If each of the values of inductance, capacitance and resistance of a series LCR circuit are doubled, the Q-factor of the circuit would

Understanding the Q-Factor Formula

The Quality Factor (Q-factor) in a series LCR circuit is a measure of its sharpness or selectivity. It indicates how efficiently the circuit oscillates at its resonant frequency. The formula for the Q-factor in a series LCR circuit is given by:

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

Where:

- $L$ is the inductance
- $C$ is the capacitance
- $R$ is the resistance

Initial Q-Factor Calculation

Let the initial inductance, capacitance, and resistance be $L_1$, $C_1$, and $R_1$ respectively. The initial Q-factor, denoted as $Q_1$, is:

$$Q_1=\frac{1}{R_1}\sqrt{\frac{L_1}{C_1}}$$

Calculating the New Q-Factor

The problem states that each of the values (inductance, capacitance, and resistance) is doubled. So, the new values are:

- New Inductance: $L_2=2L_1$
- New Capacitance: $C_2=2C_1$
- New Resistance: $R_2=2R_1$

Now, we calculate the new Q-factor, $Q_2$, using the formula with these new values:

$$Q_2=\frac{1}{R_2}\sqrt{\frac{L_2}{C_2}}$$

Substitute the new values:

$$Q_2=\frac{1}{2R_1}\sqrt{\frac{2L_1}{2C_1}}$$

Simplify the expression under the square root:

$$Q_2=\frac{1}{2R_1}\sqrt{\frac{L_1}{C_1}}$$

Comparing Q-Factors and Conclusion

We can see the relationship between the new Q-factor ($Q_2$) and the initial Q-factor ($Q_1$):

$$Q_2=\frac{1}{2}\left(\frac{1}{R_1}\sqrt{\frac{L_1}{C_1}}\right)$$

Since $Q_1=\frac{1}{R_1}\sqrt{\frac{L_1}{C_1}}$, we can write:

$$Q_2=\frac{1}{2}{Q}_1$$

This shows that the new Q-factor is half the original Q-factor. Therefore, the Q-factor reduces by a factor of 2.

The correct option is that the Q-factor would reduce by a factor 2.