Question 21.2: A Purely Capacitive AC Circuit Goal Perform basic AC circuit...
A Purely Capacitive AC Circuit
Goal Perform basic AC circuit calculations for a capacitive circuit.
Problem An 8.00- \mu \mathrm{F} capacitor is connected to the terminals of an AC generator with an rms voltage of 1.50 \times 10^{2} \mathrm{~V} and a frequency of 60.0 \mathrm{~Hz}. Find the capacitive reactance and the rms current in the circuit.
Strategy Substitute values into Equations 21.5 and 21.6.
X_{C}\equiv{\frac{1}{2 \pi f C}} (21.5)
\Delta V_{\mathrm{C,rms}}=I_{\mathrm{rms}}X_{\mathrm{C}} (21.6)
Substitute the values of f and C into Equation 21.5:
X_{C}=\frac{1}{2 \pi f C}=\frac{1}{2 \pi(60.0 \mathrm{~Hz})\left(8.00 \times 10^{-6} \mathrm{~F}\right)}=332 \Omega
Solve Equation 21.6 for the current, and substitute X_{C} and the rms voltage to find the rms current:
I_{\mathrm{rms}}=\frac{\Delta V_{C, \mathrm{rms}}}{X_{C}}=\frac{1.50 \times 10^{2} \mathrm{~V}}{332 \Omega}=0.452 \mathrm{~A}
Remark Again, notice how similar the technique is to that of analyzing a DC circuit with a resistor.