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 $60 \,V$ and $40 \,V$ 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

At resonance,
$\Rightarrow I=\frac{V}{R}=\frac{60}{120}=\frac{1}{2} A $
and as $ V_{L}=I X_{L}=I \omega L $
$\Rightarrow L=\frac{V_{L}}{I \omega}=\frac{40}{\frac{1}{2} \times 4 \times 10^{5}}=0.2\, mH$
Also, $\omega_{r}=\frac{1}{\sqrt{L C}}, C=\frac{1}{L \omega_{r}^{2}}$
$\Rightarrow C=\frac{1}{0.2 \times 10^{-3} \times\left(4 \times 10^{5}\right)}=\frac{1}{32}\, \mu F$
In case of series $L-C-R$ circuit
$\frac{X_{L}-X_{C}}{R}=\tan \phi$
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} $
$\therefore k=8$