Electric Currents and Resistance

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Homework Help Overview

The discussion revolves around calculating power loss in a high voltage transmission line due to its resistance. The problem involves understanding the relationship between voltage, current, and resistance in the context of electrical power transmission.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss how to calculate the resistance of the transmission line and the power loss using relevant formulas. Questions arise regarding the interpretation of potential and the application of Ohm's law. There is also a focus on verifying calculations and ensuring the correct values are used in the power loss formula.

Discussion Status

The discussion includes various attempts to calculate the power loss, with some participants providing guidance on the formulas to use. There are indications of differing results from calculations, prompting further questioning and verification of values used in the computations.

Contextual Notes

Participants note that the homework is past due, which may influence the urgency and approach to resolving the problem. There is also mention of an online homework service providing feedback on the correctness of the calculations.

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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 the 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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