Question

A capacitor with capacitance \(C=5.00 \cdot 10^{-6} \mathrm{~F}\) is connected to an AC power source having a peak value of \(10.0 \mathrm{~V}\) and \(f=100 . \mathrm{Hz} .\) Find the reactance of the capacitor and the maximum current in the circuit.

Short answer

Answer: The reactance of the capacitor is approximately 318.310 Ohms, and the maximum current in the circuit is approximately 0.0314 Amps.

Step-by-step solution

Calculate the reactance of the capacitor

To calculate the reactance of the capacitor, we can use the formula \(X_C = \frac{1}{2 \pi fC}\). Plugging in the given values for frequency (f) and capacitance (C), we get: \(X_C = \frac{1}{2 \pi (100 \mathrm{Hz})(5.00 \cdot 10^{-6} \mathrm{F})}\) Now, calculate the reactance: \(X_C \approx 318.310 \Omega\) The reactance of the capacitor is approximately 318.310 Ohms.

Calculate the maximum current in the circuit

Now that we have the reactance of the capacitor, we can find the maximum current in the circuit using Ohm's Law: \(I_{\mathrm{max}} = \frac{V_{\mathrm{max}}}{X_C}\) Substitute the given peak voltage value (10.0 V) and the calculated reactance (318.310 Ω): \(I_{\mathrm{max}} = \frac{10.0 \mathrm{V}}{318.310\Omega}\) Now, calculate the maximum current: \(I_{\mathrm{max}}\approx 0.0314\mathrm{A}\) The maximum current in the circuit is approximately 0.0314 Amps.