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Chapter 30: Problem 34

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

Expert verified
Answer: The reactance of the capacitor is approximately 318.31 Ω, and the maximum current in the circuit is approximately 0.0314 A.

Step by step solution

01

Calculate the angular frequency

The angular frequency (ω) of the AC power source can be calculated using the frequency (f) given by the formula: ω = 2πf Plugging in the given value of frequency f = 100 Hz, we have: ω = 2π(100) = 200π rad/s
02

Calculate the reactance of the capacitor

The reactance (X_C) of the capacitor can be calculated using the capacitance (C) and angular frequency (ω) with the formula: X_C = 1/(ωC) Plugging in the given value of capacitance C = 5.00 × 10^{-6} F and the calculated angular frequency ω = 200π rad/s, we have: X_C = 1/(200π × 5.00 × 10^{-6}) = 1000/π Ω ≈ 318.31 Ω
03

Calculate the maximum current in the circuit

With the reactance of the capacitor (X_C) and the peak voltage (V_peak) of the AC power source, we can now calculate the maximum current (I_max) in the circuit using Ohm's law: I_max = V_peak / X_C Plugging in the given value of peak voltage V_peak = 10.0 V and the calculated reactance X_C ≈ 318.31 Ω, we have: I_max = 10.0 V / 318.31 Ω ≈ 0.0314 A Thus, the reactance of the capacitor is approximately 318.31 Ω, and the maximum current in the circuit is approximately 0.0314 A.

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