Skip to main content
Back to ALTERNATING CURRENT

JEE PHYSICS MCQ: ALTERNATING CURRENT

1 Medium
1 of 1
Analyzing +4 | -1
An LCR circuit contains resistance of 100Ω100 \Omega and a supply of 200 V200 \mathrm{~V} at 300 rad300 \mathrm{~rad} angular frequency. If only capacitance is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by 60∘60^{\circ}. If, on the other hand, only inductor is taken out, the current leads by 60∘60^{\circ} with the applied voltage. The current flowing in the circuit is
View answer and explanation
Correct option: C
2 A2 \mathrm{~A}
Option C is correct because at resonance, the impedance is equal to the resistance, and the current is simply the voltage divided by the resistance, which is 2A.

More Questions from ALTERNATING CURRENT

Practice All Chapter MCQs →
In a series RC circuit with an AC source, R=300 ΩR=300 \, \Omega, C=25 μF,ε0=50 VC=25 \, \mu \mathrm{F}, \varepsilon_0=50 \mathrm{~V} and v=50/π Hzv=50 / \pi \, \mathrm{Hz}. Find the peak current and the average power dissipated in the circuit. In a series circuit R=300Ω,L=0.9 H,C=2.0R=300 \Omega, L=0.9 \mathrm{~H}, C=2.0 μF\mu F and ω=1000 rad/sec\omega=1000 \mathrm{~rad} / \mathrm{sec}. The impedance of the circuit is: An L-C-R series circuit with a resistance of 100Ω100 \Omega is connected to 200 V200 \mathrm{~V} (AC source) and angular frequency 300 rad/s300 \, \mathrm{rad} / \mathrm{s}. When only the capacitor is removed, then the current lags behind the voltage by 60∘60^{\circ}. When only the inductor is removed the current leads the voltage by 60∘60^{\circ}. The average power dissipated in original L-C-R circuit is Assertion (A) : In series L−C⋅R,ACL-C \cdot R, \mathrm{AC} circuit, current and voltage are in same phase at resonance.

Reason (R) : In series L−C−R,ACL-C-R, \mathrm{AC} circuit, resonant frequency does not depend on the value of resistance. Hence, current at resonance does not depend on resistance.
A series L−C−RL-C-R circuit is connected to an AC\mathrm{AC} voltage source. When LL is removed from the circuit, the phase difference between current and voltage is π4\frac{\pi}{4}. If instead CC is removed from the circuit, the phase difference is again π4\frac{\pi}{4} between current and voltage. The power factor of the circuit is A capacitor in an ideal LC circuit is fully charged by a DC source. Then it is disconnected from DC source, the current in the circuit
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.