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 $90^{0}$.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.

$R = \frac{V_{0}}{I_{0}} = \frac{200V}{5A} = 40\Omega$

As current lags behind the voltage by $90^{o}$, 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,

$Z = \sqrt{R^{2} + X_{L}^{2}} = \sqrt{40^{2} + 40^{2}} = 40\sqrt{2}\Omega$

$I_{rms} = \frac{V_{rms}}{Z} = \frac{V_{0}}{\sqrt{2}Z} = \frac{200}{\sqrt{2}(40\sqrt{2})} = \frac{5}{2}A$