Question

A series L-C-R circuit with R = 120 $\Omega$ has an angular frequency of $4\times {10}^5$ rad/sec. At resonance, voltages across resistance and inductor are 60 $V$ and 40 $V$ respectively. If frequency at which the circuit current lags behind the voltage by a phase of π/4 radian is
$k\times {10}^5$rad/sec, then the value of $k$ is
(answer upto two decimal places)

Answer 1

At resonance,
$\Rightarrow I=\frac{V}{R}=\frac{60}{120}=\frac{1}{2}A$
and as $V_L=I{X}_L=IwL$
$\Rightarrow L=\frac{V_L}{I\omega }=\frac{40}{0.5\times 4\times {10}^5}=0.2mH$
Also, ${\omega }_r=\frac{1}{\sqrt{LC}},C=\frac{1}{L{\omega }_r^2}$
$\Rightarrow C=\frac{1}{{0.2\times {10}^{-3}\times (4\times {10}^5)}^2}=\frac{1}{32}\mu F$

In case of series $L-C-R$ circuit
$\frac{X_L-X_C}{R}=\tan \varphi$
As, current lags behind voltage by $\frac{\pi }{4}rad.$
$\Rightarrow L\omega -\frac{1}{C\omega }=R$
$\Rightarrow \omega \times 2\times {10}^{-4}-\frac{1}{\omega \times \left(\frac{1}{32}\right)\times {10}^{-6}}=120$
$\Rightarrow {\omega }^2-6\times {10}^5\omega -16\times {10}^{10}=0$
$\Rightarrow \omega =\frac{6\times {10}^5\pm \sqrt{{(6\times {10}^5)}^2+(64\times {10}^{10})}}{2}=8\times {10}^{5}rad/s$
$\therefore k=8$