A 60-Hz ac generator with a peak voltage of 110 V drives a series RC circuit with R=10.0 and C=300mF The peak current in the circuit is

To find the peak current in the circuit, we need to follow these steps:

1. Determine the angular frequency (ω): The angular frequency is given by: $\omega = 2\pi f$ where $f$ is the frequency of the AC generator. Given $f = 60$ Hz, $\omega = 2\pi \times 60 = 120\pi \, \text{rad/s}$

2. Calculate the capacitive reactance (X_C): The capacitive reactance is given by: $X_C = \frac{1}{\omega C}$ where $C = 300$ mF = 0.300 F. Substituting the values, $X_C = \frac{1}{120\pi \times 0.300} = \frac{1}{36\pi} \approx 0.00884 \, \Omega$

3. Determine the impedance (Z) of the series RC circuit: The impedance is given by: $Z = \sqrt{R^2 + X_C^2}$ where $R = 10.0 \, \Omega$. Substituting the values, $Z = \sqrt{10.0^2 + 0.00884^2} \approx \sqrt{100 + 0.000078} \approx \sqrt{100} = 10.0 \, \Omega$

4. Calculate the peak current (I_0): The peak current is given by: $I_0 = \frac{V_0}{Z}$ where $V_0 = 110$ V (peak voltage). Substituting the values, $I_0 = \frac{110}{10.0} = 11.0 \, \text{A}$

Therefore, the peak current in the circuit is $11.0$ A.