Answer
See answers.
Work Step by Step
a. To find the voltage drop across the resistor at 60 Hz, multiply the current by the resistance.
Divide the source voltage by the impedance to determine the magnitude of the current in the circuit. Equation 21-15 gives the impedance of the circuit.
$$V_R=IR=\frac{V_{input}}{\sqrt{R^2+\frac{1}{(2\pi f C)^2}}}R$$
$$V_R= \frac{(130mV)}{\sqrt{(520\Omega)^2+\frac{1}{(2\pi (60Hz) (1.2\times10^{-6}F))^2}}}(520\Omega)\approx 30 mV$$
b. Do the same thing with a frequency of 6.0 kHz.
$$V_R= \frac{(130mV)}{\sqrt{(520\Omega)^2+\frac{1}{(2\pi (6000Hz) (1.2\times10^{-6}F))^2}}}(520\Omega)\approx 130 mV$$
Yes, the capacitor acts to eliminate 60-Hz AC but passes the higher frequency.
