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

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The discussion centers on calculating the peak current of a 0.03μF capacitor connected to a 2.0-kV (rms), 720-Hz line. The reactance of the capacitor is determined to be 7.4kΩ, leading to an rms current of 0.27 A. To find the peak current, it is clarified that the rms value must be multiplied by √2. Therefore, the peak current is approximately 0.38 A. The confusion between rms and peak values is a common issue in electrical calculations.
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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 0.03\mu F capacitor connected to a 2.0-kV(rms) 720-Hz line?

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

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

so is this the right answer, or do i have to multiply it by \sqrt{2} 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 0.03\mu F capacitor connected to a 2.0-kV(rms) 720-Hz line?
...
I_{rms} = 2.7 * 10^{-1} A

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


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


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