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_{\mathrm{C}}}\equiv\,{\frac{1}{2\pi fC}} [21.5]
\Delta V_{\mathrm{C,rms}}= I_{\mathrm{rms}}X_{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})(8.00~\times~10^{-6}\mathrm{~F})}}=\ 332\,{\Omega}
Solve Equation 21.6 for the current and substitute the values for X_{C} and the rms voltage to find the rms current:
I_{\mathrm{rms}}={\frac{\Delta V_{\mathrm{C,rms}}}{X_{C}}}={\frac{1.50~\times~10^{2}\,\mathrm{V}}{332\,\Omega}}=\ {0.452\,\mathrm{A}}
REMARKS Again, notice how similar the technique is to that of analyzing a DC circuit with a resistor.