An inductor 20 mH, a capacitor 100 μF and a resistor 50 Ω are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is

1. 0.79 W

2. 2.74 W

3. 0.43 W

4. 1.13 W

Doubtly Asked 30 Oct 2026 NEET UG 2018 Practice test neet Save WhatsApp Report

1 answer

Best answer

Answer: 1. 0.79 W

Power is lost only in the resistor: $P=I_{rms}^2R$, with $I_{rms}=V_{rms}/Z$.

$X_L=\omega L=314\times0.02=6.28\ \Omega$; $X_C=\dfrac{1}{\omega C}=\dfrac{1}{314\times10^{-4}}=31.85\ \Omega$.

$Z=\sqrt{50^2+(31.85-6.28)^2}=\sqrt{2500+654}\approx56.2\ \Omega$.

$V_{rms}=10/\sqrt2=7.07$ V, so $I_{rms}=0.126$ A.

$$P=(0.126)^2\times50\approx0.79\ \mathrm{W}$$

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?