Electric Current Power Plant problem

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Homework Help Overview

The problem involves calculating the current in a power transmission line and the power loss during transmission from a power plant to a city. The context includes parameters such as power output, distance, resistance per unit length, and voltage levels.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods to calculate current using different equations related to electric power and current. There are attempts to clarify the appropriate use of voltage and resistance in these calculations.

Discussion Status

Some participants have provided guidance on the correct interpretation of voltage in relation to resistance and current. There is ongoing exploration of how to accurately calculate the current and power loss, with multiple interpretations being considered.

Contextual Notes

Participants note the complexity of the problem due to the distinction between line-to-ground voltage and the voltage drop along the transmission line. There are also references to the assumptions made regarding resistance and power calculations.

xaer04
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[SOLVED] Electric Current

Homework Statement


"A power plant produces 1000 MW to supply a city 40km away. Current flows from the power plant on a single wire of resistance .050\Omega/km[/tex], through the city, and returns via the ground, assumed to have negligible resistance. At the power plant the voltage between the wire and ground is 115kV. a.) What is the current in the wire? b.) What fraction of the power is lost in transmission?&quot;<br /> <br /> Initial Power = P_0 = 1.0x10^9<br /> Wire length = L = 40km<br /> Resistance per unit length = R/L = .050 \Omega /km<br /> -&gt;Resistance for 40km wire = 2\Omega<br /> Voltage between wire and ground = 115x10^3V<br /> <br /> <h2>Homework Equations</h2><br /> Electric Power<br /> P = IV<br /> P = I^2 R<br /> P = \frac{V^2}{R}<br /> <br /> Electric Current<br /> I=\frac{V}{R}<br /> <br /> <h2>The Attempt at a Solution</h2><br /> I tried several of these and got all different answers.<br /> I = P/V = \frac{1x10^9W}{1.15x10^5V} = 8695.7 A<br /> I = \sqrt{P/R} = \sqrt{(1x10^9W)/(2\Omega)} = 22360.7 A<br /> I = V/R = (1.15x10^5V)/(2\Omega) = 57500 A<br /> <br /> I couldn&#039;t even get the power to work out using the remaining Power equation, reliant on Volts and Resistance.<br /> P = (V^2)/(R) = (1.15x10^5V)^2/(2\Omega) = 6.61x10^9W<br /> <br /> Please help me understand this:(
 
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OK, one cannot use the line-to-ground voltage V (115 kV) with R in I=V/R because the voltage drop along the line is not 115 kV. The voltage drop along the line is IR << V. The V (115 kV) is the voltage drop across the load, not along the line.

Similarly P is not the power dissipated along the line. P = VI is the power generated and supplied (with some loss) to the load.

So find I from P/V and then use that I to find the voltage drop along the line IR, and compare to 1000 MW (1 GW). The power dissipated by the line is then I2R.
 
Last edited:
ah... thank you very much:)
 


I still cannot figure out how to do this problem.
I have:
L=40 km
R = 0.050 ohm/km
P = 1000 MW
I = P/V ; at 40 km R=2 ohm; V= 115kV

I = 1000/115?

Am I going about this wrong? What should I do next?
 

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