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JEE • PHYSICS • ALTERNATING CURRENT

Alternating current practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 121-150 of 246 questions

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121 Medium
121 of 246
Application +4 | -1
A 16  μF16\,\,\mu F capacitor is charged to a 20  volt\mathrm{20}\,\,\mathrm{volt} potential. The battery is then disconnected and pure 40  mH40\,\,mH coil is connected across the capacitor so that LC\mathrm{LC} oscillations are setup. The maximum current in the coil is
View answer and explanation
Correct option: D
0.4  A0.4\,\,A
Option D (0.4 A0.4\,A) is correct because it represents the maximum current in the LC circuit, calculated using energy conservation.
122 Medium
122 of 246
Application +4 | -1
For a series L−C−RL - C - R circuit with R=100 ΩR = 100\,\Omega, L=0.5mHL = 0.5\mathrm{ mH} and C=0.1 pFC = 0.1\,\mathrm{pF} connected across 220V−50Hz AC\mathrm{220V} - 50\mathrm{ Hz\, AC} supply, the phase angle between current and supplied voltage and the nature of the circuit is
View answer and explanation
Correct option: D
≈90o\approx {90^\mathrm{o}}, predominantly capacitive circuit
Option D (≈90o\approx {90^\mathrm{o}}, predominantly capacitive circuit) is correct because the capacitive reactance is much greater than the inductive reactance, making the circuit predominantly capacitive.
123 Medium
123 of 246
Analyzing +4 | -1
Assertion: In series LCR circuit phase difference between current and voltage is never zero.

Reason: Voltage and current are never in phase.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D is correct because both the assertion and reason are false; the phase difference can be zero at resonance, and voltage and current can be in phase.
124 Medium
124 of 246
Application +4 | -1
A capacitor of capacitance 100 μF100 \, \mu \mathrm{F} and a coil of resistance 50 Ω50 \, \Omega and inductance 0.5 H0.5 \, \mathrm{H} are connected in series with a 110 V,50 Hz110 \mathrm{~V}, 50 \mathrm{~Hz} AC source. Find the rms value of the current.
View answer and explanation
Correct option: A
0.82 A0.82 \mathrm{~A}
Option A (0.82 A) is correct because the RMS current is calculated by dividing the RMS voltage by the impedance of the series RLC circuit.
125 Medium
125 of 246
Application +4 | -1
A 500μH500 \mu \mathrm{H} inductor, 80π2pF\frac{80}{\pi^2} \mathrm{pF} capacitor and a 628 Ω628 ~\Omega resistor are connected to form a series LCR circuit. Calculate the resonant frequency and QQ-factor of this circuit at resonance
View answer and explanation
Correct option: B
ff=2500 kHz,Q=12.5f_f=2500 \mathrm{~kHz}, Q=12.5
Option A (ff=2500 kHz,Q=10f_f=2500 \mathrm{~kHz}, Q=10) is incorrect. Option B (ff=2500 kHz,Q=12.5f_f=2500 \mathrm{~kHz}, Q=12.5) is correct because the resonant frequency and Q-factor are calculated using the standard formulas for a series LCR circuit.
126 Medium
126 of 246
Analyzing +4 | -1
Voltage and current in an ac circuit are given by V=5sin⁡(100πt−π6)V=5 \sin \left(100 \pi t-\frac{\pi}{6}\right) and I=4sin⁡(100πt+π6)I=4 \sin \left(100 \pi t+\frac{\pi}{6}\right)
View answer and explanation
Correct option: C
Current leads the voltage by 60∘60^{\circ}
Option C (Current leads the voltage by 60∘60^{\circ}) is correct because the phase angle of the current is π6\frac{\pi}{6} and the phase angle of the voltage is −π6-\frac{\pi}{6}, so the current leads by π3\frac{\pi}{3} which is 60∘60^{\circ}.
127 Medium
127 of 246
Application +4 | -1
AA radio set operates at 6 V6 \mathrm{~V} DC. A transformer with 18 turns in the secondary coil is used to step down the input 220 V AC emf to 6 V AC emf. This AC emf is then rectified by another circuit to give 6 V DC which is fed to the radio. Find the number of turns in the primary.
View answer and explanation
Correct option: A
660
Option A (660) is correct because the number of turns in the primary coil is calculated using the transformer equation.
128 Medium
128 of 246
Understanding +4 | -1
The power in ac circuit is given by P=ErmsIrmscos⁡ϕ.P = {E_{rms}}{I_{rms}}\cos \phi . The value of cos⁡ϕ\cos \phi in series LCR circuit at resonance is
View answer and explanation
Correct option: B
1
Option B (1) is correct because at resonance in a series LCR circuit, the impedance is equal to the resistance, making the power factor (cos⁡ϕ\cos \phi) equal to 1.
129 Medium
129 of 246
Understanding +4 | -1
Power factor of series LCR circuit at resonance is
View answer and explanation
Correct option: C
1
Option C (1) is correct because at resonance, the impedance is purely resistive, and the phase angle between voltage and current is zero, resulting in a power factor of 1.
130 Medium
130 of 246
Application +4 | -1
A transformer steps up an A.C. voltage from 230 v to 2300 v. If the number of turns in the secondary coil is 1000, the number of turns in the primary coil will be
View answer and explanation
Correct option: A
100
Option A (100) is correct because the number of turns in the primary coil is inversely proportional to the voltage in a transformer.
131 Medium
131 of 246
Application +4 | -1
By what percent the impedance in an series LCR, AC circuit should be increased so that the power factor changes from (1/2)\left( {1/2} \right) to (1/4) ?\left( {1/4} \right)\,? (assume resistance RR remains constant)
View answer and explanation
Correct option: B
100  %100\,\,\%
Option B (100  %100\,\,\% ) is correct because the impedance doubles when the power factor changes from 1/2 to 1/4, indicating a 100% increase.
132 Medium
132 of 246
Analyzing +4 | -1
Assertion: When capacitive reactance is smaller then the inductive reactance in series LCR circuit, voltage leads the current.

Reason: In series LCR circuit inductive reactance is always greater then capacitive reactance.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C is correct because the assertion is true (voltage leads current when capacitive reactance is smaller), but the reason is false (inductive reactance is not always greater).
133 Medium
133 of 246
Understanding +4 | -1
A resistance R, inductance L and capacitor C are connected in series to an oscillator of frequency ff. If resonant frequency is ff, then current will lag the voltage when
View answer and explanation
Correct option: D
f>frf > {f_r}
Option D is correct. Current lags the voltage when the frequency is greater than the resonant frequency, indicating an inductive circuit.
134 Medium
134 of 246
Understanding +4 | -1
The ratio of secondary to the primary turns in a transformer is 3:2\mathrm{3 : 2}. If the power output be PP, then the input power neglecting all losses must be equal to
View answer and explanation
Correct option: C
PP
Option C (PP) is correct because, in an ideal transformer (neglecting losses), the input power is equal to the output power.
135 Medium
135 of 246
Application +4 | -1
An ac ammeter is used to measure current in a circuit. When a given direct current passes through the circuit, the ac ammeter reads 3 A3 \mathrm{~A}. When another alternating current passes through the circuit, the ac ammeter reads 4 A4 \mathrm{~A}. Then find the reading of this ammeter (in A\mathrm{A}), if dc and ac flow through the circuit simultaneously.
View answer and explanation
Correct option: A
5
Option A (5) is correct because the AC ammeter reads the RMS value of the combined DC and AC currents, which is calculated to be 5A.
136 Medium
136 of 246
Application +4 | -1
A solenoid having inductance 4.0 H4.0 \mathrm{~H} and resistance 10 Ω10 \, \Omega is connected to a 4.0 V4.0 \mathrm{~V} battery at t=0\text{t}=0. Find

(a) the time constant
(b) the time elapsed before the current reaches 0.63 of its steady-state value
View answer and explanation
Correct option: A
(a) 0.40 s;0.40 \mathrm{~s}; (b) 0.40 s0.40 \mathrm{~s}
Option A ((a) 0.40 s;0.40 \mathrm{~s}; (b) 0.40 s0.40 \mathrm{~s}) is correct because the time constant and the time to reach 0.63 of steady-state current are both equal to L/R.
137 Medium
137 of 246
Understanding +4 | -1
Assertion (A) : Resonance frequency will decrease in L−C−RL-C-R series circuit if a dielectric slab is inserted in between the plates of the capacitor.

Reason (R) : By doing so, capacity of capacitor will increase.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.
Option A is correct because inserting a dielectric increases capacitance, which decreases the resonance frequency in an LCR circuit.
138 Medium
138 of 246
Remembering +4 | -1
The power factor of good choke coil is
View answer and explanation
Correct option: A
Nearly zero
Option A (Nearly zero) is correct because a good choke coil is designed to have very low resistance and high inductance, resulting in a power factor close to zero.
139 Medium
139 of 246
Analyzing +4 | -1
The value of current in two series LCRL C R circuits at resonance is same. Then
View answer and explanation
Correct option: C
For both the circuits XL/XCX_{{L}} / X_{{C}} must be same at that frequency
Option C is correct. At resonance, the inductive reactance (XLX_L) and capacitive reactance (XCX_C) are equal, so their ratio must be 1 for both circuits.
140 Medium
140 of 246
Application +4 | -1
In the circuit with R=6  ΩR = 6\,\,\Omega and impedance 10   Ω10\,\,\,\Omega. The power factor of circuit is
View answer and explanation
Correct option: B
0.6
Option B (0.6) is correct because the power factor is the ratio of resistance to impedance in an AC circuit.
141 Medium
141 of 246
Application +4 | -1
Find the effective value of current i=2+4cos⁡i=2+4 \cos 100πt100 \pi \mathrm{t}.
View answer and explanation
Correct option: A
232 \sqrt{3}
Option A (232 \sqrt{3}) is correct because it is the RMS value calculated by considering both the DC and AC components of the current.
142 Medium
142 of 246
Analyzing +4 | -1
Given below are two statements:

Statement I : An AC circuit undergoes electrical resonance if it contains either a capacitor or an inductor.

Statement II: An AC circuit containing a pure capacitor or a pure inductor consumes high power due to its non-zero power factor.

In the light of above statements, choose the correct answer from the options given below:
View answer and explanation
Correct option: C
Both Statement I and Statement II are false
The correct answer is C because electrical resonance requires both an inductor and a capacitor, and a pure capacitor or inductor has a power factor of zero, thus consuming no power.
143 Medium
143 of 246
Analyzing +4 | -1
Assertion (A) : In an AC\mathrm{AC} circuit, potential difference across the capacitor may be greater than the applied voltage.

Reason (R) : VC=IXCV_C=I X_C, whereas V=IZV=I Z and XCX_C can be greater than ZZ also.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.
Option A is correct because the potential difference across a capacitor in an AC circuit can be greater than the applied voltage due to resonance, and the reason correctly explains this with the impedance relationship.
144 Medium
144 of 246
Analyzing +4 | -1
A capacitor, an inductor and an electric bulb are connected in series to an AC supply of variable frequency. As the frequency of supply increased gradually, then the electric bulb is found to be
View answer and explanation
Correct option: C
Increase in brightness, reach a maximum and then decrease in brightness
Option C (Increase in brightness, reach a maximum and then decrease in brightness) is correct because the brightness is related to the power, which is maximum at resonance.
145 Medium
145 of 246
Analyzing +4 | -1
Power factor of an L-R series circuit is 0.6 and that of a C-R series circuit is 0.5 . If the elements ( L,CL, C and RR ) of the two circuit are joined in series the power factor of this circuit is found to be 1. The ratio of the resistance is L-R circuit to the resistance in the C-R circuit is
View answer and explanation
Correct option: D
334\frac{3 \sqrt{3}}{4}
Option D (334\frac{3 \sqrt{3}}{4}) is correct. When the power factor is 1, XL = XC. Using the given power factors, the ratio of resistances can be found.
146 Medium
146 of 246
Understanding +4 | -1
Assertion: A sinusoidal alternating current does not show any average magnetic effects.

Reason: The average value of an alternating current is zero.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because a sinusoidal alternating current has zero average value, resulting in no average magnetic effect.
147 Medium
147 of 246
Application +4 | -1
A coil having an inductance of 1π\frac{1}{\pi} henry is connected in series with 300Ω300 \Omega.If 20 V20 \mathrm{~V} from a 200 cycle source is impressed across the combination, then the phase angle between, V and I\mathrm{I} is-
View answer and explanation
Correct option: D
tan⁡−1(43)\tan ^{-1}\left(\frac{4}{3}\right)
Option D (tan⁡−1(43)\tan ^{-1}\left(\frac{4}{3}\right)) is correct because the phase angle is calculated using the inductive reactance and resistance.
148 Medium
148 of 246
Understanding +4 | -1
The instantaneous value of current and emf in an A.C. circuit are I=12sin⁡314tI = \frac{1}{{\sqrt 2 }}\sin 314t amp and E=2sin⁡(314t−π6)V,E = \sqrt 2 \sin \left( {314t - \frac{\pi }{6}} \right)V, with respect to I) will be
View answer and explanation
Correct option: A
−π6rad- \frac{\pi }{6}\mathrm{ rad}
Option A (−π6rad- \frac{\pi }{6}\mathrm{ rad}) is correct because the phase difference is directly observed from the given equations.
149 Medium
149 of 246
Understanding +4 | -1
There is a current of 20 mA20 \mathrm{~mA} passing through an ideal inductor when connected to a 100 V,20 Hz100 \mathrm{~V}, 20 \mathrm{~Hz} supply. The power consumed is
View answer and explanation
Correct option: D
Zero
Option D is correct because an ideal inductor does not dissipate power; it only stores energy in its magnetic field.
150 Medium
150 of 246
Application +4 | -1
An alternating current of rms value 10 A passed through 12 ohm resistor. The maximum potential difference across the resistor is [in Volts]
View answer and explanation
Correct option: C
169.98
Option C (169.98) is correct because the maximum potential difference is calculated by multiplying the maximum current by the resistance, where the maximum current is 2\sqrt{2} times the RMS current.
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