Question

A resistor of resistance R, capacitor of capacitance C and inductor of inductance L are connected in parallel to AC power source of voltage ${ϵ}_0\sin \omega t$. The maximum current through the resistance is half of the maximum current through the power source. Then value of R is

• A. $\frac{\sqrt{3}}{|\omega C-\frac{1}{\omega L}|}$
• B. $\sqrt{3}|\frac{1}{\omega C}-\omega L|$
• C. $\sqrt{5}|\frac{1}{\omega C}-\omega L|$
• D. None of these

Answer 1

The correct option is A $\frac{\sqrt{3}}{|\omega C-\frac{1}{\omega L}|}$

The phasor diagram for the situation is shown in figure.

$\frac{i_{R_0}}{\sqrt{(i_{R_0}{)}^2+{({ϵ}_0\omega C-\frac{{ϵ}_0}{\omega L})}^2}}=\frac{1}{2}$
or,
$\frac{i_{R_0}}{\sqrt{{\left(i_{R_0}\right)}^2+{(i_{C_0}-i_{L_0})}^2}}=\frac{1}{2}$ $\Rightarrow \frac{\frac{{\in }_0}{R}}{\sqrt{{\left(\frac{{\in }_0}{R}\right)}^2+{({\in }_0\omega C-\frac{{\in }_0}{\omega L})}^2}}=\frac{1}{2}$
$\Rightarrow R=\frac{\sqrt{3}}{|\omega C-\frac{1}{\omega L}|}$