
#1
Apr605, 01:57 AM

P: 86

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 wellinsulated [tex]0.03\mu F[/tex] capacitor connected to a 2.0kV(rms) 720Hz 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? 



#2
Apr605, 02:49 AM

HW Helper
Thanks
P: 9,818

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



#3
Apr605, 03:29 AM

P: 49

Yes he is correct. Since current can be described in the form of a sin fn, the maximum amplitude obtained is known as peak value. rms value is the root mean square value of all individual values
For eg, peak value of sin(x) is 1. But it takes values from 1 to 1. rms value=1/sqrt(2). It is the net effect. 


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