# Power Factor Correction w/ Cap.

#### jegues

Joined Sep 13, 2010
733
A 120-V, 60Hz source is applied across an inductive load of (100 + j700)Ω. What is the power factor of the load? What must be the size of the capacitor that can be placed in parallel with the load in order to improve the power factor to 0.95 lagging?

See figure attached for my attempt, I can't figure out what the impeadance of the capictor has to be.

I'm thinking I could do,

$$Z_{new} = Z_{c} // Z_{l} = x +jy$$

and select C such that,

$$arctan(\frac{y}{x}) = -18.195^{o}$$

but the math and algebra gets pretty ugly taking this approach.

Is there another way?

#### t_n_k

Joined Mar 6, 2009
5,455
Your work is fine so far.

The uncompensated inductor current

I_L=120/(100+j700)=0.024 - j0.168 Amps

With the same real term of 0.024 Amp a pf of 0.95 would equate to a new total reactive current 0.024tan(acos(0.95))=0.007888 Amp

The change in reactive current is therefore 0.168-0.007888=0.1601A

The required Xc is then Xc=120/0.1601=749.48Ω

Or C=1/(ωXc)=3.539uF

Last edited:
• nyasha

#### jegues

Joined Sep 13, 2010
733
equate to a new total reactive current 0.024tan(acos(0.95))=0.007888 Amp
Where are you getting this from? I know the acos(0.95) is going to be the angle of the new current phasor, but why is it multiplied by the real portion of the current phasor?

EDIT: Nevermind I see it now, from the power triangle,

$$tan \alpha = \frac{Q}{P} \quad \rightarrow \quad Ptan \alpha = Q$$

Last edited:
• nyasha

#### nyasha

Joined Mar 23, 2009
90
I was about to post a thread on this question. Jegues, l guess he didn't use the power triangle here since he is talking about reactive current, but l guess the same concept from the power triangle can be adopted to the current also.

#### bujang senang

Joined Sep 11, 2011
2
I have see the True, Reactive, and Apparent power note.....it use 1500 as the watt and 9.165 s the amp.if i us about 2500 watt, how would i get the amp that shuld be use