$P=IV$ and $P=I^2R$. Does this mean that $P$ is proportional to both $I$ and $I^2$? I don't understand this concept. And when should I use the various formulas for power? When another variable is constant is one thing I've heard, but sometimes I can just plug everything in different formulas which represent power and get the same answer?
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2 Answers
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It's important to understand where and when the laws are applicable:
$P=IV$
This relationship is true for any "two-terminal" device that consumes electric power. The instantaneous power consumption* of the device is equal to the product of the current flowing through it and the voltage between its two terminals in that same instant.
$I=V/R$
Ohm's Law is valid for a special class of devices, which may be called "resistors" or "conductors." Which name we choose usually depends on the $R$ (a.k.a., "resistance") that characterizes the device. If $R$ is much less than 1, and especially if the "device" really is just a hunk of metal or a length of metal wire, then we'll probably call the thing a conductor. Otherwise, we'll call it a resistor...
...Or, maybe we'll call it something else altogether like, "a heater" or "an incandescent lamp." There are many devices that have resistor-like behavior (i.e., they obey, or at least, come close to obeying Ohm's Law), and then there are other devices that don't come anywhere close to obeying Ohm's Law.
$P=I^2R$
This one can be derived from the first two by simple algebra. Substitute one equation into the other to get rid of the $V$, and there you have it. But Note! The fact that you derived it from Ohm's Law, and the fact that it has $R$ in it should tell you that, like Ohm's Law, it's only meaningful when you use it to describe a resistor-like device that has a fixed $R$.
Like Bob D. said, because of Ohm's Law, the voltage across a resistor-like device, and the current through the device are not independent of each other. In fact, they vary in proportion to each other. If you want to double the current, you must also double the voltage, and according to $P=IV$, that means you've quadrupled the power. And of course, if you double the current, then $P=I^2R$ gives you the same answer.
* Actually, $P$, $I$, and $V$ all can be treated as signed quantities, and if you do it right, then the sign of $P$ tells you whether the device is consuming power from the circuit, or supplying power to the circuit.
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$\begingroup$ P=IV is also true for devices that provide power, if you keep track of your signs correctly. (Okay, now I see your footnote, but it's not clear where the footnote is meant to be attached to the text) $\endgroup$ Commented Apr 15 at 5:03
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P=IV and P=I^2R. Does this mean that P is proportional to both I and I^2? I don't understand this concept.
No. in the first equation $V$ is not independent of I. They are related by Ohms law by $V=IR$. Substituting $IR$ for $V$ in the equation $P=VI$ gives you $P=I^{2}R$.
Hope this helps.
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2$\begingroup$ Re, "They are related by Ohm's Law" That's true if $V$ is the voltage dropped across a resistor, and $I$ is the current through the resistor. It is not generally true of all possible "loads" though. E.G., Some modern electronic devices will draw approximately constant power from their supply, and if the supply voltage goes up, the current draw will go down. $\endgroup$ Commented Apr 14 at 18:52
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$\begingroup$ @SolomonSlow I am only trying to show the OP how $P=I^{2}R$ is derived from $P=VI$ for the ordinary resistive loads. Given the apparently basic level of understanding I don't think it would be helpful to get into electronic device exceptions. Don't you agree? $\endgroup$– Bob DCommented Apr 14 at 18:58