Question

problem 21.59
what is the reactance of a well - insulated 0.033 μf capacitor connected to a 3.5 kv (rms) 720 hz line?
express your answer using two significant figures.

part b
what will be the peak value of the current?
express your answer using two significant figures.

Explanation:

Step1: Calculate the capacitive reactance

The formula for capacitive reactance is \(X_{C}=\frac{1}{2\pi fC}\).
Given \(C = 0.033\ \mu F=0.033\times10^{- 6}\ F\) and \(f = 720\ Hz\).
Substitute the values into the formula:
\(X_{C}=\frac{1}{2\pi\times720\times0.033\times10^{-6}}\)

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Step2: Calculate the rms current

The rms voltage \(V_{rms}=3.5\ kV = 3500\ V\).
Using Ohm's law for AC \(I_{rms}=\frac{V_{rms}}{X_{C}}\), so \(I_{rms}=\frac{3500}{670}\approx5.22\ A\)

Step3: Calculate the peak current

The relationship between peak current \(I_{peak}\) and rms current \(I_{rms}\) is \(I_{peak}=\sqrt{2}I_{rms}\)
\(I_{peak}=\sqrt{2}\times5.22\approx7.4\ A\)

Answer:

\(X_{C} = 670\ \Omega\), \(I_{peak}=7.4\ A\)