Hi,

I'm currently doing the AC power worksheet on this website, found at url
http://www.allaboutcircuits.com/worksheets/acpower.html

When calculating the capacitor value I keep getting exactly double the value.

I calculated the initial power factor to be 0.707 (lagging), which agrees with the solution.

The reactance of the inductor is Xl = 1.0824∏.
Xc = 1 / (2∏fC)
therefore C = 1 / (1.0824∏*2∏*60)
C = 7 . 8 * 10^-4
C = 780 uF which is exactly double the required value of 390uF

Thanks,
Steve

 

You must be missing the final C=(C*.5) in the equation

What 'Question #' in the sheet are you referring to?

 

as stated in the title, Question #39 ... thanks

 

Jeez, I cant read today..or yesterday.. I apologize. Gimme a minute.

 

If you were placing C in series with the load then 780uF would be correct. In the parallel placement you only have to compensate for the reactive part of the load current - 390uF is correct in that case..

 

Thanks, "only have to compensate for the reactive part of the load current". Was doing this wrong, silly mistake. I agree with the answer now