a 1 : 8 turn ratio. The load in the secondary is \( 10^4\ \Omega \). The current in the secondary is:
(a) 96 mA
(b) 9.6 A
(c) 96 A
(d) 0.96 A
Show Answer & Explanation

Answer: (a) 96 mA

Explanation:
1. \( V_2=8\times120=960\ \text{V} \).
2. \( I_2=960/10^4=0.096\ \text{A}=96\ \text{mA} \).

Teacher's Note:
1. Secondary voltage first, then Ohm's law.
2. Do not confuse mA with A.

Question 2: For the lamp of the previous problem (20 V, 5 W, on 200 V mains), a pure resistor is used in place of the reactive element. Its value, and the more economical choice, are:
(a) 180 ohm; the capacitor only
(b) 720 ohm; the inductor (or capacitor), as it consumes no power
(c) 720 ohm; the resistor, as it is cheaper
(d) 800 ohm; the inductor
Show Answer & Explanation

Answer: (b) 720 ohm; the inductor (or capacitor), as it consumes no power

Explanation:
1. Required total resistance \( =200/0.25=800\ \Omega \), so the extra resistor is \( 800-80=720\ \Omega \).
2. The resistor dissipates \( I^2R=0.0625\times720=45\ \text{W} \) as heat, whereas a pure inductor or capacitor dissipates none, so it is more economical.

Teacher's Note:
1. Reactive elements drop voltage without dissipating power.
2. A series resistor wastes energy as heat.

Question 3: In an ac circuit the current is \( i=i_0\sin\omega t \). The time taken for the current to change from its maximum value to the rms value is:
(a) \( \dfrac{\pi}{2\omega} \)
(b) \( \dfrac{\pi}{\omega} \)
(c) \( \dfrac{\pi}{4\omega} \)
(d) \( \dfrac{\pi}{8\omega} \)
Show Answer & Explanation

Answer: (c) \( \dfrac{\pi}{4\omega} \)

Explanation:
1. Maximum is at \( \omega t=\pi/2 \); rms value \( i_0/\sqrt2 \) occurs at \( \omega t=\pi/4 \).
2. Time taken \( =\frac{\pi/2-\pi/4}{\omega}=\frac\pi{4\omega} \) (that is T/8).

Teacher's Note:
1. rms value corresponds to 45 degrees.
2. Quarter of a quarter-cycle is T/8.

Question 4: A transformer has 50 turns in the primary and 100 turns in the secondary. If the primary is connected to a 220 V dc supply, the voltage across the secondary is:
(a) 440 V
(b) 220 V
(c) 110 V
(d) Zero
Show Answer & Explanation

Answer: (d) Zero

Explanation:
1. A transformer works on mutual induction, which needs a changing flux.
2. Direct current produces a constant flux, so no emf is induced in the secondary.

Teacher's Note:
1. Transformers cannot step dc voltages.
2. Constant flux means zero induced emf.

Question 5: An electric bulb and a capacitor are in series with an ac source. On increasing the source frequency, the brightness of the bulb:
(a) remains unchanged
(b) increases
(c) decreases
(d) sometimes increases and sometimes decreases
Show Answer & Explanation

Answer: (b) increases

Explanation:
1. \( X_C=\frac1{\omega C} \) decreases as the frequency rises.
2. So the impedance falls, the current rises and the bulb glows brighter.

Teacher's Note:
1. A capacitor opposes low frequencies more.
2. Less impedance means more current.

Question 6: In a series LCR circuit with a 50 V source, R = 300 ohm and frequency \( 50/\pi \) Hz, the rms current is 0.10 A, and the average electric and magnetic field energies are 25 mJ and 5 mJ. The capacitance, the inductance and the sum of the rms voltages across the three elements are:
(a) 20 microfarad, 1 H and 50 V
(b) 40 microfarad, 0.5 H and 90 V
(c) 10 microfarad, 2 H and 120 V
(d) 20 microfarad, 1 H and 90 V
Show Answer & Explanation

Answer: (d) 20 microfarad, 1 H and 90 V

Explanation:
1. \( \omega=100;\ \ U_B=\tfrac12LI^2=5\ \text{mJ}\Rightarrow L=1\ \text{H} \).
2. \( X_L=100\ \Omega;\ Z=50/0.1=500\Rightarrow X_C=500\ \Omega\Rightarrow C=\frac1{100\times500}=20\ \mu\text{F} \).
3. \( V_R=30,\ V_L=10,\ V_C=50\Rightarrow\text{sum}=90\ \text{V} \).

Teacher's Note:
1. The sum of rms drops is an arithmetic sum, not the source voltage.
2. Use U_B to find L, then Z to find X_C.

Question 7: If the frequency of the source emf in an ac circuit is n, the instantaneous power varies with frequency:
(a) n
(b) n/2
(c) zero
(d) 2n
Show Answer & Explanation

Answer: (d) 2n

Explanation:
1. Power \( p=e\,i \) contains products like \( \sin\omega t\sin(\omega t-\phi) \), which contain a term in \( \cos(2\omega t-\phi) \).

Teacher's Note:
1. Power oscillates at twice the supply frequency.
2. Its average is the constant part.

Question 8: In a step-up transformer the primary voltage is 220 V and the primary current is 5 A. The secondary voltage is 22000 V. The current in the secondary (neglecting losses) is:
(a) 5 A
(b) 50 A
(c) 500 A
(d) 0.05 A
Show Answer & Explanation

Answer: (d) 0.05 A

Explanation:
1. Input power = output power: \( 220\times5=22000\times I_2\Rightarrow I_2=0.05\ \text{A} \).

Teacher's Note:
1. Step-up in voltage means step-down in current.
2. Power is conserved.

Question 9: An ac voltage source \( V=200\sqrt2\sin100t \) is connected across an ac ammeter (reading rms) in series with a 1 microfarad capacitor. The ammeter reads:
(a) 40 mA
(b) 20 mA
(c) 80 mA
(d) 10 mA
Show Answer & Explanation

Answer: (b) 20 mA

Explanation:
1. \( X_C=\frac1{100\times10^{-6}}=10^4\ \Omega \).
2. \( I_{rms}=200/10^4=0.02\ \text{A}=20\ \text{mA} \).

Teacher's Note:
1. The rms voltage is 200 V.
2. Use X_C = 1/(omega C).

Question 10: A capacitor is a perfect insulator for:
(a) alternating current
(b) variable direct current
(c) constant direct current
(d) direct as well as alternating current
Show Answer & Explanation

Answer: (c) constant direct current

Explanation:
1. Steady dc cannot pass through the dielectric once the capacitor is charged.
2. Ac, and varying dc, repeatedly charge and discharge it.

Teacher's Note:
1. X_C is infinite at zero frequency.
2. Only steady dc is blocked.

Question 11: In an ac circuit a resistance R is in series with an inductance L. If the phase angle between voltage and current is 45 degrees, the inductive reactance is:
(a) R
(b) cannot be found with the given data
(c) R/2
(d) R/4
Show Answer & Explanation

Answer: (a) R

Explanation:
1. \( \tan\phi=X_L/R=\tan45^\circ=1 \), so \( X_L=R \).

Teacher's Note:
1. tan 45 degrees = 1.
2. Reactance equals resistance.

Question 12: Three parallel branches are connected to a 100 V (rms) source. Branch 1: 4 ohm resistor in series with an inductive reactance of 3 ohm. Branch 2: 6 ohm resistor in series with a capacitive reactance of 8 ohm. Branch 3: 10 ohm resistor in series with an inductive reactance of 10 ohm and a capacitive reactance of 10 ohm. The currents in the three branches are:
(a) 20 A, 10 A and 5 A
(b) 20 A, 10 A and 10 A
(c) 25 A, 14 A and 10 A
(d) 33 A, 10 A and 10 A
Show Answer & Explanation

Answer: (b) 20 A, 10 A and 10 A

Explanation:
1. \( Z_1=\sqrt{4^2+3^2}=5,\ Z_2=\sqrt{6^2+8^2}=10,\ Z_3=\sqrt{10^2+(10-10)^2}=10\ \Omega \).
2. Currents: 100/5 = 20 A, 100/10 = 10 A, 100/10 = 10 A.

Teacher's Note:
1. In parallel each branch sees the full voltage.
2. Branch 3 is at resonance.

Question 13: The potential difference across an instrument is \( V=5\cos\omega t \) V and the current is \( I=2\sin\omega t \) A. The power dissipated is:
(a) 5 W
(b) Zero
(c) 10 W
(d) 2.5 W
Show Answer & Explanation

Answer: (b) Zero

Explanation:
1. Since \( \sin\omega t=\cos(\omega t-90^\circ) \), V and I differ in phase by 90 degrees.
2. \( P=\tfrac12(5)(2)\cos90^\circ=0 \).

Teacher's Note:
1. A 90 degree phase difference means a wattless current.
2. The instrument is purely reactive.

Question 14: When 100 V dc is applied across a solenoid, a steady current of 1 A flows. When 100 V ac is applied, the current drops to 0.5 A. If the ac frequency is \( 150/\sqrt3\,\pi \) Hz, the impedance and inductance of the solenoid are:
(a) \( 1100\ \Omega\text{ and }\tfrac3{117}\ \text{H} \)
(b) \( 200\ \Omega\text{ and }\tfrac1\pi\ \text{H} \)
(c) \( 200\ \Omega\text{ and }1.0\ \text{H} \)
(d) \( 100\ \Omega\text{ and }\tfrac1{16}\ \text{H} \)
Show Answer & Explanation

Answer: (b) \( 200\ \Omega\text{ and }\tfrac1\pi\ \text{H} \)

Explanation:
1. From dc: \( R=100\ \Omega \). From ac: \( Z=100/0.5=200\ \Omega \).
2. \( X_L=\sqrt{200^2-100^2}=100\sqrt3\ \Omega,\ \omega=2\pi f=\frac{300\pi}{\sqrt3} \).
3. \( L=\frac{100\sqrt3\cdot\sqrt3}{300\pi}=\frac1\pi\ \text{H} \).

Teacher's Note:
1. The dc current gives only the resistance.
2. Then X_L follows from Z and R.

Question 15: A 100 V ac source of angular frequency 500 rad/s is connected to an LCR circuit with L = 0.8 H, C = 5 microfarad and R = 10 ohm in series. The potential difference across the resistance is:
(a) \( 50\sqrt3\ \text{V} \)
(b) \( 100/\sqrt2\ \text{V} \)
(c) 50 V
(d) 100 V
Show Answer & Explanation

Answer: (d) 100 V

Explanation:
1. \( X_L=500\times0.8=400\ \Omega,\ X_C=\frac1{500\times5\times10^{-6}}=400\ \Omega \).
2. The circuit is at resonance, so \( Z=R \) and the whole source voltage appears across R.

Teacher's Note:
1. Check whether X_L equals X_C before computing anything else.
2. At resonance V_R equals the source voltage.

Question 16: A 55 W, 110 V bulb is run from a 220 V, 50 Hz line through a choke coil in series. The inductance needed so that the bulb gets its rated voltage is:
(a) 0.70 H
(b) 2.43 H
(c) 1.40 H
(d) 1.21 H
Show Answer & Explanation

Answer: (d) 1.21 H

Explanation:
1. Bulb current \( I=55/110=0.5\ \text{A} \), resistance \( R=110/0.5=220\ \Omega \).
2. Total impedance \( Z=220/0.5=440\ \Omega \), so \( X_L=\sqrt{440^2-220^2}=381\ \Omega \).
3. \( L=X_L/\omega=381/(100\pi)=1.21\ \text{H} \).

Teacher's Note:
1. The bulb needs 0.5 A, so the total impedance must be 440 ohm.
2. Voltages across R and L add in quadrature, not directly.

Question 17: A bulb designed to operate on 12 V direct current gives the same brightness when connected to an ac source. The peak voltage of that source is:
(a) 12 V
(b) 24 V
(c) 16.9 V
(d) 8.5 V
Show Answer & Explanation

Answer: (c) 16.9 V

Explanation:
1. Equal brightness means equal power, so the rms value of the ac must be 12 V.
2. \( V_0=\sqrt2\times12=16.9\ \text{V} \).

Teacher's Note:
1. The dc voltage equals the rms value of the equivalent ac.
2. Peak = sqrt2 times rms.

Question 18: For the series circuit with R = 300 ohm, C = 20 microfarad, L = 1.0 H and a 50 V rms source of angular frequency 100 rad/s, the average electric field energy in the capacitor and the average magnetic field energy in the inductor are:
(a) 25 mJ and 5 mJ
(b) 12.5 mJ and 2.5 mJ
(c) 50 mJ and 10 mJ
(d) 5 mJ and 25 mJ
Show Answer & Explanation

Answer: (a) 25 mJ and 5 mJ

Explanation:
1. Current 0.10 A rms and \( V_C=50\ \text{V} \) rms (from the previous analysis).
2. \( U_E=\tfrac12CV_{C,rms}^2=\tfrac12(20\times10^{-6})(2500)=25\ \text{mJ} \).
3. \( U_B=\tfrac12LI_{rms}^2=\tfrac12(1)(0.01)=5\ \text{mJ} \).

Teacher's Note:
1. Average stored energy uses rms values, so the factor 1/2 stays.
2. Because X_C is larger than X_L here, the capacitor stores more energy.

Question 19: The maximum values of the alternating voltage and current in a circuit on a 50 Hz supply are 400 V and 20 A. At an instant when both are increasing positively, the voltage is \( 200\sqrt2 \) V and the current is 10 A. The average power consumed is:
(a) 3464 W
(b) 3864 W
(c) 4000 W
(d) 2000 W
Show Answer & Explanation

Answer: (b) 3864 W

Explanation:
1. \( \sin\theta_v=\frac{200\sqrt2}{400}\Rightarrow\theta_v=45^\circ;\ \sin\theta_i=\frac{10}{20}\Rightarrow\theta_i=30^\circ \).
2. Phase difference \( \phi=15^\circ \), so \( P=\tfrac12(400)(20)\cos15^\circ=3864\ \text{W} \).

Teacher's Note:
1. Both rising means the phase angles are in the first quadrant.
2. Use the phase difference in cos phi.

Question 20: A 20 V, 5 W lamp is used on 200 V ac mains of frequency \( \frac{50}{\pi}\sqrt{11} \) Hz. The inductance to be put in series with the lamp so that it works at its rated voltage is:
(a) 1.2 H
(b) 2.4 H
(c) 24 H
(d) 0.24 H
Show Answer & Explanation

Answer: (b) 2.4 H

Explanation:
1. Lamp current \( I=5/20=0.25\ \text{A} \), \( R=80\ \Omega \); total \( Z=200/0.25=800\ \Omega \).
2. \( X_L=\sqrt{800^2-80^2}=240\sqrt{11}\ \Omega,\ \omega=100\sqrt{11} \), so \( L=X_L/\omega=2.4\ \text{H} \).

Teacher's Note:
1. The lamp must carry its rated current of 0.25 A.
2. sqrt(800^2 - 80^2) = 80 sqrt(99) = 240 sqrt(11).

Question 21: An electromagnetic wave of wavelength 300 m is to be received using a 2.5 microfarad capacitor in a resonant circuit. Using \( \pi^2=10 \), the inductance required is:
(a) \( 2.5\times10^{-9}\ \text{H} \)
(b) \( 1\times10^{-8}\ \text{H} \)
(c) \( 1\times10^{-6}\ \text{H} \)
(d) \( 4\times10^{-8}\ \text{H} \)
Show Answer & Explanation

Answer: (b) \( 1\times10^{-8}\ \text{H} \)

Explanation:
1. \( f=c/\lambda=3\times10^8/300=10^6\ \text{Hz} \).
2. \( L=\frac{1}{4\pi^2f^2C}=\frac{1}{40\times10^{12}\times2.5\times10^{-6}}=10^{-8}\ \text{H} \).

Teacher's Note:
1. Resonance condition: 2 pi f = 1/sqrt(LC).
2. Use c = 3 x 10^8 m/s.

Question 22: The average power delivered to a series ac circuit is given by (symbols have their usual meaning):
(a) \( E_{rms}I_{rms}\sin\phi \)
(b) \( E_{rms}I_{rms} \)
(c) Zero
(d) \( E_{rms}I_{rms}\cos\phi \)
Show Answer & Explanation

Answer: (d) \( E_{rms}I_{rms}\cos\phi \)

Explanation:
1. Average of \( e\,i \) over a cycle gives \( E_{rms}I_{rms}\cos\phi \).

Teacher's Note:
1. cos phi is the power factor.
2. It is 1 for a resistor and 0 for a pure L or C.

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