Question

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

• A. $2\times {10}^{5}rad/s$
• B. $6\times {10}^{5}rad/s$
• C. $8\times {10}^{5}rad/s$
• D. $10\times {10}^{5}rad/s$

Answer 1

The correct option is C $8\times {10}^{5}rad/s$

$w=4\times {10}^5,R=120,I=\frac{60}{120}=0.5A$

at resonance $V_R=60V,$ [at resonance, whole voltage is dropped on resistor]

$V_L=40V$

$V_C=-40V$

For inductor, $\omega L=\frac{40}{0.5}$

$L=\frac{80}{4\times {10}^5}=0.2mH$

for capacitor $\frac{1}{wC}=\frac{40}{0.5}$

$C=3.125\times {10}^{-8}F$

$cos\frac{\pi }{4}=\frac{R}{Z}$

$\Rightarrow Z=\sqrt{2}R$

$Z=\sqrt{R^2+(\omega L-\frac{1}{wC}{)}^2}=\sqrt{2}R$

squaring on both sides

$R^2+(\omega L-\frac{1}{\omega C}{)}^2=2R^2$

$\Rightarrow {(\omega L-\frac{1}{\omega C})}^2=R^2$

Now putting the values of $L,C$ and $R$ we can find the value of $\omega$

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