What is the phase angle of an AC series circuit that is constructed of a 14.5-Ω resistor along with 16.5-Ω inductive reactance and 9.41-Ω capacitive reactance?
A)26.1°
B)11.9°
C)43.1°
D)35.0°
E)38.7°

Question

Quizplus · Accepted Answer

The phase angle ($\theta$) in an AC circuit can be calculated using the formula $\theta = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)$, where $X_L$ is the inductive reactance, $X_C$ is the capacitive reactance, and $R$ is the resistance. Plugging in the given values: $\theta = \tan^{-1}\left(\frac{16.5 - 9.41}{14.5}\right) = \tan^{-1}(0.489)$, which gives a phase angle of approximately 26.1°.

What Is the Phase Angle of an AC Series Circuit