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 $$\frac{\pi}{2}$$. If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere?

We have element X which passes peak current 5 A in phase with the peak voltage 100 V, so X is a resistor with resistance $$R = \frac{100}{5} = 20\;\Omega$$.

Element Y passes the same peak current of 5 A but lagging the voltage by $$\frac{\pi}{2}$$, so Y is an inductor with inductive reactance $$X_L = \frac{100}{5} = 20\;\Omega$$.

When X and Y are connected in series, the total impedance is $$Z = \sqrt{R^2 + X_L^2} = \sqrt{20^2 + 20^2} = 20\sqrt{2}\;\Omega$$.

The peak current in the series circuit is $$I_0 = \frac{V_0}{Z} = \frac{100}{20\sqrt{2}} = \frac{5}{\sqrt{2}}$$ A.

The rms current is $$I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{5}{\sqrt{2} \cdot \sqrt{2}} = \frac{5}{2}$$ A.

Hence, the correct answer is Option 4.