An inductor 20   mH , a capacitor 50  μ F and a resistor 40   Ω are connected in series across a source of emf V = 10   sin   ⁡ 340   t . The power loss in AC . circuit is:

Step-by-step solution

Given : R=40Ω , L= 20 mH , C=50 μF Comparing the given equation for voltage with general equation V = V 0   sin   ωt ω = 340   rad   s - 1 and peak voltage V 0 = 10   V Inductive reactance X L ω L = 340 × 20 × 10 - 3 = 68 × 10 - 1 = 6.8   Ω Capacitive reactance X C 1 ω C = 1 340 × 50 × 10 - 6 = 10 4 34 × 5 = 2 34 × 10 3 = 0.0588 × 10 3 = 58.82   Ω   Z= X L -