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JEE PHYSICS MCQ: ALTERNATING CURRENT

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Statement (A): Maximum power is dissipated in a circuit (through R ) in resonance.

Statement (B): At resonance in a series LCR circuit, the voltage across indcutor and capacitor are out of phase.

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Correct option: A

Both (A) and (B) are true

Option A (Both (A) and (B) are true) is correct because at resonance, maximum power is dissipated, and the voltage across the inductor and capacitor are out of phase in a series LCR circuit.

More Questions from ALTERNATING CURRENT

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In a series RC circuit with an AC source,

R=300 \, \Omega

C=25 \, \mu \mathrm{F}, \varepsilon_0=50 \mathrm{~V}

v=50 / \pi \, \mathrm{Hz}

. Find the peak current and the average power dissipated in the circuit. In a series circuit

R=300 \Omega, L=0.9 \mathrm{~H}, C=2.0

\mu F

\omega=1000 \mathrm{~rad} / \mathrm{sec}

. The impedance of the circuit is: An L-C-R series circuit with a resistance of

100 \Omega

is connected to

200 \mathrm{~V}

(AC source) and angular frequency

300 \, \mathrm{rad} / \mathrm{s}

. When only the capacitor is removed, then the current lags behind the voltage by

60^{\circ}

. When only the inductor is removed the current leads the voltage by

60^{\circ}

. The average power dissipated in original L-C-R circuit is Assertion (A) : In series

L-C \cdot R, \mathrm{AC}

circuit, current and voltage are in same phase at resonance.

Reason (R) : In series

L-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-R

circuit is connected to an

\mathrm{AC}

voltage source. When

L

is removed from the circuit, the phase difference between current and voltage is

\frac{\pi}{4}

C

is removed from the circuit, the phase difference is again

\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