The quality factor (\( Q \)) of a series RLC circuit is given by: \[ Q = \frac{1}{R} \sqrt{\frac{L}{C}}, \] where:
\( R = 100 \, \Omega \),
\( L = 1 \, \text{H} \),
\( C = 6.25 \, \mu\text{F} = 6.25 \times 10^{-6} \, \text{F} \).
Substitute the values: \[ Q = \frac{1}{100} \sqrt{\frac{1}{6.25 \times 10^{-6}}}. \]
Simplify: \[ Q = \frac{1}{100} \sqrt{1.6 \times 10^5} = \frac{1}{100} \cdot 400 = 4. \]
Final Answer: The quality factor is: \[ \boxed{4}. \]
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: