Electric Currents and Resistance

AI Thread Summary
The discussion focuses on calculating power loss due to resistance in a high voltage transmission line. The line has a resistance of 0.06 ohms/km, carries a current of 1337A, and spans 191km. Participants clarify that the potential refers to voltage, which is essential for applying Ohm's law and calculating power loss. After calculating the total resistance as 11.46 ohms, one participant initially miscalculates the power loss as 648.01 MW, while another correctly arrives at 20.5 MW using the formula P = I^2 R. The conversation emphasizes the importance of using accurate values and formulas for correct results.
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A high voltage transmission line with a resistance of 0.06 ohms/km carries a current of 1337A. The line is at a potential of 500kV at the power station and carries the current to a city lovated 191km from the power station.
What is teh power loss due to resistance in the line? Answer in units of MW.

I know the formulas, but don't know how to start it or what the potential means.
 
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First find the resistance of the line. You have all the information you need.

The potential is the voltage, use that with Ohms law to find the current.

Apply the same power equation as in the last problem.
 
OK, this is what I have done so far.

I used the .06 ohms/km and multiplied it bye 191km, to get the resistance. I also found the original power by multiplying the 1337A by the 500000V. So...I have 11.46 ohms of resistance and the original power is 6.685e8. Now I think I should plug the 11.46 ohms and the 1337A into the P = I^2 R formula. Any corrections?
 
That sounds good to me.
 
It's not working... the online homework service said it was wrong. I took the original number and the end number and subtracted them, then converted to MW and got 648.01, but that is not right apparently. Any help?
 
How did you get to 648.01MW? I get 20.5MW using RI^2. Are you sure you are using the right values?
 
Thanks for the help but the homework is passed due, I tried it again and got 20.5MW
 
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