JEE Main 2017 · Physics · Electromagnetic Induction and Alternating Current · medium

A series circuit consists of a resistor of 30 Ω, an inductor of reactance 80 Ω and a capacitor of reactance 40 Ω connected to an ac source of fixed…

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A series circuit consists of a resistor of

30\ \Omega

, an inductor of reactance

80\ \Omega

and a capacitor of reactance

40\ \Omega

connected to an ac source of fixed rms voltage and frequency. The capacitor is then adjusted until the circuit is in resonance. The ratio of the average power dissipated after adjustment to that before is

A. $\dfrac{25}{9}$
B. $\dfrac{5}{3}$
C. $\dfrac{9}{25}$
D. $\dfrac{3}{5}$

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Answer: (A) $\dfrac{25}{9}$

Before:

Z = \sqrt{30^2 + (80-40)^2} = 50\ \Omega

. The average power is

P = \dfrac{V^2R}{Z^2}

. At resonance

X_L = X_C

, so

Z = R = 30\ \Omega

and

P' = \dfrac{V^2}{R}

. Therefore

\dfrac{P'}{P} = \dfrac{Z^2}{R^2} = \dfrac{2500}{900} = \dfrac{25}{9}

. The ratio of currents is

5/3

, which is the first distractor, but power goes as

I^2

. Equivalently

\cos^2\phi = 9/25

before resonance.

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