An inductor of reactance 100 Ω, a capacitor of reactance 50 Ω, and a resistor of resistance 50 Ω are connected in series with an AC source of 10 V, 50 Hz. Average power dissipated by the circuit is ____ (a) 1 W (b) 2 W (c) 3 W (d) 4 W

Answer: (a). Z = √(R² + (XL - XC)²) = √(50² + 50²) = 50√(2) Ω. Irms = (10)/(50√(2)) = (1)/(5√(2)) A. P = Irms²R = 1/50 × 50 = 1 W. So the answer is (a) 1 W.

An inductor of reactance 100 Ω, a capacitor of reactance 50 Ω, and a… | MHT CET 2026 (11 Apr, Shift 1)

Q10MHT CET 2026 · 11 APR · SHIFT 1Medium

Power, Power Factor & Wattless Current

An inductor of reactance $100\ \Omega$ , a capacitor of reactance $50\ \Omega$ , and a resistor of resistance $50\ \Omega$ are connected in series with an AC source of $10$ V, $50$ Hz. Average power dissipated by the circuit is ____

Show answer and solution

Answer: (a)

$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{50^2 + 50^2} = 50\sqrt{2}\ \Omega$ .
$I_{rms} = \frac{10}{50\sqrt{2}} = \frac{1}{5\sqrt{2}}\text{ A}$ .
$P = I_{rms}^2R = \frac{1}{50} \times 50 = 1\text{ W}$ .
So the answer is (a) 1 W.

Report a mistake

Power, Power Factor & Wattless Current question from MHT CET 2026 (11 Apr, Shift 1)

More questions on Power, Power Factor & Wattless Current