Question

A series $\text{LCR}$ circuit containing a resistance of $120\Omega$ has angular resonance frequency ${\omega }_0=4\times {10}^5{\text{rad s}}^{–1}$. At resonance, the voltage across resistance and inductance are $60\text{V}$ and $40\text{V}$ respectively. When angular frequency $\omega =n\times {10}^5{\text{rads}}^{–1}$, the current in the circuit lags the voltage by ${45}^{\circ }.$ The value of $n$ is (integer only).

• A. 8.00
• B. 8.0
• C. 8

Answer 1

Answer: (A), (B), (C)

Given,

$R=120\Omega ;{\omega }_0=4\times {10}^5{\text{rad s}}^{–1}{V}_R=60\text{V};V_L=40\text{V}$

At resonance, impedance $Z=R$

$I=\frac{V}{R}=\frac{60}{120}=\frac{1}{2}\text{A}$

$V_L=I\times X_L=I\times \omega L$

$\therefore L=\frac{V_L}{I{\omega }_0}=\frac{40\times 2}{4\times {10}^5}=2\times {10}^{-4}\text{H}$

The Resonance frequency is given by

${\omega }_0=\frac{1}{\sqrt{LC}}$

$C=\frac{1}{{\omega }_0^{2}L}=\frac{1}{2\times {10}^{-4}\times 16\times {10}^{10}}=\frac{1}{32\times {10}^6}$

Given, current in the circuit lags the applied voltage by ${45}^{\circ }$

$\therefore \tan {45}^{\circ }=\frac{\omega L-\frac{1}{\omega C}}{R}$

$R=\omega L-\frac{1}{\omega C}$

$120=\omega \times 2\times {10}^{-4}-\frac{32\times {10}^6}{\omega }$

${\omega }^2\times 2\times {10}^{-4}-120\omega -32\times {10}^6=0$

$\omega =\frac{120\pm \sqrt{(120{)}^2+4\times 2\times {10}^{-4}\times 32\times {10}^6}}{2\times 2\times {10}^{-4}}$

$\omega =\frac{120\pm 200}{4\times {10}^{-4}}$

On taking (+)

$\omega =8\times {10}^5{\text{rad s}}^{–1}$

$\therefore n=8$