A circuit element 'X' when connected to peak voltage of 200 V, a peak current of 5 A flows which lags behind the voltage by $\frac{\pi}{2}$. A circuit element Y when connected to same peak voltage, same peak current flows which is in phase with the voltage. Now X and Y are connected in series with same peak voltage. The rms value of current through the circuit will be:

2.5 A

The correct answer is Option (3) → 2.5 A

Using Ohm's law for AC circuits,

$V_p=I_pX_L$

$⇒X_L=\frac{V_p}{I_p}=\frac{200}{5}=40Ω$

for element y (Resistor),

$V_p=I_pR$

$⇒R=\frac{V_p}{I_p}=\frac{200}{5}=40Ω$

Net impedance, $Z=\sqrt{R^2+{X_L}^2}

$=\sqrt{(40)^2+(40)^2}$

$=40\sqrt{2}Ω$

$V_{rms}=\frac{V_p}{\sqrt{2}}=\frac{200}{\sqrt{2}}=100\sqrt{2}V$

Using Ohm's law,

$I_{rms}=\frac{V_{rms}}{Z}=\frac{100\sqrt{2}}{40\sqrt{2}}$

$=2.5A$