The impedance of an $R C$ circuit containing a $40.0 - \mu \mathrm { F }$ capacitor is $800 \Omega$. If the frequency of the applied ac voltage is $18.0 \mathrm {~Hz}$, what is the resistance of the resistor?
A) $769 \Omega$
B) $800 \Omega$
C) $569 \Omega$
D) $669 \Omega$
E) $869 \Omega$

Question

Quizplus · Accepted Answer

The impedance $Z$ of an $RC$ circuit is given by $Z = \sqrt{R^2 + (1/\omega C)^2}$, where $R$ is the resistance, $\omega = 2\pi f$ is the angular frequency, and $C$ is the capacitance. Given $Z = 800 \Omega$, $C = 40.0 \mu F = 40.0 \times 10^{-6} F$, and $f = 18.0 Hz$, we can solve for $R$. First, calculate the reactance of the capacitor $X_C = 1/\omega C = 1/(2\pi f C)$. Substituting the given values, $X_C = 1/(2\pi \times 18.0 \times 40.0 \times 10^{-6}) \approx 221 \Omega$. Then, using the impedance formula, $R = \sqrt{Z^2 - X_C^2} = \sqrt{800^2 - 221^2} \approx 769 \Omega$.

The Impedance of An RCR CRC Circuit Containing A 40.0−μF40.0 - \mu \mathrm { F }40.0−μF

The Impedance of An $R C$ Circuit Containing A $40.0 - \mu \mathrm { F }$