An e.m.f. e = E ₀ t is applied to a circuit containing L , C and R in series where X_L=3 R and X_C=R . The average power dissipated in the circuit is

Step-by-step solution

The average power dissipated is determined by the power dissipated in the resistor. For a sinusoidal AC source e = E₀ t , the average power is P_ avg = I_ rms ² R . The RMS voltage is V_ rms = E₀ 2 . The impedance Z is computed from the resistances: Z = R² + (X_ L - X_ C )² = R² + (3R - R)² = R² + (2R)² = 5R² = R 5 . The RMS current follows as I_ rms = V_ rms Z = E₀ 2 R 5 = E₀ R 10 . Substituting into the power formula: P_ avg = ( E₀ R 10 )² R = E₀² R² 10 R = E₀² 10R . Final answer: E₀² 10R