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Electical Power

  • Thread starter jgens
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  • #1
jgens
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Homework Statement



A 300 MW power station produces electricity at 80 kV which is then supplied to consumers along cabes of total resistance 5.0 ohms. What percentage of the power is lost in the cables.

Homework Equations



P = VI = V2/R

The Attempt at a Solution



I said the total power was PT = 300 MW

Power dissipated in the cables: Pd = V2/R = (80 kV)2/(5.0 ohms) = 1280 MW

This didn't make any sense because that power dissipated exceeds the power produced. So here's another attempt:

Power remaining: PR = 300 MW

Power dissipated: PD = V2/R = (80 kV)2/(5.0 ohms) = 1280 MW

Total Power: PT = PR + PD = (1280 + 300) MW = 1580 MW

Percentage Lost: PD/PT = 1280/1580 = 81.0%

I'm not sure if this is right but it's my best guess. Any help is appreciated. Thanks.
 
Last edited:

Answers and Replies

  • #2
mgb_phys
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In P=IV the V is the voltge drop accross the resistor (in this case the power line) which is not 80kV!
What other formula can you get from P=IV?
 
  • #3
Redbelly98
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The 80 kV is the potential across the 5.0 Ω plus the resistance of whatever the consumers have connected to the circuit. These two resistances may be considered to be in series. The point is, we do not have 80 kV applied directly across the 5.0Ω.

What other relations for power (other than V2/R) do you know about?
 
  • #4
jgens
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Well, here's another go:

Total Power: PT = 300 MW = VI => I = PT/V = (300 MW)/(80 kV) = 3750 A

Power Dissipated thought Cables: PD = I2RC = (3750 A)2(5 ohms) = 70.3 MW

PD/PT = 70.3/300 = 23.4%

Thanks for the help by the way!
 
  • #5
Redbelly98
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Looks good ... you're welcome.
 

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