(a) Calculate the capacitive reactance of a 5.00 mF capacitor when 60.0 Hz and 10.0 kHz AC voltages are applied. (b) What is the rms current if the applied rms voltage is 120 V?
Strategy
The capacitive reactance is found directly from the expression in X_{C}=\frac{1}{2 \pi f C}. Once X_{C} has been found at each frequency, Ohm’s law stated as I=V / X_{C} can be used to find the current at each frequency.
Solution for (a)
Entering the frequency and capacitance into X_{C}=\frac{1}{2 \pi f C} gives
X_{C}=\frac{1}{2 \pi f C} (23.59)
=\frac{1}{6.28(60.0 / s )(5.00 \mu F )}=531 \Omega \text { at } 60 Hz.
Similarly, at 10 kHz,
X_{C}=\frac{1}{2 \pi f C}=\frac{1}{6.28\left(1.00 \times 10^{4} / s \right)(5.00 \mu F )} (23.60)
= 3.18 Ω at 10 kHz.
Solution for (b)
The rms current is now found using the version of Ohm’s law in I=V / X_{C}, given the applied rms voltage is 120 V. For the first frequency, this yields
I=\frac{V}{X_{C}}=\frac{120 V }{531 \Omega}=0.226 A \text { at } 60 Hz. (23.61)
Similarly, at 10 kHz,
I=\frac{V}{X_{C}}=\frac{120 V }{3.18 \Omega}=37.7 A \text { at } 10 kHz. (23.62)
Discussion
The capacitor reacts very differently at the two different frequencies, and in exactly the opposite way an inductor reacts. At the higher frequency, its reactance is small and the current is large. Capacitors favor change, whereas inductors oppose change. Capacitors impede low frequencies the most, since low frequency allows them time to become charged and stop the current. Capacitors can be used to filter out low frequencies. For example, a capacitor in series with a sound reproduction system rids it of the 60 Hz hum.