Question

Consider an $L-C-R$ circuit as shown in figure, with an $AC$ source of peak value $V_0$ and angular frequency $\omega$. Then the peak value of current through the $AC$ source is :

• A. $\frac{V_0}{\sqrt{\frac{1}{R^2}+{(\omega L-\frac{1}{\omega C})}^2}}$
• B. $V_0\sqrt{1/R^2+{(\omega C-\frac{1}{\omega L})}^2}$
• C. $\frac{V_0}{{\sqrt{R^2+(\omega L-\frac{1}{\omega C})}}^2}$
• D. None of these.

Answer 1

Answer: (B)

$V_0\sqrt{1/R^2+{(\omega C-\frac{1}{\omega L})}^2}$

As R, L , C are in parallel, equivalent impedance is given by:

$\frac{1}{Z}=\frac{1}{Z_R}+\frac{1}{Z_L}+\frac{1}{Z_C}=\frac{1}{R}+\frac{1}{jL\omega }+jC\omega$
$\left|\frac{1}{Z}\right|=|\frac{1}{R}+\frac{1}{jL\omega }+jC\omega |=\sqrt{\frac{1}{R^2}+{(\frac{1}{L\omega }-C\omega )}^2}$

Peak value of current is given by:

$I_{peak}=\left|\frac{V_0}{Z}\right|=V_0\sqrt{\frac{1}{R^2}+{(\frac{1}{L\omega }-C\omega )}^2}$