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

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1. May 12, 2013

### pietastesgood

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. May 12, 2013

### ehild

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

3. May 12, 2013

### pietastesgood

Aha! I understand now, thanks!