Target Exam

CUET

Subject

Physics

Chapter

Alternating Current

Question:

An inductor of reactance 100Ω and resistor of resistance 100Ω are joined in series and connected by AC source of emf 220 V. Power dissipated in the circuit is

Options:

Zero

242 W

346 W

141 W

Correct Answer:

242 W

Explanation:

The correct answer is Option (2) → 242 W

The total impedance (Z),

$Z=\sqrt{R^2+{X_L}^2}$

$=\sqrt{100^2+100^2}$

$=\sqrt{20000}=141.42Ω$

Power factor, $\cos\phi=\frac{R}{Z}=\frac{100}{141.42}=0.707$

$I_{rms}=\frac{V_{rms}}{Z}=\frac{220}{141.42}=1.555A$

Power, $P=V_{rms}×I_{rms}×\cos\phi$

$=220×1.555×0.707=242.0W$