Question

A voltage $V_{PQ}=V_0\cos \omega t$ (where $V_0$ is a real amplitude) is applied between the points P and Q in the network shown in the figure. The values of capacitance and inductance are $C=\frac{1}{\omega R\sqrt{3}}$ and $L=\frac{R\sqrt{3}}{\omega }$
Then, the total impedance between P and Q is

• A. 1.5 R
• B. 2R
• C. 3R
• D. 4R
• E. 2.5 R

Answer 1

Answer: (C)

$\because$ Given that, $C=\frac{1}{\omega R\sqrt{3}},L=\frac{R\sqrt{3}}{\omega }$

In the above figure,
$Z_1=R+j\omega L=R+j\omega \left(\frac{R\sqrt{3}}{\omega }\right)=R+jR\sqrt{3}$
$Z_2=R-j\frac{1}{\omega c}=R-j\frac{1}{\omega \left(\frac{1}{\omega R\sqrt{3}}\right)}=R-jR\sqrt{3}$
$\because$ Impedance $Z_1$ and $Z_2$ are in parallel,
So, $Z_{\text{eq }}=\frac{Z_1{Z}_2}{Z_1+Z_2}$
$=\frac{(R+jR\sqrt{3})(R-jR\sqrt{3})}{R+jR\sqrt{3}+R-jR\sqrt{3}}$
$=\frac{R^2+R^2(\sqrt{3}{)}^2}{2R}$
$\left(\because (a-b)(a+b)=a^2-b^2\right)$
$Z_{eq}=\frac{4R^2}{2R}=2R$
So, total impedance between $P$ and $Q$ is
$Z_{PQ}=R+Z_{eq}$
$Z_{PQ}=R+2R=3R$