1. For a series LCR resonant circuit, the quality factor is $Q = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}$ where $\omega_0 = \dfrac{1}{\sqrt{LC}}$.

2. Initially, $Q_1 = 100$

3. Now the changes are:

• inductance doubled: $L' = 2L$
• resistance halved: $R' = \dfrac{R}{2}$
• capacitance remains same: $C' = C$

4. New quality factor: $Q_2 = \frac{1}{R'}\sqrt{\frac{L'}{C}}$ Substitute $R' = \dfrac{R}{2}$ and $L' = 2L$: $Q_2 = \frac{1}{R/2}\sqrt{\frac{2L}{C}}$ $Q_2 = \frac{2}{R}\cdot \sqrt{2}\sqrt{\frac{L}{C}}$ $Q_2 = 2\sqrt{2}\left(\frac{1}{R}\sqrt{\frac{L}{C}}\right)$ $Q_2 = 2\sqrt{2}\, Q_1$

5. Using $Q_1=100$: $Q_2 = 2\sqrt{2}\times 100 = 200\sqrt{2}$ $Q_2 \approx 200 \times 1.414 = 282.8$

6. Since this is an integer-type answer, $Q_2 \approx 283$

Therefore, the new quality factor is 283.