Back to ALTERNATING CURRENT

JEE PHYSICS MCQ: ALTERNATING CURRENT

1 Medium

1 of 1

Analyzing +4 | -1

Assertion: With increase in frequency of an A.C supply, the impedance of an L-C-R series circuit, decreases at first, becomes minimum and then increases.

Reason: In L-C-R series circuit, impedance (z)

z=\sqrt{R^2+\left(X_L-X_C\right)^2}

.

View answer and explanation

Correct option: A

If both assertion and reason are true and the reason is the correct explanation of the assertion.

Option A (If both assertion and reason are true and the reason is the correct explanation of the assertion) is correct because the impedance of an LCR circuit changes with frequency due to the reactances, and the given formula accurately describes the impedance.

More Questions from ALTERNATING CURRENT

Practice All Chapter MCQs →

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