# Power Factor Help.

Discussion in 'Homework Help' started by g.akshay, Nov 21, 2012.

1. ### g.akshay Thread Starter New Member

Nov 21, 2012
1
0
First of I would just like to mention, whoever created this site, you sir/mam have done an awesome job.
Now to the point here is the question.
An impedance coil having a 0.2 lagging power factor is connected in series with a 300-W lamp in order to supply the lamp with 120 V from a 208-V 60Hz source. Find the voltage across the terminals of the impedance coil.
Based on what I have been taught I say the answer MUST be 88 volts because of KVL. But using equations to find the voltage gives me a different answer can someone please explain how. This is what I am doing.
E = 208
Vx = ?
Powerfactor (pf) = 0.2

Pf = P/S
P = I*Vx
S = I* E
0.2 = I*Vx/I*E
0.2 = Vx/E
Vx = E*0.2
Vx =208*0.2 = 41.6V. That answer makes the power factor = 0.2 any other answer the power factor doesn't equal 0.2 so anyone mind helping me out.
Another way of doing it. Pr = 300 Watts. Vr = 120 V I = P/V I = 2.5
S = EI = 2.5*208 = 520. PF = P/S. P = S*PF = 520*0.2 = 104. P = I^2*Rx. Rx = P/I^2. Rx = 104/2.5*2.5 = 16.64. Vx = IRx = 2.5*16.64 = 41.6.
So I'm pretty much lost. Any help/explanation would be great.