Peak Current of 0.03\mu F Capacitor on 2kV, 720Hz Line

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SUMMARY

The peak current of a 0.03μF capacitor connected to a 2.0-kV (rms) 720-Hz line is calculated using the formula for reactance, resulting in a reactance of 7.4kΩ. The rms current is determined to be 0.27 A. To find the peak current, the rms value must be multiplied by √2, yielding a peak current of approximately 0.38 A. This calculation clarifies the relationship between rms and peak values in AC circuits.

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leolaw
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I am always confused with rms value and peak value, so here is the problem:

What will be the peak value of the current of a well-insulated [tex]0.03\mu F[/tex] capacitor connected to a 2.0-kV(rms) 720-Hz line?

So first I find the reactance of the capacitor, which is:
[tex]X_c=\frac{1}{2\pi fC}[/tex]
[tex]7.4k \Omega[/tex]

and then [tex]V_{rms} = I_rms * X_c[/tex]
[tex]2*10^3 = I_{rms} (7.4 * 10^3)[/tex]
[tex]I_{rms} = 2.7 * 10^{-1} A[/tex]

so is this the right answer, or do i have to multiply it by [tex]\sqrt{2}[/tex] to get the peak value?
 
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leolaw said:
I am always confused with rms value and peak value, so here is the problem:

What will be the peak value of the current of a well-insulated [tex]0.03\mu F[/tex] capacitor connected to a 2.0-kV(rms) 720-Hz line?
...
[tex]I_{rms} = 2.7 * 10^{-1} A[/tex]

so is this the right answer, or do i have to multiply it by [tex]\sqrt{2}[/tex] to get the peak value?


The peak value is sqrt(2) times the rms value, and the peak was asked...


ehild
 

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