An L-C-R series circuit with a resistance of 100 is connected to 200 V (AC source) and angular frequency 300 rad / s . When only the capacitor is removed, then the current lags behind the voltage by 60^ . When only the

Correct answer: D. 400 W Given, R=100 , V=200 ~V, θ=60^ , ω=300 rad / s Phase angle, θ=(X_L)/(R)=(X_C)/(R) 60^ =(X_L)/(R)=(X_C)/(R) X_L=X_C= 3 R i.e. Z= R^2+( 3 R- 3 R)^2 =R So average power, P=(V^2)/(R)=(200 × 200)/(100)=400 ~W

JIPMER 2016

Exam: JIPMER
Year: 2016
Subject: Physics
Chapter: Alternating Current

JIPMER 2016 · Physics · Alternating Current

An L-C-R series circuit with a resistance of 100 is connected to...

L-C-R

series circuit with a resistance of

100 \Omega

is connected to 200 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

Options

A50 W
B100 W
C200 W
D400 W

Correct answer

D. 400 W

Step-by-step solution

Given,

R=100 \Omega, V=200 \mathrm{~V}, \theta=60^{\circ}, \omega=300 \mathrm{rad} / \mathrm{s}

\therefore

Phase angle,

\tan \theta=\frac{X_L}{R}=\frac{X_C}{R}

\Rightarrow \quad \tan 60^{\circ}=\frac{X_L}{R}=\frac{X_C}{R}

\therefore \quad X_L=X_C=\sqrt{3} R

i.e.

Z=\sqrt{R^2+(\sqrt{3} R-\sqrt{3} R)^2}=R

So average power,

P=\frac{V^2}{R}=\frac{200 \times 200}{100}=400 \mathrm{~W}

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