JEE Main: Find Average Power in RLC Circuit | Easy Trick Explained ⚡
❓ Question An inductor of reactance X L = 100 Ω X_L = 100~\Omega , a capacitor of reactance X C = 50 Ω X_C = 50~\Omega , and a resistor of resistance R = 50 Ω R = 50~\Omega are connected in series with an AC source of 10 V , f = 50 Hz f = 50~\text{Hz} . The average power dissipated by the circuit is W W . Find the value of W W . 🖼️ Question Image ✍️ Step-by-Step Solution Step 1 — Given data R = 50 Ω , X L = 100 Ω , X C = 50 Ω , V rms = 10 V . Step 2 — Find net reactance The total reactance in a series RLC circuit is the difference between inductive and capacitive reactances: X = X L − X C = 100 − 50 = 50 Ω . Step 3 — Find impedance The total impedance Z Z is given by Z = R 2 + X 2 = 50 2 + 50 2 = 5000 = 70.71 Ω . Step 4 — Find current (rms value) I rms = V rms Z = 10 70.71 = 0.1414 A . Step 5 — Find power factor (cos φ) Power factor for a series RLC circuit: cos ϕ ...