A capacitor of reactance 4√(3) Ω and a resistor of resistance 4 Ω are connected in series with an ac source of peak value 8√(2) V. The power dissipation in the circuit is __________ W. Numerical answer type. Work it out, then open the answer.

Answer: 4. Z = √(R² + XC²) = √(16 + 48) = 8 Ω. Vrms = (8√(2))/(√(2)) = 8 V, so Irms = 8/8 = 1 A. P = Irms²R = 1 × 4 = 4 W.

A capacitor of reactance 4√(3) Ω and a resistor of resistance 4 Ω are… | JEE Main 2024 (9 Apr, Shift 2)

Q38JEE Main 2024 · 9 APR · SHIFT 2Medium

Power, Power Factor & Wattless Current

A capacitor of reactance $4\sqrt{3}\ \Omega$ and a resistor of resistance $4\ \Omega$ are connected in series with an ac source of peak value $8\sqrt{2}\text{ V}$ . The power dissipation in the circuit is __________ W.

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

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Answer: 4

$Z = \sqrt{R^2 + X_C^2} = \sqrt{16 + 48} = 8\ \Omega$ .
$V_{rms} = \frac{8\sqrt{2}}{\sqrt{2}} = 8\text{ V}$ , so $I_{rms} = \frac{8}{8} = 1\text{ A}$ .
$P = I_{rms}^2R = 1 \times 4 = 4\text{ W}$ .

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