JEE Main 2016 · Physics · Electromagnetic Induction and Alternating Current · hard

A series circuit has a resistor of 30 Ω, an inductor of reactance 80 Ω and a capacitor of reactance 40 Ω across a 100 V (rms) ac source. A second…

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A series circuit has a resistor of

30\ \Omega

, an inductor of reactance

80\ \Omega

and a capacitor of reactance

40\ \Omega

across a

100\ \mathrm{V}

(rms) ac source. A second capacitor of reactance

40\ \Omega

is now added in series. The ratio of the average power consumed after the change to that before is

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

Initially

Z_1=\sqrt{30^2+(80-40)^2}=50\ \Omega

and

P_1=\dfrac{V^2R}{Z^2}=\dfrac{10^4\times30}{2500}=120\ \mathrm{W}

. Capacitors in series have reactances that add, so the net capacitive reactance becomes

80\ \Omega

, equal to

X_L

. The circuit is now at resonance with

Z_2=R=30\ \Omega

and

P_2=\dfrac{V^2}{R}=\dfrac{10^4}{30}=333.3\ \mathrm{W}

. The ratio is

\dfrac{333.3}{120}=\dfrac{25}{9}

. Equivalently

P\propto \dfrac{R}{Z^2}

gives

\dfrac{(50/30)^2}{1}

Original question written in the JEE Main 2016 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake