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Why do we use I^2R instead of V^2/R when dealing with power loss in transformers?

  1. Jun 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Ok so i was doing my revision for my exams and i was reading on transformers and they said that stepped-up Voltages and stepped-down currents minimizes for tranformers I^2R power losses in the transmission cable. I know that current is decreased, but a higher voltage, by V^2/R, would also mean a greater power loss?

    3. The attempt at a solution
    I know there have been similar questions like this one below.

    physicsforums. com/showthread.php?t=227834( theres a space at the .com cause i cant post links yet)

    But i do not understand their reasoning. For example stewartcs said that problem with this line of reasoning is that P = IV is the power delivered by an electrical energy source not the power dissipated by it.

    Also the person who asked the question said that he was mixing up the supplied nominal V of the power lines with the V lost by the resistance.

    i do not really understand what both of them are saying. can anyone explain it to me? thanks!
     
  2. jcsd
  3. Jun 3, 2012 #2
    V2/R means the voltage drop at the resistor/appliances NOT the supply voltage.
    Less current means less voltage drop whereas the resistance remained.
     
  4. Jun 3, 2012 #3
    no but if u increase the supply voltage wont the voltage drop across the wires also increase? Like for example you increase the voltage of a battery the voltage drop across the resistor increases
     
  5. Jun 3, 2012 #4

    gneill

    User Avatar

    Staff: Mentor

    No, not if you step up the voltage for transmission and step it down again at the other end for delivery to the load; The majority of the voltage drop appears across the transformer windings at each end of the transmission loop. The power lost is the power dissipated in the transmission wires due to the current flowing through the resistance of the transmission wires.

    Yes this current produces a voltage drop in that transmission wire resistance, but due to Ohm's Law that voltage drop is proportional to the current: V = IR. If the current is smaller, then the voltage drop will be smaller. You could use this transmission line voltage drop to calculate the power loss with V2/R, but since this voltage drop is due to the CURRENT flowing in the transmission lines, it's more natural to work with the I2R form.
     
  6. Jun 3, 2012 #5
    oh ok thanks for your help i get it now!
     
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