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JEE • PHYSICS • ALTERNATING CURRENT

Alternating current practice · Medium

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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 246 questions
61 Medium
61 of 246
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.
62 Medium
62 of 246
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.
63 Medium
63 of 246
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.
64 Medium
64 of 246
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).
65 Medium
65 of 246
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.
66 Medium
66 of 246
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.
67 Medium
67 of 246
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.
68 Medium
68 of 246
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.
69 Medium
69 of 246
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).
70 Medium
70 of 246
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.
71 Medium
71 of 246
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.
72 Medium
72 of 246
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.
73 Medium
73 of 246
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.
74 Medium
74 of 246
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.
75 Medium
75 of 246
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.
76 Medium
76 of 246
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.
77 Medium
77 of 246
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.
78 Medium
78 of 246
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}).
79 Medium
79 of 246
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.
80 Medium
80 of 246
Application +4 | -1
What is the self-inductance of a coil which produces 5 V when the current changes from 3 A to 2 A in one millisecond?
View answer and explanation
Correct option: B
5 mH
The correct answer is B (5 mH) because the self-inductance is calculated using the formula L=edi/dtL = \frac{e}{di/dt}.
81 Medium
81 of 246
Understanding +4 | -1
Statement (A): The reactance offered by an inductance in A.C. circuit decreases with increase of A.C frequency.

Statement (B): The reactance offered by a capacitor in A.C circuit increases with increase of A.C frequency.
View answer and explanation
Correct option: D
Both (A) and (B) are false
Option D is correct because the inductive reactance increases with frequency, and the capacitive reactance decreases with frequency.
82 Medium
82 of 246
Analyzing +4 | -1
A series LCR circuit is connected to an ac voltage source. When L\mathrm{L} is removed from the circuit, the phase difference
between current and voltage is π3\frac{\pi }{3}. If instead C\mathrm{C} is removed from the circuit, the phase difference is again π3\frac{\pi }{3} between
current and voltage. The power factor of the circuit is:
View answer and explanation
Correct option: C
1
Option C (1) is correct because when the circuit is in resonance, the impedance is equal to the resistance, making the power factor equal to 1.
83 Medium
83 of 246
Application +4 | -1
In AC circuit a resistance of R Ω is connected in series with an inductance L. If the phase difference between the current and voltage is 45°, the inductive reactance will be
View answer and explanation
Correct option: C
R
Option C (R) is correct because the inductive reactance equals the resistance when the phase difference is 45 degrees in an RL circuit.
84 Medium
84 of 246
Understanding +4 | -1
Statement (A): In high current low voltage windings of a transformer thick wire is used to minimize energy loss due to heat produced.

Statement (B): The core of any transformer is laminated so as to reduce the energy loss due to eddy currents.
View answer and explanation
Correct option: A
Both (A) and (B) are true
Option A (Both (A) and (B) are true) is correct because both statements regarding transformer windings and core lamination are factually accurate.
85 Medium
85 of 246
Understanding +4 | -1
Choke coil is used to control
View answer and explanation
Correct option: A
AC
Option A (AC) is correct because a choke coil's inductive reactance controls current in AC circuits, while having negligible resistance for DC.
86 Medium
86 of 246
Analyzing +4 | -1
An AC source producing emf

e=e0[cos⁡(100πs−1)t+cos⁡(500πs−1)t]e=e_0\left[\cos \left(100 \pi \mathrm{s}^{-1}\right) t+\cos \left(500 \pi \mathrm{s}^{-1}\right) t\right]

is connected in series with a capacitor and resistor. The steady state current in the circuit is found to be

I=I1cos⁡[(100πs−1)t+ϕ1]+I2cos⁡[(500πs−1)t+ϕ2]I=I_1 \cos \left[\left(100 \pi \mathrm{s}^{-1}\right) t+\phi_1\right]+I_2 \cos \left[\left(500 \pi \mathrm{s}^{-1}\right) t+\phi_2\right]
View answer and explanation
Correct option: C
I1<I2I_1 < I_2
Option C (I1<I2I_1 < I_2) is correct because the impedance is lower at higher frequencies in a capacitor circuit, leading to a larger current.
87 Medium
87 of 246
Application +4 | -1
A 9/100π9 / 100 \pi inductor and a 12Ω12 \Omega resistance are connected in series to a 225 V,50 Hz225 \mathrm{~V}, 50 \mathrm{~Hz} ac source. Calculate the current in the circuit and the phase angle between the current and the source voltage.
View answer and explanation
Correct option: A
37∘37^{\circ}
Option A (37∘37^{\circ}) is correct because it is the phase angle calculated using the arctangent of the ratio of inductive reactance to resistance.
88 Medium
88 of 246
Understanding +4 | -1
In an LCR series AC circuit, the voltage across each of the components, L, C and R is 50 V. the voltage across the LC combination will be
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because in a series LCR circuit, the voltage across the inductor and capacitor are 180 degrees out of phase and cancel each other out when they have equal magnitudes.
89 Medium
89 of 246
Analyzing +4 | -1
Assertion: With increase in frequency of an A.C supply, the impedance of an L-C-R series circuit, decreases at first, becomes minimum and then increases.

Reason: In L-C-R series circuit, impedance (z)
z=R2+(XL−XC)2z=\sqrt{R^2+\left(X_L-X_C\right)^2}.
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 the impedance of an LCR circuit changes with frequency due to the reactances, and the given formula accurately describes the impedance.
90 Medium
90 of 246
Understanding +4 | -1
In a step-up transformer, the turn ratio is 1:2\mathrm{1 : 2}. A Leclanche cell (e.m.f. 1.5  V\mathrm{1}\mathrm{.5}\,\,\mathrm{V}) is connected across the primary. The voltage developed in the secondary would be
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because a transformer does not work with direct current (DC).
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