An inductor of inductance L and a resistor of resistance R are joined in series and connected to an a.c. source of frequency ω. The power dissipated in the circuit is (a) ((R² + ω² L²))/(V) (b) (V)/((R² + ω² L²)) (c) (V² R)/((R² + ω² L²)) (d) √((R² + ω² L²)/(V²))

Answer: (c). Z = √(R² + ω²L²), I = V/Z (r.m.s.). Power P = I²R = (V²R)/(R² + ω²L²). So the answer is (c).

An inductor of inductance L and a resistor of resistance R are joined… | MHT CET 2026 (19 May, Shift 1)

Q7MHT CET 2026 · 19 MAY · SHIFT 1Medium

Power, Power Factor & Wattless Current

An inductor of inductance $L$ and a resistor of resistance $R$ are joined in series and connected to an a.c. source of frequency $\omega$ . The power dissipated in the circuit is

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Answer: (c)

$Z = \sqrt{R^2 + \omega^2L^2}$ , $I = \frac{V}{Z}$ (r.m.s.).
Power $P = I^2R = \frac{V^2R}{R^2 + \omega^2L^2}$ .
So the answer is (c).

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Power, Power Factor & Wattless Current question from MHT CET 2026 (19 May, Shift 1)

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