This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: + P 4Ω + P 10Ω = 551.9W NOTE: In this case, all the secondary loads were resistive. If there were reactive loads as well, we would have to find the corresponding reactive powers. This could be done similar to the above process, where we find the individual reactive powers and then sum them up. E.g.: If there was a capacitor in series with the 4Ω resistor, we would use Qc = I 2 Xc, where ‘I’ is the current going through the capacitor, (I2 – I3). Also, if there was an inductor in series with the 10Ω resistor, we would use Q L = I 2 X L , where ‘I’ would be the current going through the inductor, I3. The total reactive power is then simply Qc+Q L . The total power would then be the phasor sum of real and reactive powers, called the apparent power measured in VA: S Total = 2 2 total total Q P +...
View
Full Document
 Fall '10
 Khan
 reactive powers, 11.7A I3

Click to edit the document details