A $10 - \Omega$ resistor is in series with a $100 - \mu \mathrm { F }$ capacitor at $120 \mathrm {~Hz}$. What is the phase angle?
A) $- 37 ^ { \circ }$
B) $- 53 ^ { \circ }$
C) $- 82 ^ { \circ }$
D) $- 4.7 ^ { \circ }$
E) $+ 37 ^ { \circ }$

Question

Quizplus · Accepted Answer

The phase angle $\theta$ in a series RC circuit is given by $\theta = -\arctan\left(\frac{X_C}{R}\right)$, where $X_C$ is the capacitive reactance and $R$ is the resistance. The capacitive reactance $X_C = \frac{1}{2\pi fC}$, where $f$ is the frequency and $C$ is the capacitance. Plugging in the given values, $f = 120 \, \text{Hz}$ and $C = 100 \, \mu\text{F} = 100 \times 10^{-6} \, \text{F}$, we get $X_C = \frac{1}{2\pi \times 120 \times 100 \times 10^{-6}} \, \Omega$. Calculating $X_C$ yields approximately $13.26 \, \Omega$. Then, $\theta = -\arctan\left(\frac{13.26}{10}\right)$, which gives a phase angle close to $-53^\circ$.

A 10−Ω10 - \Omega10−Ω Resistor Is in Series with A 100−μF100 - \mu \mathrm { F }100−μF

A $10 - \Omega$ Resistor Is in Series with A $100 - \mu \mathrm { F }$