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To double the power dissipated by a resistor you can

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data
    1) double the potential difference, ΔV
    2) double the current, I
    3) reduce the resistance, R, by 1/2

    2. Relevant equations

    3. The attempt at a solution
    These problems are usually straightforward but I'm a little confused this time because I keep seeing all of the choices work out.
    I know P=IV, so if I double V or double I, then I will get twice the power.
    I also know P=V^2/R so reducing R by 1/2 will also double the power.
    However, if R is reduced by 1/2, in the equation P=I^2*R the power will be halved instead of doubled? And if I or V are doubled, then for P=V^2/R and P=I^2*R power will increase by 4?

    It's late and I may very likely be overlooking something obvious, but I'm not sure what I'm missing here...
  2. jcsd
  3. Feb 19, 2012 #2


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    Gold Member

    Well, looking at [itex]P=IV[/itex], the problem with thinking you can simply change either value independently is that you can't. Consider a super simple case: [itex] 1\Omega[/itex] resistor attached to a 1V power source. By [itex]V=IR[/itex], this would result in a 1A current and a 1 W power draw.

    Okay, good.

    Now, if you looked at [itex]P=IV[/itex], you would think "what if I just double the voltage?". However, by changing the voltage, considering the resistance is constant, by [itex]V=IR[/itex] you can't have the current stay the same! It must be cut in half. Conceptually, this should make sense because by applying a larger voltage across something, you should expect to draw a higher current. So as you can tell, increasing the voltage will be compensated by a decrease in current. Changing the resistance is the only option. Changing the resistance will change the current flow, but the voltage is fixed (changing the load on a battery doesn't change the fact that the battery is 1.5V or whatever it is). So that is the only choice you have. Halving the resistance will double the current without anything changing with the voltage.
  4. Feb 19, 2012 #3
    Ahhh I see. Thank you very much for clarifying that.
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