Question

A series $L-C-R$ circuit with $R=120\Omega$ has an angular frequency $4\times 10^{5}rad{s}^{-1}$. At resonance, voltages across resistance and inductor are $60V$ and $40V$ respectively. If frequency at which the circuit current lags behind the voltage by a phase of $\frac{\pi }{4}$ radian is $k\times 10^{5}rad{s}^{-1}$, then the value of $k$ is___

Answer 1

Answer: 8

At resonance,
$\Rightarrow I=\frac{V}{R}=\frac{60}{120}=\frac{1}{2}A$
and as $V_L=I{X}_L=I\omega L$
$\Rightarrow L=\frac{V_L}{I\omega }=\frac{40}{\frac{1}{2}\times 4\times 10^5}=0.2mH$
Also, ${\omega }_r=\frac{1}{\sqrt{LC}},C=\frac{1}{L{\omega }_r^2}$
$\Rightarrow C=\frac{1}{0.2\times 10^{-3}\times \left(4\times 10^5\right)}=\frac{1}{32}\mu F$
In case of series $L-C-R$ circuit
$\frac{X_L-X_C}{R}=\tan \varphi$
As, current lags behind voltage by $\frac{\pi }{4}rad$.
$X_L-X_C=R$
$\Rightarrow L\omega -\frac{1}{C\omega }=R$
$\Rightarrow \omega \times 2\times 10^{-4}-\frac{1}{\omega \times \left(\frac{1}{32}\right)\times 10^{-6}}=1\times 120$
$\Rightarrow {\omega }^2-6\times 10^5\omega -16\times 10^{10}=0$
$\omega =\frac{6\times 10^5\pm \sqrt{{\left(6\times 10^5\right)}^2\left(64\times 10^{10}\right)}}{2}=8\times 10^{5}rad{s}^{-1}$
$\therefore k=8$