A capacitor of capacitance ( 10 F ) is connected to an AC source and an AC Ammeter. If the source voltage varies as ( V=50 2 100 t ), the reading of the ammeter is

Step-by-step solution

Given, capacitance =10 F=10 10⁻⁶ F ; source voltage =50 2 100 t and =100 We know, I_ rms = V_ rms X_ C and V_ rms = V_ 2 Now, V_ =50 2 ; X_ C = 1 C Therefore, I_ rms = V_ 2 C= 50 2 2 100 10 10⁻⁶I_ rms =5 10⁴ 10⁻⁶=50 10⁻³ Therefore, average value of ac current over a cycle is 50 ~mA