An alternating voltage E =100 2 (50 t ) is connected to a 2 F capacitor through an a.c. ammeter. The ammeter reading will be

Step-by-step solution

The alternating voltage is given by E = 100 2 (50t) , corresponding to a peak voltage E₀ = 100 2 V and angular frequency = 50 rad/s . For the capacitor with capacitance C = 2 10⁻⁶ F , the capacitive reactance is X_C = 1 C = 1 50 2 10⁻⁶ = 10^4 . The peak current is I₀ = E₀ X_C = 100 2 10^4 = 0.01 2 A . Since the ammeter measures the RMS current, we compute I_ rms = I₀ 2 = 0.01 2 2 = 0.01 A . Converting to milliamperes gives I_ rms = 10 mA . The ammeter reading is 10 mA .