A capacitor of capacitance 10 μ F is connected to an AC ammeter. If the source voltage varies as V = 50 2 sin ⁡ 100 t , the reading of the ammeter is

Step-by-step solution

C = 10  μ F = 10 × 1 0 - 6  F V = 50 2 sin ⁡ 100 t Reading of ammeter I = V r m s X C X C = impendence of capacitor X C = 1 ω C I = V r m s 1 ω C V r m s = V 2 = 50 2 2 = 50   V Reading of ammeter I = 50 × 100 × 10 × 1 0 - 6 A = 50 × 10 - 3   A ⇒ 50   mA