Question

A pure resistive circuit element $X$ when connected to an ac supply of peak voltage $200V$ gives a peak current of $5A$ 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. $5A$

Answer 1

Answer: (C)

As current is in phase with the applied voltage, $X$ must be $R$.
$R=\frac{V_0}{I_0}$
$=\frac{200V}{5A}$
$=40\Omega$
As current lags behind the voltage by $90^{\circ }$, $Y$ must be an inductor.
$X_L=\frac{V_0}{I_0}$
$=\frac{200V}{5A}$
$=40\Omega$
In the series combination of $X$ and $Y$,
$Z=\sqrt{R^2+X_L^2}$
$=\sqrt{40^2+40^2}$
$=40\sqrt{2}\Omega$
$I_{rms}=\frac{V_{rms}}{Z}$
$=\frac{V_0}{\sqrt{2}Z}$
$=\frac{500}{\sqrt{2}\left(40\sqrt{2}\right)}$
$=\frac{5}{2}A$