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: \[ \boxed{4}. \]
A wire of resistance $ R $ is bent into a triangular pyramid as shown in the figure, with each segment having the same length. The resistance between points $ A $ and $ B $ is $ \frac{R}{n} $. The value of $ n $ is:


Let \( a \in \mathbb{R} \) and \( A \) be a matrix of order \( 3 \times 3 \) such that \( \det(A) = -4 \) and \[ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} \] where \( I \) is the identity matrix of order \( 3 \times 3 \).
If \( \det\left( (a + 1) \cdot \text{adj}\left( (a - 1) A \right) \right) \) is \( 2^m 3^n \), \( m, n \in \{ 0, 1, 2, \dots, 20 \} \), then \( m + n \) is equal to: