An inductor 20 mH, capacitor 100   μF and a resistor 50  Ω are connected in series across a source of emf, V =   10 sin ⁡( 314 t ⁡) . The power loss in the circuit is

Step-by-step solution

V 0 = 10 V , ω = 314 rad/s P = V r m s I r m s cos ⁡ ϕ P = V r m s V r m s Z R Z = ( V r m s ) 2 R Z 2 X L = ω L = 314 20 × 10 - 3 = 6.280 X C = 1 ω C = 1 314 × 100 × 10 - 6 = 31.84 Ω R = 50 Ω Z = ( X C − X L ) 2 + R 2 = 31.84 - 6.28 2 + 50 2 = 56 Ω ⇒ P = ( 10 2 ) 2 × 50 ( 56 ) 2 = 0.79 W