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AP Physics E&M P=I^2*R vs P=IV

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  1. May 12, 2013 #1
    1. The problem statement, all variables and given/known data

    [Broken]
    Part f. Calculated current = 0.5 A. Voltage = 30 V. Resistance = 40 Ω

    2. Relevant equations

    P=IV
    P=I^2*R
    P=V^2/R

    3. The attempt at a solution

    I've been curious about this for a while. Sometimes when I am asked to find the power dissipated through a resistor, and I use P=IV, it doesn't yield the same answer as if I used P=I^2*R, which is generally the correct answer. This is such an example. P=IV yields 15 W, while P=I^2*R yields 10 W, which is the correct answer. Can anyone explain this? And under which circumstances should I use P=I^2*R, and under which circumstances should I use P=IV? Thanks in advance!
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 12, 2013 #2

    ehild

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    The power dissipated in a resistor is P=IV where V is the voltage across the resistor and I is the current through the resistor. Ohm's Law states that the voltage across a resistor of resistance R is V=IR. So P=IV can also be written as P=I(IR)=I2R.

    30 V is the voltage across the chain of the 40 Ω and 20 Ω resistors. It is 20 V across the 40 Ω resistor when the switch is closed for a long time.

    ehild
     
  4. May 12, 2013 #3
    Aha! I understand now, thanks!
     
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