A circuit element X when connected to an AC supply of peak voltage 100 V gives a peak current of 5 A which is in phase with the voltage. A second element Y when connected to the same AC supply also gives the same value of peak current which lags behind the voltage by π 2 . If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere?

Step-by-step solution

As current is in phase with the applied voltage, element X should be resistive with R = V 0 I 0 = 100 5 = 20   Ω . As current lags behind voltage by 90 ° , element Y should be inductive with X L = V 0 I 0 = 100 5 = 20   Ω When X and Y are connector in series, Impedance, Z = X L 2 + R 2 = 20 2 + 20 2 = 20 2   Ω Now, peak current, I 0 = V 0 Z = 100 20 2 = 5 2   A Thus, the rms value of current is I rms = I 0 2 = 5 2   A