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-µF capacitor is connected to the terminals of an AC generator with an rms voltage of 1.50 × 10² V and a frequency of 60.0 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}
\Delta V_{C, rms }=I_{ rms } X_{C}
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 Hz )\left(8.00 \times 10^{-6} 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_{ rms }=\frac{\Delta V_{C, rms }}{X_{C}}=\frac{1.50 \times 10^{2} V }{332 \Omega}=0.452 A
Remark Again, notice how similar the technique is to that of analyzing a DC circuit with a resistor.