A series AC circuit containing an inductor (20 mH), a capacitor (120 μ F) and a resistor (60 Ω) is driven by an AC source of 24V/50 Hz. The energy dissipated in the circuit in 60 s is: (a) 5.65 × 102J (b) 2.26 × 103J (c) 5.17 × 102 J (d) 3.39 × 103 J

Answer: (c). ω = 100π: XL = 100π× 0.02 ≈ 6.28 Ω, XC = 1100π× 120×10⁻⁶ ≈ 26.5 Ω Z = √(60² + 20.2²) ≈ 63.3 Ω, I = 24/63.3 ≈ 0.379 A P = I² R ≈ 8.62 W E = P t = 8.62 × 60 ≈ 5.17× 10² J, so the answer is (c).

A series AC circuit containing an inductor (20 mH), a capacitor (120… | JEE Main 2019 (9 Jan, Shift 2)

Q63JEE Main 2019 · 9 JAN · SHIFT 2Difficult

Power, Power Factor & Wattless Current

A series AC circuit containing an inductor (20 mH), a capacitor (120 $\mu$ F) and a resistor (60 $\Omega$ ) is driven by an AC source of 24V/50 Hz. The energy dissipated in the circuit in 60 s is:

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Answer: (c)

$\omega = 100\pi$ : $X_L = 100\pi\times 0.02 \approx 6.28\ \Omega$ , $X_C = \frac{1}{100\pi\times 120\times10^{-6}} \approx 26.5\ \Omega$
$Z = \sqrt{60^2 + 20.2^2} \approx 63.3\ \Omega$ , $I = 24/63.3 \approx 0.379$ A
$P = I^2 R \approx 8.62$ W
$E = P t = 8.62 \times 60 \approx 5.17\times 10^2$ J, so the answer is (c).

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