Question

If the reading of voltmeter $V$ shown in the figure at resonance is $200\text{V}$, then the quality factor of the circuit is,

• A. $2$
• B. $4$
• C. $1$
• D. $3$

Answer 1

The correct option is B $4$

Let the current flowing in the circuit at the given instant is $i$.

The quality factor is given by:

$Q=\frac{{\omega }_{0}L}{R}=\frac{X_L}{R}$

$\Rightarrow Q=\frac{X_L}{R}\times \frac{i}{i}=\frac{V_L}{V_R}$

At resonance, the voltage across the resistor is equal to the applied voltage.

$\Rightarrow V_R=50\text{V}$

At resonance, the magnitude of voltage across the inductor is equal to the magnitude of voltage across the capacitor.

$\Rightarrow V_L=V_C=200V$ (Given)

$\therefore Q=\frac{200}{50}=4$

Hence, $\left(B\right)$ is the correct answer.