A series LCR circuit containing a resistance of $120\Omega$ has angular resonance frequency  4$\times {10}^5 rad/sec$. At Resonance the voltages across resistance and inductance are $60 \mathrm{V}$ and $40 \mathrm{V}$ respectively. Find the values of L and C. At what frequency the current in the circuit lags the voltage by $45°$ ?

 

At resonance as $X=0$,
$I=\frac{V}{R}=\frac{60}{120}=\frac{1}{2} \mathrm{A}$
and as $V_L=I{X}_L=I\omega L, L=\frac{V_L}{I\omega }$
So, $L=\frac{40}{(1/2)\times 4\times {10}^5}=0.2\mathrm{mH}$
and as ${\omega }_0=\frac{1}{\sqrt{LC}},C=\frac{1}{L{\omega }_0^2}$
i.e., $C=\frac{1}{0.2\times {10}^{-3}\times {\left(4\times {10}^5\right)}^2}=\frac{1}{32}\mu \mathrm{F}$
Now in case of series LCR circuit,
$\tan \varphi =\frac{X_L-X_C}{R}$
So current will lag the applied voltage by $45°$ if,
$\tan 45°=\frac{\omega L-\frac{1}{\omega C}}{R}$
i.e., $1\times 120=\omega \times 2\times {10}^{-4}-\frac{1}{\omega (1/32)\times {10}^{-6}}$
i.e., ${\omega }^2-6\times {10}^5\omega -16\times {10}^{10}=0$
i.e., $\omega =\frac{6\times {10}^5\pm \sqrt{{\left(6\times {10}^5\right)}^2+64\times {10}^{10}}}{2}$
i.e.,$\omega =\frac{6\times {10}^5+10\times {10}^5}{2}=8\times {10}^5\mathrm{rad}/\mathrm{s}$