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JEE PHYSICS MCQ: ALTERNATING CURRENT

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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
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Correct option: A
Becomes zero instantaneously
Option A (Becomes zero instantaneously) is correct because if the capacitor is fully charged and then disconnected, the initial current in the circuit is zero.

More Questions from ALTERNATING CURRENT

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In a series RC circuit with an AC source, R=300 ΩR=300 \, \Omega, C=25 μF,Ξ΅0=50Β 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=300Ξ©,L=0.9Β H,C=2.0R=300 \Omega, L=0.9 \mathrm{~H}, C=2.0 ΞΌF\mu F and Ο‰=1000Β rad/sec\omega=1000 \mathrm{~rad} / \mathrm{sec}. The impedance of the circuit is: An L-C-R series circuit with a resistance of 100Ξ©100 \Omega is connected to 200Β 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 Lβˆ’Cβ‹…R,ACL-C \cdot R, \mathrm{AC} circuit, current and voltage are in same phase at resonance.

Reason (R) : In series Lβˆ’Cβˆ’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 Lβˆ’Cβˆ’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 Ο€4\frac{\pi}{4}. If instead CC is removed from the circuit, the phase difference is again Ο€4\frac{\pi}{4} between current and voltage. The power factor of the circuit is In a choke coil, the resistance XL{X_L} and resistance RR are such that
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