JEE Main 2019 · Physics · Electromagnetic Induction and Alternating Current · easy

A series circuit has a resistor of 30 Ω, an inductor and a capacitor. At the supply frequency the inductive reactance is 80 Ω and the capacitive…

Single correct option

A series circuit has a resistor of

30\ \Omega

, an inductor and a capacitor. At the supply frequency the inductive reactance is

80\ \Omega

and the capacitive reactance is

40\ \Omega

, and the supply is

200\ \mathrm{V}

(rms). If the capacitance alone is adjusted until the circuit is in resonance, the average power dissipated becomes

Show answer and solution

Answer: (D) $1.33\ \mathrm{kW}$

Initially

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

I=4\ \mathrm{A}

and

P=I^2R=480\ \mathrm{W}

. Adjusting

C

so that

X_C=X_L=80\ \Omega

(frequency unchanged) makes the net reactance zero, so

Z=R=30\ \Omega

. The rms current becomes

200/30=6.67\ \mathrm{A}

and the average power is

P=\dfrac{V^2}{R}=\dfrac{200^2}{30}\approx1333\ \mathrm{W}=1.33\ \mathrm{kW}

. The value

V^2/Z=800\ \mathrm{W}

forgets that the power is dissipated only in the resistor.

Original question written in the JEE Main 2019 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake