JEE Main 2020 · Physics · Electromagnetic Induction and Alternating Current · medium

A series circuit contains a resistor of 120 Ω, an inductor of 0.15 H and a capacitor of 10 muF, connected to a source of 260 V (rms) with angular…

Numerical answer

A series circuit contains a resistor of

120\ \Omega

, an inductor of

0.15\ \mathrm{H}

and a capacitor of

10\ \mu\mathrm{F}

, connected to a source of

260\ \mathrm{V}

(rms) with angular frequency

1000\ \mathrm{rad\,s^{-1}}

. The average power dissipated in the circuit (in watts) is

Show answer and solution

Answer: 480

X_L = \omega L = 150\ \Omega

and

X_C = 1/(\omega C) = 1/(1000\times10^{-5}) = 100\ \Omega

. Net reactance

= 50\ \Omega

, so

Z = \sqrt{120^2 + 50^2} = 130\ \Omega

. Current

I_{rms} = 260/130 = 2\ \mathrm{A}

. Average power

P = I_{rms}^2R = 4\times120 = 480\ \mathrm{W}

(equivalently

VI\cos\phi

with

\cos\phi = 120/130

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