# Power Factor - Apparent Power Calculations

#### ooxneozoo

Joined Jul 20, 2017
7
I'm in the AC text book on chapter 11 under power factor. The question I have is regarding the image/link below.

If S=Apparent Power= I²(Z), then how does that = 169.256?

Wouldn't Z = XL + R
Z = 60.319 + 60
Z = 120.319

Then S = I²(Z)
S = 1.410²(120.319)
S = 1.9881(120.319)
S = 239.206 ?

*I'm not getting 169.256 as shown in the textbook and image below. Even when applying pythagorean theory I'm unable to come up with the answer given.

Can you please tell me where I am going wrong on this?

True power, reactive power, and apparent power for a resistive/reactive load.

Last edited:

#### LesJones

Joined Jan 8, 2017
4,190
You have only added the resistance to the reactance. They are considdered as being at right angles to each other so you have to use vector addition. so Z= √60^2 + 60.319^2 =√3600 + 3638.38 = √7638.38 = 85.08

Les.

#### ooxneozoo

Joined Jul 20, 2017
7
1

Last edited:

#### ooxneozoo

Joined Jul 20, 2017
7
You have only added the resistance to the reactance. They are considdered as being at right angles to each other so you have to use vector addition. so Z= √60^2 + 60.319^2 =√3600 + 3638.38 = √7638.38 = 85.08

Les.
Thanks, i did forget to add the angles. Though, when I do, I still only get the 85.08 that you show. I don't get the 169.256VA, can you show me where I'm going wrong?

z = 60.319 < 90 | 0 + j60.319
+ z = 60 < 0 | 60 + j0
_______________________
z = 85.08 < 45.15 | 60 + j60.319

#### ooxneozoo

Joined Jul 20, 2017
7
Thanks, i did forget to add the angles. Though, when I do, I still only get the 85.08 that you show. I don't get the 169.256VA, can you show me where I'm going wrong?

z = 60.319 < 90 | 0 + j60.319
+ z = 60 < 0 | 60 + j0
_______________________
z = 85.08 < 45.15 | 60 + j60.319

NVM, forgot to add the original formula! Need more coffee

1.410^2(85.08) = 169.147

#### LesJones

Joined Jan 8, 2017
4,190
Ohms and VA are not the same thing.

Les.