Question

Consider a series rlc circuit where r = 45.0 ω, c = 15.5 μf, and l = 0.0940 h, driven at a frequency of 50.0 hz. determine the phase angle of the circuit in degrees.

Answer 1

$R = 45 \ \Omega. \newline C = 15.5 \ \mu F = 15.5 \cdot 10^{-6} \ F.\newline l = 0.0940 \ H.\newline \nu = 50 \ Hz. \newline \newline x_{l} = 2 \pi \nu l = 2 \pi \cdot 50 \cdot 0.0940 = 29.5309709437 \approx 29.5 \ \Omega. \newline x_{c} = \frac{1}{2 \pi C \nu} = \frac{1}{2 \pi \cdot 15.5 \cdot 10^{-6} \cdot 50} = 0.00486946861 \approx 0.005 \ \Omega. \newline Z = \sqrt{R^{2}+(x_{l}-x_{c})^{2}} = \sqrt{45^{2}+(29.5-0.005)^{2}} = 53.8047862648 \approx 54 \ \Omega. \newline \newline \cos(\theta) = \frac{R}{Z} = \frac{45}{54} = 0.8\bar{3} \Rightarrow \arccos(0.8\bar{3}) = 33.5573098 \approx 33.5 \textdegree.$