For a series $L - C - R$ circuit with $R = 100\,\Omega $, $L = 0.5{\rm{ mH}}$ and $C = 0.1\,{\rm{pF}}$ connected across ${\rm{220V}} - 50{\rm{ Hz\, AC}}$ supply, the phase angle between current and supplied voltage and the nature of the circuit is

$ \approx {90^{\rm{o}}}$, predominantly capacitive circuit

${{\rm{0}}^{\rm{o}}}$, resistive circuit

$ \approx {90^{\rm{o}}}$, predominantly inductive circuit

${0^{\rm{o}}}$, resonance circuit

Back to ALTERNATING CURRENT

JEE PHYSICS MCQ: ALTERNATING CURRENT

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Application +4 | -1

For a series

L - C - R

circuit with

R = 100\,\Omega

L = 0.5\mathrm{ mH}

and

C = 0.1\,\mathrm{pF}

connected across

\mathrm{220V} - 50\mathrm{ Hz\, AC}

supply, the phase angle between current and supplied voltage and the nature of the circuit is

View answer and explanation

Correct option: D

\approx {90^\mathrm{o}}

, predominantly capacitive circuit

Option D (

\approx {90^\mathrm{o}}

, predominantly capacitive circuit) is correct because the capacitive reactance is much greater than the inductive reactance, making the circuit predominantly capacitive.

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}

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