Question

An $\text{LCR}$ series circuit containing a resistance of $120\Omega$ has angular resonance frequency $4\times {10}^5\text{rad/s}$. At resonance, the voltage across resistance and inductance are $60\text{V}$ and $40\text{V}$ respectively. At what frequency the current in the circuit lags the voltage by ${45}^{\circ }$. Give answer in $\times {10}^5\text{rad/sec}$.

Answer 1

Given , $R=120\Omega ,{\omega }_0=4\times {10}^5{\text{rads}}^{-1},V_R=60\text{V};V_L=40\text{V}$

At resonance , $V_{rms}=V_R=60\text{V}$

$\therefore I_{rms}=\frac{V_R}{R}=\frac{60}{120}=0.5\text{A}$

Thus, $V_L=V_C$

$V_L=I_{rms}\times L\omega =0.5\times L\times 4\times {10}^5$

$40=2\times {10}^5\times L$

$\Rightarrow L=2\times {10}^{-4}\text{H}=0.2\text{mH}$

$V_C=I_{rms}\times \frac{1}{C\omega }$

$40=0.5\times \frac{1}{4\times {10}^{-5}C}$

$\therefore C=\frac{0.5}{16}\times {10}^{-6}=\frac{1}{32}\mu \text{F}$

Given that current will lag the applied voltage by ${45}^{\circ }$

$\tan {45}^{\circ }=\frac{\omega L-\frac{1}{\omega C}}{R}$
$\omega L-\frac{1}{\omega C}=R$
$\Rightarrow \omega (2\times {10}^{-4})-\frac{32\times {10}^6}{\omega }=120$
$\Rightarrow {\omega }^2(2\times {10}^{-4})-32\times {10}^6-120\omega =0$
$\Rightarrow \omega =8\times {10}^5\text{rad/sec}$