Question

A pure resistive circuit element $X$ when connected to an $AC$ supply of peak voltage $200V$ gives a peak current of $5A$. A second current element $Y$ when connected to same $AC$ supply gives the same value of peak current but the current lags behind by ${90}^{\circ }$. If series combination of $X$ and $Y$ is connected to the same supply, what is the impendance of the cirucit?

• A. $40\Omega$
• B. $80\Omega$
• C. $40\sqrt{2}\Omega$
• D. $2\sqrt{40}\Omega$

Answer 1

The correct option is C $40\sqrt{2}\Omega$

Pure resistive, $R=\frac{E_0}{l_0}=\frac{200}{5}=40\Omega$
AS current lags behind the applied voltage by ${90}^{\circ }$, therefore element $Y$ must be pure inductor.
$X_L=\frac{E_0}{l_0}=\frac{200}{5}=40\Omega$
$\therefore$ Total impedance,
$Z=\sqrt{R^2+X_L^2}=\sqrt{{40}^2+{40}^2}$
$==40\sqrt{2}\Omega$.