Question

31.3 forced oscillations of three simple circuits
a 25 - μf capacitor is connected to an ac source of emf with a frequency of 500.0 hz and a maximum emf of 15 v. the maximum current is
3700 a.
0 a.
1.2 a.
0.19 a.
94 a.

Explanation:

Step1: Calculate the capacitive reactance \(X_C\)

The formula for capacitive reactance is \(X_C=\frac{1}{2\pi fC}\).
Given \(f = 500.0\space Hz\), \(C=25\times10^{- 6}\space F\).

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Step2: Calculate the maximum current \(I_{max}\)

The formula for maximum current in an AC - capacitive circuit is \(I_{max}=\frac{\mathcal{E}_{max}}{X_C}\).
Given \(\mathcal{E}_{max} = 15\space V\), \(X_C\approx12.74\space\Omega\)

$$ I_{max}=\frac{15}{12.74}\approx1.2\space A $$

Answer:

1.2 A.