1. EET 325 Oct. 14, 2005 Lecture Power factor correction

2. Recall from ac circuits P = VICos(  ) Q( for the circuit) = VISin(  ) jQ (for an inductor) = jI 2 X L =jV 2 /X L -jQ (for a capacitor) = -jI 2 X C = -jV 2 /X C

3. Example Number 1 Given : a 240 volt supply voltage 13 kW load 0.65 lagging power factor

4.  = Cos -1 (0.65) = 49.458 o I L = I L = 83.333/ -49.458 o P = VICos(  ) = 13 kW Q = VISin(  ) = j15.198 KVars

5. Determine the capacitor Value for Unity Power Factor Let’s Add a capacitor as shown -jQ = -j15.198 KVars = -jV 2 /X C X C = 3.789 = 1/(  C) Where  = 2  f = 2  60 = 377

6. X C = 3.789 = 1/(377 C) C = 700 uf

7. I C = 240/(-jX C ) = 63.327 /90 o I L = 83.333 / -49.458 o I S = I L + I C I S = 54.17/0 o