JEE PHYSICS MCQ: ALTERNATING CURRENT

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In a pure inductive circuit the induced emf

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Correct option: B

Is maximum at the time of switching off the resistance

Option B (Is maximum at the time of switching off the resistance) is correct because the induced EMF opposes the change in current, and the change is most rapid when switching off.

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}

and

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

and

\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}

. If instead

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