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.