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Power, electricity, multiple choice Q

  1. Mar 20, 2012 #1
    1. The problem statement, all variables and given/known data

    If you step up the voltage in a wire by a factor of 100, how is heat affected?
    A. No change
    B. Heat loss increases by a factor of 10000
    C. Heat loss decreases by a factor of 10000
    D. Not enough information to tell

    Answer: "C, heat loss decreases by a factor of 10000 because the joule heat generated is I2 * R and since P=IV, a 100fold increase in V is a 100fold decrease in I."

    Another short question, please:
    Regarding the V in P=IV. Does V correspond to the voltage drop of the circuit element (resistor, light bulb, etc) or the voltage of the emf source?

    2. Relevant equations
    P=IV, V=IR


    3. The attempt at a solution
    My most recent attempt was to use P=V^2/R so we can only focus on V and P. BUT, this tells me that power increases, not decreases.

    Is the book wrong?
     
  2. jcsd
  3. Mar 20, 2012 #2

    cepheid

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    P = VI is kind of a general formula. In the case of a resistor, the power dissipated as heat in the resistor is the voltage across the resistor multiplied by the current through the resistor.

    In the case of a battery or voltage source, the power provided by the source is equal to the voltage across the source multiplied by the current through the source.

    If you think about where this equation comes from in the first place it makes sense. The voltage across the source is the potential energy gained per unit of charge passing across the source. The current through the source is the rate at which charge passes across the source. So for the product of the two, you have:

    (energy per unit charge)*(charge per unit time) = energy per unit time

    The product of voltage and current therefore tells you the rate at which energy is imparted to or taken away from charges flowing in a circuit.
     
  4. Mar 20, 2012 #3

    cepheid

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    This answer in bold basically makes no sense at all. First of all, the power is dissipated in the wire as heat. So P = heat, and to figure out the change in heat generated, you figure out the change in P.

    Assuming you have a wire with a constant resistance, and you increase the voltage across it by a factor of 100, then you'll ALSO increase the current in it by a factor of 100 (assuming R is constant -- this just comes from Ohm's law). Since P = IV, you'll increase P by a factor of 100*100 = 10,000.

    BOTH P = I2R and P = V2/R give you this result.
     
  5. Mar 20, 2012 #4
    This makes perfect sense! thank you.
     
  6. Mar 20, 2012 #5
    i'm glad. this has bothered me for a long time.

    thank you very much for the very helpful responses.
     
  7. Mar 20, 2012 #6

    cepheid

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    Okay alternatively, here is the situation that the book MIGHT be talking about, and it is different from what I originally envisioned, which was just connecting a wire (resistor) across a voltage source.

    Maybe the book is talking about a situation like where your wire is a power transmission line that connects a power plant to a city. In this case, the power output P of the power plant is constant, P = const. Also, in this case, P is not the heat dissipated in the transmission lines. That heat is only a small fraction of the total power generated. The rest of it goes into doing useful work at the load (i.e. powering people's washing machines etc.)

    So, in this case, if P = const, then increasing V decreases the current being carried by the transmission lines. This reduces the I2R (joule heating) losses in those transmission lines.

    So the idea is that P = VplantI

    There is a small voltage drop Vwire across the power line, and Vwire = IR, where R is the resistance of the power line. Vwire << Vplant. Most of the voltage develops across the load at the other end of the transmission line. In any case, the heat generated in the wire is VwireI = I(IR) = I2R.

    If you increase Vplant by factor of 100, then the new current (call it "I prime") I' = I/100, hence the heat developed in the wire is I'(I'R) = (I/100)(IR/100) = I2R/10,000

    That's the scenario the book might have been referring to.
     
    Last edited: Mar 20, 2012
  8. Mar 20, 2012 #7
    Voltage is increasing, power is constant. Therefore, current must decrease inversely proportional to the V increase. Yes?

    But if P=IV, then P=(I/100)*(100V) = P. It looks like it works with the equation that you used, P=I^2/R, but how would I make it work with P=IV? Seems very unintuitive to not use a power equation with voltage in it given that the question gives you information regarding voltage.
     
  9. Mar 20, 2012 #8

    cepheid

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    Pay close attention. There are two different voltages present in this situation. THat's why I gave them different subscripts. The voltage across the transmission line is what is relevant here, and it is only a small fraction of the voltage supplied by the power plant. Furthermore, it is always equal to IR, so if I decreases, so does Vwire. Read over what I wrote again, and if you still have questions let me know.
     
  10. Mar 20, 2012 #9
    edit: hold on i'm trying to prove to myself why current decreased.
     
  11. Mar 20, 2012 #10
    Ok i'm sorry about missing the details initially.

    P=IV, voltage increases 100x so I must necessarily decrease 100x in order to keep power constant.

    Current decreased in wire and therefore voltage also decreased proportionally.

    P = IV = (I *1/100) * (V * 1/100) = 1/10000

    or the equation that you wrote
    or P=V^2 / R

    this seems in line with what you wrote. If it's correct, thanks again for your help :-3
     
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