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)

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 $Y$,
$Z=\sqrt{R^{2}+X^{2}_{L}}$
$=\sqrt{40^{2}+40^{2}}$
$=40\sqrt{2}\,\Omega$
$I_{rms}=\frac{V_{rms}}{Z}$
$=\frac{V_{0}}{\sqrt{2}Z}$
$=\frac{500}{\sqrt{2}\left(40\sqrt{2}\right)}$
$=\frac{5}{2}A$