Question

A capacitor is rated at \(0.025 \mu \mathrm{F}\). How much rms current flows when the capacitor is connected to a \(110-\mathrm{V}\) rms, \(60.0-\mathrm{Hz}\) line?

Short answer

Based on the given information, the problem required finding the rms current flowing through a capacitor connected to an AC voltage source. To solve this, we first calculated the capacitive reactance using the formula \(X_C = \frac{1}{2\pi fC}\) and found it to be approximately \(106.22 \Omega\). Then, we used Ohm's law for AC circuits (\(I = \frac{V}{X_C}\)) to find the rms current, which was approximately \(1.036 A\).

Step-by-step solution

Calculate the capacitive reactance

We need to determine the capacitive reactance, denoted as \(X_C\), which can be calculated using the formula: \(X_C = \frac{1}{2\pi fC}\). Here, \(f\) represents the frequency of the AC source and \(C\) represents the capacitance. Plugging in the given values, we have: \(X_C = \frac{1}{2\pi (60.0Hz)(0.025\mu F)}\) Convert the capacitance in farads: \(0.025\mu F = 0.025 \times 10^{-6} F\) Now, calculate \(X_C\): \(X_C = \frac{1}{2\pi (60Hz)(0.025 \times 10^{-6} F)} \approx 106.22 \Omega\)

Calculate the rms current using Ohm's law for AC circuits

Ohm's law states that \(I = \frac{V}{X_C}\), where \(I\) is the rms current, \(V\) is the rms voltage, and \(X_C\) is the capacitive reactance calculated in step 1. Plugging in the values, we get: \(I = \frac{110V}{106.22 \Omega} \approx 1.036 A\) Therefore, the rms current flowing through the capacitor is approximately \(1.036 A\).