An inductor ($X_L$ = 2$\Omega$) a capacitor ($x_c$ = 8$\Omega$) and a resistance (8$\Omega$) is connected in series with an ac source. The voltage output of A.C source is given by v = 10 cos 100$\pi$t. The instantaneous p.d. between A and B is equal to x $\times {10}^{-1}$ volt, when it is half of the voltage output from source at that instant Find out value of x.

impedance of circuit $=\sqrt{R^2+{\left(X_C-X_L\right)}^2}$

$Z=\sqrt{8^2+(8-2{)}^2}=10\Omega$

The current leads in phase by $\varphi$ = 37°

$\therefore \mathrm{i}=\frac{10\cos \left(100\pi \mathrm{t}+37°\right)}{\mathrm{Z}}=\cos \left(100\pi \mathrm{t}+37°\right)$      $\left(X_C>X_L\right)$

The instantaneous potential difference across A B is

AB=$I_m\left(X_C-X_L\right)\cos \left(100\pi t+37°-90°\right)$

$=6\cos \left(100\pi t-53°\right)$

The instantaneous potential difference across A B is half of source voltage.

AB $\Rightarrow 6\cos \left(100\pi t-53°\right)=5\cos 100\pi t$

solving we get $\cos 100\pi t=\frac{1}{\sqrt{1+(7/24{)}^2}}=\frac{24}{25}$

 $\therefore$ instantaneous potential difference $$\begin{gathered} =5\times \frac{24}{25}=\frac{24}{5}\text{ volts }=48\times {10}^{-1} \mathrm{V} \\ x=48 \end{gathered}$$