Voltage, current, and power wasted

  • Thread starter cuttlefish
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  • #1
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Homework Statement


a power station delivers 750 kW of power at 12,000 V to a factory through wires with total resistance of 3.0 ohms. How much less power is wasted if the electricity is delivered at 50,000 V rather than 12,000 Volts?


Homework Equations


V=IR
P=V^2/R=I^2R


The Attempt at a Solution


I'm having some basic problems conceptualizing what's going on in this problem. The way I understand it, there's a potential difference of 12,000 V from plant to factory. The wires connecting them act as a resistor of 3ohms. At the factory, 750 kW of power arrives and I'm supposed to find how much power dissipates in the wires and what the difference is if the potential is 50,000 instead.
So I decided to find the potential at the factory, since I assume we don't count the factory as V=0. So:
P=V^2/R so V=sqrt(PR)=sqrt(75kW*3 ohms)=1500 V
So there's been a Voltage drop of 12,000-1500=10500V, so the power dissipated is 3.6 x 10^7 W.
In the case of 50,000 being the potential the potential drop is 48500 W. So power dissipated is 7.8 x 10^8 W. Now when I take the difference of these I get 7.4 x 10^8.
Now, I admit that none of this makes sense because the power lost is greater at 50,000 V than at 12,000, which it obviously shouldn't be. Also, the correct answer is 11kW, so this is horribly incorrect. Can someone steer me towards the right way to think of this problem?
 

Answers and Replies

  • #2
Delphi51
Homework Helper
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P=V^2/R so V=sqrt(PR)=sqrt(75kW*3 ohms)=1500 V
No. These electrical formulas apply to one component. In this case, you are doing the resistance. But the power lost in the resistor is NOT 75 kW.

Draw the circuit diagram: a source connected to a load through a 3 ohm resistor.
It says "delivered to the load" so write 12 kV and 750 KW on the LOAD side.
You have two quantities for the load, so you can calculate something else about it - current. The current will be the same for the resistor, so you can then make some progress there - calculate the power for it. About 12 kW I think.
 
  • #3
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That makes a lot more sense. I was confused as to where to view the potential difference. Thanks for the advice. Incidentally, the 75 kW was a typo, but thanks for pointing it out, too!
 
  • #4
Delphi51
Homework Helper
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the 75 kW was a typo
Oh, I didn't notice that it was 75 and not 750!
It is wrong because that is the power for the load, not the power for the resistance in the line. You must be very aware of which numbers apply to which components. And when using a formula for a component, only use the numbers that apply to it. That is the key to all electrical calcs. Good luck!
 

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