In the circuit shown in the figure, the ratio of the quality factor and the band width is _____ s.

R=10Ω,L=3H,C=27μF

Step 1: Bandwidth

$$\Delta \omega = \frac{R}{L}$$​

Step 2: Quality factor

Q = $$\frac{\omega_0}{\Delta \omega} = \frac{\omega_0 L}{R}$$​

Step 3: Required ratio

$$\frac{Q}{\Delta \omega} = \frac{\omega_0 L}{R} \cdot \frac{L}{R}$$

= $$\omega_0 \frac{L^2}{R^2}$$

$$\omega_0 = \frac{1}{\sqrt{LC}}$$1​:

$$\frac{Q}{\Delta \omega} = \frac{1}{\sqrt{LC}} \cdot \frac{L^2}{R^2}$$

= $$\frac{L^2}{R^2 \sqrt{LC}}$$

Step 4: Substitute values

$$\sqrt{LC} = \sqrt{3 \times 27 \times 10^{-6}} = \sqrt{81 \times 10^{-6}} = 9 \times 10^{-3}$$

$$\frac{Q}{\Delta \omega} = \frac{3^2}{10^2 \times 9 \times 10^{-3}}$$

=$$ \frac{9}{100 \times 9 \times 10^{-3}}$$

= $$\frac{1}{10^{-1}} = 10\,\text{s}$$

Final Answer:

10