# AC Power Worksheet - Q39

Discussion in 'Homework Help' started by Spoon, May 16, 2010.

1. ### Spoon Thread Starter Member

May 18, 2009
12
0
Hi,

I'm currently doing the AC power worksheet on this website, found at url

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

2. ### retched AAC Fanatic!

Dec 5, 2009
5,201
315
You must be missing the final C=(C*.5) in the equation

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

3. ### Spoon Thread Starter Member

May 18, 2009
12
0
as stated in the title, Question #39 ... thanks

4. ### retched AAC Fanatic!

Dec 5, 2009
5,201
315
Jeez, I cant read today..or yesterday.. I apologize. Gimme a minute.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
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..

Spoon likes this.
6. ### Spoon Thread Starter Member

May 18, 2009
12
0
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