Question

A pure resistive circuit element $X$ when connected to an ac supply of peak voltage $200\,V$ gives a peak current of $5\,A$ which is in phase with the voltage. A second circuit element $Y$, when connected to the same ac supply also gives the same value of peak current but the current lags behind by $90^?$. If the series combination of $X$ and $Y$ is connected to the same supply, what will be the rms value of current?

• A. $\frac{10}{\sqrt{2}}A$
• B. $\frac{5}{\sqrt{2}}A$
• C. $\frac{5}{2}A$
• D. $5\,A$

Answer 1

Answer: (C)

$\frac{5}{2}A$

Answer 2

As current is in phase with the applied voltage, $X$ must be $R$. $R=\frac{V_{0}}{I_{0}}$ $=\frac{200\,V}{5\,A}$ $=40\,\Omega$ As current lags behind the voltage by $90^{\circ}$, $Y$ must be an inductor. $X_{L}=\frac{V_{0}}{I_{0}}$ $=\frac{200\,V}{5\,A}$ $=40\,\Omega$ In the series combination of $X$ and $Y$, $Z=\sqrt{R^{2}+X^{2}_{L}}$ $=\sqrt{40^{2}+40^{2}}$ $=40\sqrt{2}\,\Omega$ $I_{rms}=\frac{V_{rms}}{Z}$ $=\frac{V_{0}}{\sqrt{2}Z}$ $=\frac{500}{\sqrt{2}\left(40\sqrt{2}\right)}$ $=\frac{5}{2}A$