Question

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

Answer 1

At resonance as $X=0,I=\frac{V}{R}=\frac{60}{120}=\frac{1}{2}A$
and $V_L=I{X}_L=I\omega L,L=\frac{V_L}{I\omega }\Rightarrow L=\frac{40}{0.5\times 4\times {10}^5}=0.20mH$
i.e., $C=\frac{1}{L{\omega }_0^2}=\frac{1}{0.2\times {10}^{-3}\times {(4\times {10}^5)}^2}=\frac{1}{32}\mu F$
Now in case of series LCR circuit
$\tan \varphi =\frac{X_L-X_C}{R}$
So current will lag the applied voltage by ${45}^o$ if $\tan {45}^o=\frac{\omega L-\frac{1}{\omega C}}{R}$
ie., $1\times 120=\omega \times 2\times {10}^{-4}-\frac{1}{\omega \left(1/32\right)\times {10}^{-6}}$
ie., ${\omega }^2-6\times {10}^5\omega -16\times {10}^{10}=0$
ie., $\omega =\frac{6\times {10}^5+10\times {10}^5}{2}=8\times {10}^{5}rad/s$