Question

For an $RLC$ circuit driven with a voltage of amplitude$V_m$ and frequency\ ${\omega }_0=\frac{1}{\sqrt{LC}}$ the current exhibits resonance. The quality factor,$Q$is given by:

• A. $\frac{R}{\left({\omega }_{0}C\right)}$
• B. $\frac{CR}{{\omega }_0}$
• C. $\frac{{\omega }_{0}L}{R}$
• D. $\frac{{\omega }_{0}R}{L}$

Answer 1

The correct option is C

$\frac{{\omega }_{0}L}{R}$

Step 1: Given data

It is an $RLC$ circuit it means Resistor, Inductor, Capacitor circuit

Amplitude $=V_m$

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

We know quality factors, $Q=\frac{1}{R}\sqrt{\frac{L}{C}}$

Where,

$L=$Inductance

$R=$Resistance

$C=$Capacitance

And also the question is given that current exhibits resonance, therefore

$$\begin{gathered} X_L=X_C \\ \omega L=\omega C \end{gathered}$$

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

$$\begin{gathered} {\omega }_{0}L=\frac{L}{\sqrt{LC}} \\ {\omega }_{0}L=\sqrt{\frac{L}{C}} \end{gathered}$$

Step 2. To find the quality factor,

Quality factor, $Q=\frac{1}{R}\sqrt{\frac{L}{C}}$

In the question, it is given that current exhibits resonance. So, the condition for resonance is

$$\begin{gathered} {\omega }_{0}L=\sqrt{\frac{L}{C}} \\ \sqrt{\frac{L}{C}}=\omega L \\ Q=\frac{1}{R}\sqrt{\frac{L}{C}} \\ =\frac{{\omega }_{0}L}{R} \end{gathered}$$

Hence the quality factor $Q=\frac{{\omega }_{0}L}{R}$

Therefore Option C is the correct answer.