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Power Loss In Transmission Lines (Again?)

  1. May 18, 2009 #1
    Edit: My question is better expressed in my next post so ignore this post.

    1. The problem statement, all variables and given/known data
    A power station delivers 890 kW of power at 12 kV to a factory through wires with total resistance 5.0 [tex]\Omega[/tex]

    How much less power is wasted if the electricity is delivered at 50 kV rather than 12 kV?



    2. Relevant equations

    eq-1) P = IV
    eq-2) P = I2R

    Ohm's Law: V = IR

    3. The attempt at a solution

    I know the problem can be solved by solving for the current value in each case from eq-1, and plugging it in to eq-2 to calculate power loss.

    My real question is why can't we apply Ohm's law to calculate I, and then plug it into eq-2? Why must we use eq-1 to get the current? Doesn't Ohm's law apply for all ohmic conductors, and since we have both the resistance and voltage of and across the conductor, can't we get the current that way?


    That was all copied and pasted from a previous thread. I've also googled around and found a similar answer: http://in.answers.yahoo.com/question/index?qid=20090220223426AAe2nSs

    I think I understand the distinction between voltage supplied and voltage drop. However, I still don't understand why the voltage loss through the wire can't be calculated using V=IR.

    For example, In a series circuit with Load1 and Load2,
    the Voltage supplied is 12V
    The Resistance of Load1 is 4 [tex]\Omega[/tex] and Resistance of Load 2 is 2[tex]\Omega[/tex]

    Since V= IRtotal, I = 2A.
    Hence, Voltage drop across Load1 is Vdrop across 1 = IR = 8V
    Similarly, Voltage drop across Load 2 is= IR = 2 x 2 = 4V

    Hence, total drop across the two loads is 12V which is equal to the supplied voltage.

    Now, if you consider the wire in transmission cables to be one large load, why isn't the voltage drop across the wire be equal to voltage supplied?

    I know I'm missing or confused with something but I just don't understand. Hopefully the answer can be explained with high school level knowledge (since I'm in High School).

    Thanks
     
    Last edited: May 19, 2009
  2. jcsd
  3. May 18, 2009 #2

    LowlyPion

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    Welcome to PF.

    It is. And the power lost in the wire is I2*R

    The Power supplied is fixed. It's 890kW.

    From the power supply then there is the equivalence

    V1*I1 = V2*I2 = 890kW

    If you raise the voltage then then the current will drop inversely.

    V1/V2 = I2/I1 or I2 = I1*(V1/V2)

    In the case of the problem your new I2 would be

    I2 = I1*(12/50)

    If your current is .24 of what it was then isn't the power dissipated in the transmission (.24)2 of what it was at the lower voltage?
     
  4. May 19, 2009 #3
    Thanks for the reply.
    Sorry, my first post was all over the place. The following image should explain what I'm confused about better.

    http://img245.imageshack.us/img245/2293/image0023.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  5. May 19, 2009 #4

    diazona

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    The 2-ohm resistance only represents the wires - there will be additional resistance (probably much greater) from the factory itself.

    If the 2-ohm load was the only resistance on the circuit, the three statements at the top of the image would be inconsistent (36000 W dissipated, 12000 V, resistance 2 ohms: these cannot all be true of one circuit)
     
  6. May 19, 2009 #5
    I thought that in electricity distribution, once the voltage is stepped up, the electricity travels through wires only before it is stepped down for domestic use. Since transformers operate through the induction of emf and hence current [which is why I drew those coils in the diagram] in another circuit, I considered the circuit with the transmission wires as "one circuit".
     
  7. May 19, 2009 #6
    And even if I were to consider other resistors in that circuit, wouldn't that mean the total resistance in that circuit would need to be 4000 Ohms for the statements to be consistent?

    And if the total resistance was that, that would mean all the power would dissipate out of the circuit? Hmmm.. I'm probably confusing myself even more now.

    Any ideas?
     
  8. May 19, 2009 #7
    Ok, I thought about it for a while and I came up with this diagram.

    http://img526.imageshack.us/img526/2300/18937972.png [Broken]

    But if I was to take out the coil in the second circuit so the voltage is not step-down, that would lead to the same dilemma which I stated above. Is there something special that happens when there I add coils that can allow voltage to be step up and down into a circuit?
     
    Last edited by a moderator: May 4, 2017
  9. May 19, 2009 #8

    LowlyPion

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    I think you are making it way more complicated than it is.

    The power station is an EMF source.

    It will either source at 12kV or 50kV. But the power is a constant.

    There is a 5Ω load between the power source and the city load that it will drive. Sure there is step up and step down. But the question is asking about the power transmission in between. Since there is 890kW available either way, they just want to know the power loss to deliver.

    And this is determined simply by looking at the current sourced through the transmission impedance of the lines taken in isolation.

    This is I2R.
     
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