1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electical Power

  1. May 10, 2009 #1

    jgens

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

    P = VI = V2/R

    3. 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: May 10, 2009
  2. jcsd
  3. May 10, 2009 #2

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
  4. May 10, 2009 #3

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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?
     
  5. May 10, 2009 #4

    jgens

    User Avatar
    Gold Member

    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!
     
  6. May 10, 2009 #5

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Looks good ... you're welcome.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Electical Power
  1. Electic Field (Replies: 5)

Loading...