Why Do P=IV and P=I^2*R Yield Different Results in This Case?

AI Thread Summary
The discussion centers on the discrepancy between power calculations using P=IV and P=I^2*R in a circuit with a 30 V voltage, 0.5 A current, and 40 Ω resistance. While P=IV gives a result of 15 W, P=I^2*R correctly yields 10 W, highlighting the importance of using the appropriate formula based on the context. The power dissipated in a resistor is accurately captured by P=IV when considering the voltage across the resistor and the current through it. Ohm's Law clarifies this relationship, showing that P=IV can be transformed into P=I^2R. Understanding when to apply each formula is crucial for accurate power calculations in electrical circuits.
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Homework Statement




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

Homework Equations



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

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!
 
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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
 
Aha! I understand now, thanks!
 
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