Question

A series LCR circuit containing a resistor of $120\Omega$ has an angular resonant frequency $4\times {10}^{5}rad{s}^{-1}$. At resonance the voltages across resistor and inductor are $60\text{V}$ and $40\text{V}$ respectively.
At what frequency, the current in the circuit lags the voltage by ${45}^{\circ }$?

• A. $4\times {10}^{5}rad{s}^{-1}$
• B. $3\times {10}^{5}rad{s}^{-1}$
• C. $8\times {10}^{5}rad{s}^{-1}$
• D. $2\times {10}^{5}rad{s}^{-1}$

Answer 1

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


After resolving this phasor we get

From the above phasor, we can say
$\tan {45}^{\circ }=\frac{X_L-X_C}{R}\Rightarrow 1=\frac{\omega L-\frac{1}{\omega C}}{R}$ .....(i)
Now given that $R=120\Omega$
and
Now we know that when the circuit was given at resonance,
voltage across the inductor $V_L=40V$
So $X_L=X_C$ and also $V_L=V_C$
and the circuit will behave as purely resistive.
$\Rightarrow Z=R$
$\therefore I_{rms}=\frac{V_{rms}}{Z}=\frac{V_{rms}}{R}\Rightarrow I_{rms}=\frac{60}{120}=\frac{1}{2}\text{A}$

$\Rightarrow V_L=I_{rms}\times \omega \times L$
$\Rightarrow 40=0.5\times 4\times {10}^5\times L\Rightarrow L=0.2\text{m}H$


$V_C=i_{rms}\times \frac{1}{\omega C}\Rightarrow 40=\frac{1}{2}\times \frac{1}{\omega C}\Rightarrow C=\frac{1}{32}\mu F$

Putting these values in eq(i)
$120=\omega \times 2\times {10}^{-4}-\frac{1}{\omega \frac{1}{32}\times {10}^{-6}}$
On solving the quadratic equation we get
$\omega =8\times {10}^5\text{rad/s}$