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JEE • PHYSICS • ALTERNATING CURRENT

Alternating current practice · Mixed difficulty

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Practice format and chapter quiz

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 61-90 of 279 questions
61 Easy
61 of 279
Remembering +4 | -1
The power factor of LCR circuit at resonance is
View answer and explanation
Correct option: B
1
Option B (1) is correct because at resonance, the impedance of an LCR circuit is purely resistive, resulting in a power factor of 1.
62 Medium
62 of 279
Analyzing +4 | -1
Assertion (A): An inductance and a resistance are connected in series with an AC circuit. In this circuit, the current and the potential difference across the resistance logs behind potential difference across the inductance by an angle π2.\frac{\pi}{2}.
Reason (R): In L - R circuit, voltage leads the current by phase angle which depends on the value of inductance and resistance both.
View answer and explanation
Correct option: B
Both assertion and reason are true but reason is not the correct explanation of assertion
Option B (Both assertion and reason are true but reason is not the correct explanation of assertion) is correct because while both statements are true, the reason doesn't directly explain the phase difference between voltage across the resistor and inductor.
63 Medium
63 of 279
Application +4 | -1
A 60μF60 \mu {F} capacitor is connected to a 110 V,60 HzAC110 {~V}, 60 {~Hz} {AC} supply. Determine the rms value of the current in the circuit.
View answer and explanation
Correct option: A
2.5 A2.5 {~A}
Option A (2.5 A2.5 {~A}) is correct because it is calculated using the formula Irms=VrmsXCI_{rms} = \frac{V_{rms}}{X_C} where XC=12πfCX_C = \frac{1}{2 \pi f C}.
64 Medium
64 of 279
Understanding +4 | -1
Statement (A): Flux leakage in a transformer can be minimized by winding the primary and secondary coils one over the other.

Statement (B): Core of the transformer is made of soft iron.
View answer and explanation
Correct option: A
Both (A) and (B) are true
Option A (Both (A) and (B) are true) is correct because winding coils over each other minimizes flux leakage, and soft iron is used for the core due to its high permeability and low hysteresis loss.
65 Medium
65 of 279
Application +4 | -1
A coil has a resistance 10 Ω10\, \Omega and an inductance of 0.4 Henry. It is connected to an AC source of 6.5 V,30πHz6.5 \mathrm{~V}, \frac{30}{\pi} \mathrm{Hz}. The average power consumed in the circuit, is :-
View answer and explanation
Correct option: A
58 W\frac{5}{8} \mathrm{~W}
Option A (58 W\frac{5}{8} \mathrm{~W}) is correct because it is the average power calculated using the given resistance, voltage, and impedance.
66 Medium
66 of 279
Understanding +4 | -1
Given below are two statements:

Statement I: Maximum power is dissipated in a circuit containing an inductor, a capacitor and a resistor connected in series with an AC source, when resonance occurs.

Statement II: Maximum power is dissipated in a circuit containing pure resistor due to zero phase difference between current and voltage.

In the light of the above statements, choose the correct answer from the options given below:
View answer and explanation
Correct option: A
Statement I is false but Statement II is true
Option A (Statement I is false but Statement II is true) is correct because maximum power dissipation occurs at resonance, and in a purely resistive circuit, the power factor is unity.
67 Medium
67 of 279
Application +4 | -1
In a series L-R growth circuit, if maximum current and maximum induced emf in an inductor of inductance 3 mH3 \mathrm{~mH} are 2 A2 \mathrm{~A} and 6 V6 \mathrm{~V} respectively, then the time constant of the circuit is
View answer and explanation
Correct option: A
1 ms1 \mathrm{~ms}
Option A (1 ms1 \mathrm{~ms}) is correct because the time constant is calculated by dividing the inductance by the resistance, where resistance is derived from Ohm's law using the maximum induced EMF and current.
68 Medium
68 of 279
Understanding +4 | -1
Can we use 15 Hz15 \mathrm{~Hz} ac for lighting purposes?
View answer and explanation
Correct option: A
Yes
Option B is correct because at 15 Hz, the fluctuations in current are rapid enough (30 times per second) that the bulb appears to glow continuously due to the persistence of vision.
69 Hard
69 of 279
Application +4 | -1
AC voltage V(t) =20sin ωt= \mathrm{20 sin}\,\mathrm{\omega t} of frequency 50 Hz is applied to a parallel plate capacitor. The separation between the plates is 2 mm and the area is 1 m2.1\,{m^2}. The amplitude of the oscillating displacement current for the applied AC voltage is ….
[Take, ε0=8.85×10−12F/m]{\varepsilon _0} = 8.85 \times {10^{ - 12}}F/m]
View answer and explanation
Correct option: C
27.79μA\mathrm{27}\mathrm{.79 \mu A}
Option C (27.79μA\mathrm{27}\mathrm{.79 \mu A}) is correct because it is the calculated amplitude of the oscillating displacement current using the given AC voltage, frequency, plate separation, and area.
70 Medium
70 of 279
Understanding +4 | -1
An alternating voltage is represented as E=20sin⁡300tE=20 \sin 300 t. The average value of voltage over one cycle will be
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because the average value of a sinusoidal alternating voltage over one complete cycle is zero.
71 Medium
71 of 279
Understanding +4 | -1
In the non-resonant circuit, what will be the nature of the circuit for frequencies higher than the resonant frequency
View answer and explanation
Correct option: B
Capacitive
Option B (Capacitive) is correct because in a non-resonant circuit at frequencies higher than the resonant frequency, the capacitive reactance dominates.
72 Medium
72 of 279
Understanding +4 | -1
A 10 ohm resistance, 5 mH coil and 10 μF capacitor are joined in series. When a suitable frequency alternating current source is joined to this combination, the circuit resonates. If the resistance is halved, the resonance frequency
View answer and explanation
Correct option: C
Remains unchanged
Option C is correct because the resonance frequency in a series LCR circuit depends only on the inductance (L) and capacitance (C), not on the resistance (R).
73 Medium
73 of 279
Application +4 | -1
A coil of resistance RR and inductance LL is connected across an a.c.a.c. power supply of r.m.s.r.m.s. voltage V{V}. The power dissipated in the coil is
View answer and explanation
Correct option: C
V2R/(R2+ω2L2)V^2 R /\left(R^2+\omega^2 L^2\right)
Option C (V2R/(R2+ω2L2)V^2 R /\left(R^2+\omega^2 L^2\right)) is correct because the power dissipated in the coil is calculated using the formula involving voltage, resistance, and impedance.
74 Medium
74 of 279
Understanding +4 | -1
An LCR series circuit is compared with a damped oscillator of mass ' m{m} ' with force constant K{K} and damping coefficient ' b{b} '. The correct matching of quantities in damped oscillator analogous with those in LCR circuit is
View answer and explanation
Correct option: A
L→m,C→1k,R→bL \rightarrow m, C \rightarrow \frac{1}{k}, R \rightarrow b
Option A (L→m,C→1k,R→bL \rightarrow m, C \rightarrow \frac{1}{k}, R \rightarrow b) is correct because inductance (L) is analogous to mass (m), capacitance (C) to the inverse of the spring constant (1/k), and resistance (R) to the damping coefficient (b) in a damped oscillator.
75 Medium
75 of 279
Understanding +4 | -1
In a series L−C−RL - C - R resonance circuit, if we change the resistance only, from a lower to higher value,
View answer and explanation
Correct option: A
the bandwidth of resonance circuit will increase
Option A (the bandwidth of resonance circuit will increase) is correct because bandwidth is directly proportional to resistance in a series LCR circuit.
76 Medium
76 of 279
Application +4 | -1
An alternating current source of frequency 100 Hz is joined to a combination of a resistance, a capacitance and a coil in series. The potential difference across the coil, the resistance and the capacitor is 46, 8 and 40 volt respectively. The electromotive force of alternating current source in volt is
View answer and explanation
Correct option: C
10
Option C (10) is correct because it is the calculated EMF of the alternating current source using the given potential differences.
77 Medium
77 of 279
Remembering +4 | -1
In a series LCRL C R circuit, at resonance, the bandwidth frequency can be represented as
View answer and explanation
Correct option: D
R2πL\frac{R}{2 \pi L}
Option D (R2πL\frac{R}{2 \pi L}) is correct because the bandwidth in a series LCR circuit at resonance is given by the formula bandwidth = R / (2πL).
78 Hard
78 of 279
Application +4 | -1
A voltage of peak value 283 V and varying frequency is applied to a series L – C – R combination in which R=3  ΩR = 3\;\mathrm{\Omega } , L=25L = 25 mH and C=400μFC = 400\mu \mathrm{F} . The frequency (in Hz)of the source at which maximum power is dissipated in the above, is
View answer and explanation
Correct option: D
50.3
Option D (50.3) is correct because maximum power dissipation occurs at resonance, and the resonant frequency is calculated using the formula v = 1/(2*pi*sqrt(LC)).
79 Medium
79 of 279
Understanding +4 | -1
In a step-up transformer the turn’s ratio is 10. If the frequency of the current in the primary coil is 50Hz then the frequency of the current in the secondary coil will be……… Hz
View answer and explanation
Correct option: B
50
Option B (50) is correct because the frequency of the current remains the same in a transformer, regardless of the turns ratio.
80 Medium
80 of 279
Application +4 | -1
In R−LR-L series DCD C circuit E=20 V,L=5mHE=20 \mathrm{~V}, L=5 \mathrm{mH} and R=1ΩR=1 \Omega. If battery is connected at t=0t=0 sec. then find time when I\mathbf{I} is 12.6 A12.6 \mathrm{~A}.
View answer and explanation
Correct option: C
5 ms5 \mathrm{~ms}
Option C (5 ms5 \mathrm{~ms}) is correct because the time when the current reaches 63% of its maximum value in an R-L circuit is equal to the time constant L/R.
81 Medium
81 of 279
Understanding +4 | -1
For a series L – C – R circuit, the phase difference between current and voltage at the condition of resonance will be
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because at resonance, the inductive reactance and capacitive reactance cancel each other out, resulting in a phase difference of zero.
82 Medium
82 of 279
Application +4 | -1
In an LCR circuit capacitance is changed from CC to 2C2 C. For the resonant frequency to remain unchanged, the inductance should be changed from LL to
View answer and explanation
Correct option: C
L2\frac{L}{2}
Option C (L/2) is correct because the resonant frequency is inversely proportional to the square root of the inductance and capacitance product, so halving L compensates for doubling C.
83 Medium
83 of 279
Application +4 | -1
A resistance RR, an inductance L=0.01HL=0.01 {H} and a capacitance CC are connected in series. When a voltage V=400cos⁡(300t−10∘)VV=400 \cos \left(300 t-10^{\circ}\right) {V} is applied to the series combination, the current flowing is 102cos⁡(3000t−55∘)10 \sqrt{2} \cos \left(3000 t-55^{\circ}\right) A. Find the capacitance of capacitor in microfarads.
View answer and explanation
Correct option: A
33
Option A (33) is correct because the capacitance is calculated based on the phase difference between voltage and current, and the given values of resistance, inductance, and angular frequency.
84 Medium
84 of 279
Application +4 | -1
When 100 VDC100 \mathrm{~V} \mathrm{DC} is applied across a solenoid, a current of 1 A1 \mathrm{~A} flows in it. When 100V AC100 \mathrm{V} \mathrm{~AC} is applied across the same coil, the current drops to 0.5 A0.5 \mathrm{~A}. If the frequency of the AC\mathrm{AC} source is 50 Hz50 \mathrm{~Hz}, the impedance and inductance of the inductor for the given AC\mathrm{AC} circuit are
View answer and explanation
Correct option: A
200 Ω200 ~\Omega and 0.55 henry
Option A (200 Ω200 ~\Omega and 0.55 henry) is correct because it calculates the resistance using DC voltage and current, then calculates impedance using AC voltage and current, and finally finds the inductance.
85 Medium
85 of 279
Analyzing +4 | -1
An R-L-C series circuit with 100 Ω100\,\Omega resistance is connected to an AC source of 200 V200 {~V} and ω=300 rad/s\omega=300 \,{rad} / {s}. When only capacitor is removed, the current lags behind voltage by 60∘60^{\circ}. When only inductor is removed, the current leads voltage by 60∘60^{\circ}. The power dissipated in the R-L-C circuit is
View answer and explanation
Correct option: B
400 W400 {~W}
Option B is correct because at resonance, the impedance equals the resistance, allowing direct calculation of the current and power dissipation.
86 Easy
86 of 279
Understanding +4 | -1
At high frequency, the capacitor offer
View answer and explanation
Correct option: B
Less reactance
Option B (Less reactance) is correct because capacitive reactance is inversely proportional to frequency, so at high frequency, the capacitor offers less reactance.
87 Medium
87 of 279
Understanding +4 | -1
Statement (A): An ac emf which oscillates symmetrically about zero, the current it sustains also oscillates symmetrically about zero.

Statement (B): In any circuit element, current is always in the phase with voltage.
View answer and explanation
Correct option: B
(A) is true but (B) is false
Option B is correct because while the current oscillates symmetrically about zero with a symmetrical AC EMF, current and voltage are not always in phase in every circuit element.
88 Medium
88 of 279
Application +4 | -1
A resistance of 300 Ω300\, \Omega and an inductance of 1π\frac{1}{\pi} henry are connected in series to a ac voltage of 20 volts and 200 Hz200 {~Hz} frequency. The phase angle between the voltage and current is
View answer and explanation
Correct option: A
tan⁡−143\tan ^{-1} \frac{4}{3}
Option A (tan⁡−143\tan ^{-1} \frac{4}{3}) is correct because the phase angle in an RL circuit is given by tan⁡−1(ωLR)\tan^{-1}(\frac{\omega L}{R}).
89 Easy
89 of 279
Application +4 | -1
Write the equation of a sinusoidal voltage of rms valus 220 V220 \mathrm{~V} and frequency 50 Hz50 \mathrm{~Hz}
View answer and explanation
Correct option: B
2202sin⁡(100πt)220 \sqrt{2} \sin (100 \pi t)
Option B (2202sin⁡(100πt)220 \sqrt{2} \sin (100 \pi t)) is correct because it represents a sinusoidal voltage with the correct RMS value and frequency.
90 Medium
90 of 279
Understanding +4 | -1
Assertion: If the frequency of alternating current in an ac circuit consisting of an inductance coil is increased then current gets decreased.

Reason: The current is inversely proportional to frequency of alternating current.
View answer and explanation
Correct option: A
If both assertion and reason are true and the reason is the correct explanation of the assertion.
Option A (If both assertion and reason are true and the reason is the correct explanation of the assertion) is correct because increasing the frequency in an inductive circuit increases the inductive reactance, which reduces the current.
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