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This is a picture of a homework question that I do not understand all too well. The question is below: The power supply in Fig. P9.59 generates 50 kW the line impedance is 0.095 . If the load consumes 43 kW and the voltmeter leads 220 V rms, determine the ammeter leading and the power factor of the inductive load.

I don't know how to start. I know some functions that are most likely to be used: P = Vrms x Irms x cos(theta) with cos(theta) being the power factor. Also, I extracted the information and I don't understand how to find the current. That being the obvious first step. I tried using P = VI => 7kW / 220 = I. But, I checked the answer my professor put online and it's not the same. He somehow got Irms = 271.4 Arms. I don't understand it. please help. Thanks.

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  • \$\begingroup\$ The diagram shows a DC voltage source: a circle with a +/- polarity. :) \$\endgroup\$
    – Kaz
    Commented Mar 23, 2013 at 18:40
  • \$\begingroup\$ It also mentions 480V RMS and 90kW. Can you link to the correct diagram please? \$\endgroup\$
    – Oli Glaser
    Commented Mar 23, 2013 at 18:41
  • \$\begingroup\$ That's the correct diagram. The question above that is a different question. About 3 questions refer to the same diagram. \$\endgroup\$
    – user32011
    Commented Mar 23, 2013 at 18:47
  • \$\begingroup\$ Ah okay, that makes it clearer. \$\endgroup\$
    – Oli Glaser
    Commented Mar 23, 2013 at 19:02

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If you know the input power and the output power then the difference is the power lost.... in the feed wires.

Do you know how to calculate the power dissipated in a resistor based on what current is flowing through it? If yes, then you reverse this formula to uncover what current must be flowing to dissipate the power lost in the feed wire. When you have done this you'll find that your prof is right and he also made the assumption that the feed wire between generator and load was purely resistive.

Let's see if you can figure this out by your comments. PF is 0.72 by the way so let's see if you can figure this out too.

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The clue is that the line impedance is given. 7000 Watts is wasted in the system because the source is putting out 50000, and the load uses only 43000. The waste is dissipated in the line impedance. The impedance is given as a real number, so it is just a resistance. Use the \$P = I^2R\$ formula against 7000 watts, rather than trying to use the \$P = VI\$ formula. You can't do that because the \$V\$ across the line impedance isn't known! (It isn't 220).

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  • \$\begingroup\$ I understand you are trying to be helpful but when answering homework questions it is probably best to be a little bit more vague. This way the OP can actually claim some credit for finding the solution and will also have more opportunity for understanding by actually working through the problem themselves. \$\endgroup\$
    – bhillam
    Commented Mar 24, 2013 at 8:47
  • \$\begingroup\$ @bhillam Kindly read a key detail in the question. The teacher has already given out the numeric answer on the web. That most likely means that the assignments have been collected already. At this point, the professor him/herself will most likely gladly go over how the answer was derived. \$\endgroup\$
    – Kaz
    Commented Mar 24, 2013 at 9:55
  • \$\begingroup\$ fair enough, I'll remove my downvote. \$\endgroup\$
    – bhillam
    Commented Mar 25, 2013 at 2:17
  • \$\begingroup\$ sorry it won't let me remove the downvote as more than 72 hours have passed. Would someone else be willing to counter it? \$\endgroup\$
    – bhillam
    Commented Mar 27, 2013 at 21:46
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You are on the right track, just calculating the current incorrectly. You know the generator is sourcing 50kW, and the load tales 43kW. So the 7kW is being dissipated in the real line impedance (i.e. it is resistive).

So you have two knowns:

  • The resistance = 95mΩ

  • The power dissipated by it = 7kW.

Now all you need to do is work out the current that would cause a 95mΩ resistor dissipate this amount of power. You can then work out the apparent power and calculate the power factor also.

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