A circuit element 'X' when connected to peak voltage of 200 V, a peak current of 5A flows which lags behind the voltage by \( \displaystyle\frac{\pi}{2} \). 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:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 2.5\,\mathrm{A} \)
A current lagging the voltage by \( \pi/2 \) means X is a pure inductor; a current in phase with the voltage means Y is a pure resistor.
\( \displaystyle X_L = \frac{V_0}{I_0} = \frac{200\,\mathrm{V}}{5\,\mathrm{A}} = 40\,\Omega \)
\( \displaystyle R = \frac{200\,\mathrm{V}}{5\,\mathrm{A}} = 40\,\Omega \)
In series the impedance is
\( \displaystyle Z = \sqrt{R^{2} + X_L^{2}} = \sqrt{40^{2} + 40^{2}} = 40\sqrt{2}\,\Omega \)
\( \displaystyle I_0' = \frac{V_0}{Z} = \frac{200}{40\sqrt{2}} = \frac{5}{\sqrt{2}}\,\mathrm{A} \)
This is the peak current. The rms value is
\( \displaystyle I_{\mathrm{rms}} = \frac{I_0'}{\sqrt{2}} = \frac{5}{\sqrt{2}\times\sqrt{2}} = 2.5\,\mathrm{A} \)