Question

In a series LCR resonant circuit, the quality factor is measured as $100$. If the inductance is increased by two fold and resistance is decreased by two fold, then the quality factor after this change will be __________. (up to two decimal places).

Answer 1

Step 1: Given

Quality factor, $Q=100$

Initial inductance=$L$

Final inductance=$L\text{'}=2L$

Initial resistance=$R$

Final resistance=$R\text{'}=\frac{R}{2}$

Step 2: Determine the quality factor

The quality factor is given by,

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

(where $\omega =\frac{1}{\sqrt{LC}}$ is the frequency)

Step 3: Determine the change in quality factor

New quality factor,

$$\begin{gathered} Q\text{'}=\frac{\sqrt{L\text{'}}}{R\text{'}\sqrt{C}} \\ =\frac{\sqrt{2L}}{{\displaystyle \frac{R}{2}}\sqrt{C}} \\ =2\sqrt{2}\frac{\sqrt{L}}{R\sqrt{C}} \\ =2\sqrt{2}Q \\ =2\sqrt{2}\left(100\right) \\ =282.84 \end{gathered}$$

Therefore, the quality factor after the change will be $282.84$.