Q 12-07-159JEE MainJEE Main 2025 (7 Apr, Shift 2)Easy
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 $10\ \text{V}$, $50\ \text{Hz}$. The average power dissipated by the circuit is ______ W.
Numerical value type. Enter your answer.
Answer: 1
$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{50^2 + 50^2} = 50\sqrt2\ \Omega$, so $Z^2 = 5000\ \Omega^2$.
Taking $10\ \text{V}$ as the rms value:
$$P = I_{\text{rms}}^2R = \frac{V_{\text{rms}}^2R}{Z^2} = \frac{100\times50}{5000} = 1\ \text{W}$$
Solution by Sreeraj P, M.Sc Physics