An inductor of reactance 100 Ω, a capacitor of reactance 50 Ω and a resistor of resistance 50 Ω are connected in series to a 10 V rms AC source. Find the average power dissipated in watts.

Question

G1 Physics · Accepted Answer

For a series RLC circuit, Z = √[R² + ( X L − X C ) 2 ] . X_L-X_C)²]. X L ​ − X C ​ ) 2 ] . Given, R = 50 Ω, X L X_L X L ​ = 100 Ω, X C X_C X C ​ = 50 Ω. Therefore, Z = √[50²+(100-50)²] = √(2500+2500) = 50√2 Ω. Average power is P = V r m s 2 R / Z 2 . V_rms²R/Z². V r ​ m s 2 R / Z 2 . Thus, P = (10²)(50)/(50√2)² = 100×50/(5000) = 1 W. Answer: 1.