Question

Consider $L,C,R$ circuit as shown in figure, with a.c. source of peak value $V$ and angular frequency $\omega$. Then the peak value of current through the ac source.

• A. $\frac{V}{\sqrt{\frac{1}{R^2}+{(\omega L-\frac{1}{\omega C})}^2}}$
• B. $V\sqrt{\frac{1}{R^2}+{(\omega C-\frac{1}{\omega L})}^2}$
• C. $\frac{V}{\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}}$
• D. $\frac{VR\omega C}{\sqrt{{\omega }^2{C}^2+R({\omega }^2{C}^2-1{)}^2}}$

Answer 1

The correct option is B $V\sqrt{\frac{1}{R^2}+{(\omega C-\frac{1}{\omega L})}^2}$

$X_c=\frac{1}{j\omega c}$

$X_L=j\omega L$

And $R$ are in parallel,

Therefore, equivalent admittance of the circuit is

$Y=\frac{1}{X_c}+\frac{1}{X_L}+\frac{1}{R}.$

$Y=\frac{1}{R}+j(c\omega -\frac{1}{L\omega })$

$\left|Y\right|=\sqrt{\frac{1}{R^2}+({\omega c-\frac{1}{\omega L})}^2}$

Current $I=VY=V\sqrt{\frac{1}{R^2}+({\omega c-\frac{1}{\omega L})}^2}$