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Homework Help: Electric Current Power Plant problem

  1. Sep 18, 2007 #1
    [SOLVED] Electric Current

    1. The problem statement, all variables and given/known data
    "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[itex]\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?"

    Initial Power [itex]= P_0 = 1.0x10^9[/itex]
    Wire length [itex]= L = 40km[/itex]
    Resistance per unit length [itex]= R/L = .050 \Omega /km[/itex]
    ->Resistance for 40km wire [itex]= 2\Omega[/itex]
    Voltage between wire and ground [itex] = 115x10^3V[/itex]

    2. Relevant equations
    Electric Power
    [tex]P = IV[/tex]
    [tex]P = I^2 R[/tex]
    [tex]P = \frac{V^2}{R}[/tex]

    Electric Current

    3. The attempt at a solution
    I tried several of these and got all different answers.
    [tex]I = P/V = \frac{1x10^9W}{1.15x10^5V} = 8695.7 A[/tex]
    [tex]I = \sqrt{P/R} = \sqrt{(1x10^9W)/(2\Omega)} = 22360.7 A[/tex]
    [tex]I = V/R = (1.15x10^5V)/(2\Omega) = 57500 A[/tex]

    I couldn't even get the power to work out using the remaining Power equation, reliant on Volts and Resistance.
    [tex]P = (V^2)/(R) = (1.15x10^5V)^2/(2\Omega) = 6.61x10^9W[/tex]

    Please help me understand this:(
  2. jcsd
  3. Sep 18, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    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: Sep 19, 2007
  4. Sep 18, 2007 #3
    ah... thank you very much:)
  5. Oct 12, 2010 #4
    Re: [SOLVED] Electric Current

    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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