# Power factor correction problem

Discussion in 'Homework Help' started by testing12, Apr 11, 2011.

1. ### testing12 Thread Starter Member

Jan 30, 2011
80
2
Hello ,
I had to solve the following problem:

I did this problem using admittance. (since there in parallel) I simply took the impedance above took Y= (1/Z)

and
G (susceptance) = 2(pi)fC

i got C = 312 microfarads. My question is how did they do part b in the solution manual. Im not understanding that. Please advise thank you.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
I wonder if it shouldn't be

$C=\frac{Ptan(\theta_1-\theta_2)}{\omega V_{rms}^2}$

where

$\theta=(\theta_1-\theta_2)$

is the angular difference between supply voltage and current

and

$p.f.=cos(\theta_1-\theta_2)=cos(\theta)$

In any case that formula should be derived or stated somewhere in the text from which the problem is quoted.

It's easy enough to show that

Motor Reactive Power, Sm

$S_m=VIsin(\theta)=VIcos(\theta)\frac{sin(\theta)}{cos(\theta)}=P_mtan(\theta)$

Where Pm is the motor real power.

The capacitive reactive power, Sc is

$S_c=\frac{V^2}{X_c}=V^2\omega C$

And assuming Sm=Sc for complete compensation.

$C=\frac{P_mtan(\theta)}{\omega V^2}$