Question

An inductor $(L=\frac{1}{100\pi }H)$ a capacitor $(C=\frac{1}{500\pi }F)$ and a resistance $\left(3\Omega \right)$ is connected in series with an AC voltage source as shown in the figure. The voltage of the AC source is given as $V=10cos\left(100\pi t\right)$ volt. What will be the potential difference between A and B?

• A. $8cos(100\pi t-127)$ volt
• B. $8cos(100\pi t-53)$ volt
• C. $8cos(100\pi t-37)$ volt
• D. $8cos(100\pi t+37)$ volt

Answer 1

The correct option is B $8cos(100\pi t-53)$ volt

$V=10\cos \left(100\pi t\right)\}w=100\pi$

total impedance of circuit

$\begin{array}{c}\\ z=\sqrt{({\alpha }_C-X_L{)}^2+R^2}\\ \end{array}\}$ $\begin{array}{c}{\alpha }_C=\frac{1}{wC}=5\\ {\alpha }_L=wL=1\end{array}$

$Z=\sqrt{(5-1{)}^2+3^2}=5$

Phase difference $\varphi ={\tan }^{-1}\frac{X_C-X_L}{R}$

$={\tan }^{-1}\frac{5-1}{3}={\tan }^{-1}\frac{4}{3}$

$\varphi ={53}^{\circ }$

Current $=\frac{V}{Z}=\frac{10}{5}\cos (100\pi t-{53}^{\circ })$

$i=2\cos (100\pi t-{53}^{\circ })$

impedance between $AB$

$(X_C-X_L)=4$

Potential difference between AB

$=iR=2\cos (100\pi t-53)\times 4$

$=8\cos (100\pi t-53)volt$

= option B