Skip to main content
Back to ALTERNATING CURRENT

Your Progress

0 / 8 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • PHYSICS • ALTERNATING CURRENT

Alternating current practice · Hard

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

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.

Take a chapter quiz →
1 Hard
1 of 8
Application +4 | -1
A resistance of 20 ohms is connected to a source of an alternating potential V=220sin⁡[100 π t]V = 220\sin \left[ {100\,\pi \,t} \right]. The time taken by the current to change from its peak value to r.m.s. value is [in sec]
View answer and explanation
Correct option: D
2.5×10−32.5 \times {10^{ - 3}}
Option D (2.5×10−32.5 \times {10^{ - 3}}) is correct because it is the time calculated for the current to change from its peak value to its RMS value.
2 Hard
2 of 8
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.
3 Hard
3 of 8
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)).
4 Hard
4 of 8
Application +4 | -1
The peak voltage of 220 volt AC mains (in volt) is
View answer and explanation
Correct option: C
311
Option C (311) is correct because the peak voltage is calculated by multiplying the RMS voltage by the square root of 2.
5 Hard
5 of 8
Application +4 | -1
What is the value of inductance L for which the current is a maximum in a series L-C-R circuit with C=10μFC = 10\mu \mathrm{F} and ω=1000  s−1?\omega = 1000\;{\mathrm{s}^{ - 1}}?
View answer and explanation
Correct option: A
100 mH
Option A (100 mH) is correct because the current is maximum at resonance, where the inductive reactance equals the capacitive reactance, leading to L = 1/(ω2\omega^2C).
6 Hard
6 of 8
Understanding +4 | -1
For a series R-L-C circuit:

(a) Voltage across L and C differ by π\pi phase

(b) Current through L and R are in same phase

(c) Voltage across R and L differ by π/2\pi/2 phase

(d) Voltage across L and current through C differ by π/2\pi/2 phase
View answer and explanation
Correct option: D
All are correct
Option D (All are correct) is correct because it accurately describes the phase relationships between voltage and current in a series RLC circuit.
7 Hard
7 of 8
Application +4 | -1
In an AC circuit, V and I are given by V = 100 sin [100 t] volts, I = 100 sin [100 t+π3]\left[ {100\,t + \frac{\pi }{3}} \right] mA. The power dissipated in casuist is [in watt]
View answer and explanation
Correct option: C
2.5
Option C (2.5) is correct because the power dissipated is calculated using the RMS values of voltage and current, along with the phase difference between them.
8 Hard
8 of 8
Analyzing +4 | -1
An ac source of angular frequency ω is fed across a resistor r and a capctior C in series. The current registered is II. If the frequency of source is changed to ω/3 (maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency ω
View answer and explanation
Correct option: A
35\sqrt {\frac{3}{5}}
The ratio of reactance to resistance at the original frequency is 35\sqrt{\frac{3}{5}} (Option A) because it is derived from the given current and frequency changes in the RC circuit.
Revision notes and key takeaways

Review ALTERNATING CURRENT in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

  • ✓ This set contains 8 single-correct-answer MCQs on ALTERNATING CURRENT.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
0 /8
0✓ 0✗
0% Accuracy
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.