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

The correct option is C $\frac{5}{2}$A

Given, Peak current $5A,voltage=200V,angle={90}^0$

As the current is in the phase with the applied voltage X be R

$R=\frac{V_0}{I_0}=\frac{200v}{5A}=40\Omega$

As a current lags behind the voltage by ${90}^0$ 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:

$z=\sqrt{R^2+X_L^2}=\sqrt{{40}^2+{40}^2}=40\sqrt{2}$

$I_{rms}=\frac{V_{rms}}{z}=\frac{V_0}{z\sqrt{2}}=\frac{200}{\sqrt{2}\times 40\sqrt{2}}=\frac{5}{2}A$