Question

A series LCR circuit containing a resistance of 120 $\Omega$ has angular resonance frequency $4\times {10}^{5}rad{s}^{-1}$. At resonance the voltage across the resistance and inductance are 60 V and 40 V respectively. The angular frequency at which current in the circuit lags the voltage by ${45}^0$ is

• A. $16\times {10}^{5}rad{s}^{-1}$
• B. $8\times {10}^{5}rad{s}^{-1}$
• C. $4\times {10}^{5}rad{s}^{-1}$
• D. $2\times {10}^{5}rad{s}^{-1}$

Answer 1

The correct option is B $8\times {10}^{5}rad{s}^{-1}$

Given:

${\omega }_{res}=4\times {10}^{5}R=120\Omega V_R=60\Omega V_L=40\Omega \varphi ={45}^0$

Now, $w_{res}=\frac{1}{\sqrt{LC}}=4\times {10}^5$

$LC=\frac{1}{16}\times {10}^{-20}$.......(1)

Also, at resonance assume $I_{rms}=i$

$\Rightarrow 60=\left(i\right)\left(120\right)$

and $40=\left(i\right)\left(X_L\right)=\left(i\right)\left(\right(4\times {10}^5\left)L\right)$

Dividing both we get,

$L=2\times {10}^{-4}H$.......(2)

From (1) and (2), we get

$C=\frac{1}{32}\times {10}^{-6}F$

For $4s$ phase angle, $\tan \varphi =\tan {45}^o=1=\frac{X_L-X_C}{R}$

$\Rightarrow \omega L-\frac{1}{\omega C}=120$

$\Rightarrow {\omega }^2(2\times {10}^{-4})-\frac{1}{\frac{1}{32}\times {10}^{-6}}=120\left(w\right)$

$\Rightarrow \omega =8\times {10}^{5}rad/s$