An AC source carries a rating of 220 V , 50 Hz . When the average voltage is evaluated over a time interval of 0.01 s , this average

Step-by-step solution

Frequency of the AC source, f = 50 Hz Time period, T = 1 f = 1 50 = 0.02 s The given time interval for evaluating the average voltage is t = 0.01 s , which is exactly T 2 . The instantaneous voltage of the AC source can be represented as V = V₀ ( t + ) . The average voltage over an interval t from t₁ to t₁ + t is given by: V_ avg = 1 t _ t₁ ^ t₁ + t V₀ ( t + ) dt Since t = T 2 , the integral evaluates the area under the voltage curve over half a cycle. If the time interval starts at t₁ such that t₁ + = 0 , the aver