Q 12-07-142JEE MainJEE Main 2019 (9 Jan, Shift 2)Medium
A series AC circuit containing an inductor $(20\ \text{mH})$, a capacitor $(120\ \mu\text{F})$ and a resistor $(60\ \Omega)$ is driven by an AC source of $24\ \text{V}/50\ \text{Hz}$. The energy dissipated in the circuit in $60\ \text{s}$ is:
Answer: (A) $5.17\times10^2\ \text{J}$
$X_L = 2\pi(50)(0.02) = 6.28\ \Omega$, $X_C = \dfrac{1}{2\pi(50)(120\times10^{-6})} = 26.5\ \Omega$.
$$Z = \sqrt{60^2 + (26.5-6.28)^2} = \sqrt{3600+409} \approx 63.3\ \Omega$$
$I_{rms} = \dfrac{24}{63.3} = 0.379\ \text{A}$, so $P = I_{rms}^2R = 0.1437\times60 = 8.62\ \text{W}$.
$$E = 8.62\times60 \approx 5.17\times10^2\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics