A capacitor with a capacitive reactance of $40.0 \Omega$ is connected in series with a $100 - \Omega$ resistor across an ac source of frequency $60.0 \mathrm {~Hz}$. What is the phase angle?
A) $- 21.8 ^ { \circ }$
B) $+ 21.8 ^ { \circ }$
C) $- 68.2 ^ { \circ }$
D) $90.0 ^ { \circ }$
E) $+ 68.2 ^ { \circ }$

Question

Quizplus · Accepted Answer

The phase angle $\theta$ in a series RC circuit is given by $\theta = \tan^{-1}\left(\frac{-X_C}{R}\right)$, where $X_C$ is the capacitive reactance and $R$ is the resistance. Plugging in the given values, $\theta = \tan^{-1}\left(\frac{-40.0}{100}\right) = \tan^{-1}(-0.4)$, which gives $\theta \approx -21.8^\circ$.

A Capacitor with a Capacitive Reactance Of 40.0Ω40.0 \Omega40.0Ω Is Connected in Series with A

A Capacitor with a Capacitive Reactance Of $40.0 \Omega$ Is Connected in Series with A