Question

An inductor $(X_L=2\Omega )$ , a capacitor $(X_C=8\Omega )$ and a resistance $\left(8\Omega \right)$ are 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 potential difference 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 $x$________

Answer 1

Impedance of circuit , $Z=\sqrt{R^2+(X_C-X_L{)}^2}$

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

The current leads in phase by , $\varphi ={37}^{\circ }$ and $X_C>X_L$

$\therefore i=\frac{10\cos (100\pi t+{37}^{\circ })}{Z}=\cos (100\pi t+{37}^{\circ })$

The instantaneous potential difference across $AB$ is, $=I_m(X_C-X_L)\cos (100\pi t+{37}^{\circ }-{90}^{\circ })$
$=6\cos (100\pi t-{53}^{\circ })$

The instantaneous potential difference across $AB$ is half of source voltage.
$\Rightarrow 6\cos (100\pi t-{53}^{\circ })=5\cos 100\pi t$

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

$\therefore$ instantaneous potential difference $=5\times \frac{24}{25}=\frac{24}{5}\text{volts}\text{or}48\times {10}^{-1}\text{V}$

Thus, $x=48$