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 ___________ W. Numerical answer type. Work it out, then open the answer.

Answer: 1. Z = √(50² + (100 - 50)²) = 50√(2) Ω, so Z² = 5000 Ω². P = Vrms²RZ² = (100 × 50)/(5000). P = 1 W.

An inductor of reactance 100 Ω, a capacitor of reactance 50 Ω, and a… | JEE Main 2025 (7 Apr, Shift 2)

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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 $10\text{ V}, 50\text{ Hz}$ . Average power dissipated by the circuit is ___________ W.

Numerical answer type. Work it out, then open the answer.

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Answer: 1

$Z = \sqrt{50^2 + (100 - 50)^2} = 50\sqrt{2}\ \Omega$ , so $Z^2 = 5000\ \Omega^2$ .
$P = \frac{V_{rms}^2R}{Z^2} = \frac{100 \times 50}{5000}$ .
$P = 1\text{ W}$ .

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