Alternating Current: Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Alternating Current. Understanding the underlying formula is key to arriving at the correct answer below:

Reactances are $X_L = \omega L = 314 \times 0.020 = 6.28 \Omega$ and $X_C = \frac{1}{314 \times 100 \times 10^{-6}} \approx 31.85 \Omega$.
Impedance $Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{50^2 + (25.57)^2} \approx 56.15 \Omega$.
$V_{rms} = \frac{10}{\sqrt{2}} = 5\sqrt{2} V$, so power $P = \frac{V_{rms}^2 R}{Z^2}$.
$P = \frac{50 \times 50}{(56.15)^2} \approx 0.79 W$.