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$$ \frac{V^2}{R} = P = I^2 \cdot R $$

According to this equation, Power is inversely proportional to resistance on the left side and directly proportional to resistance on the right side.

How can power be both inversely AND directly proportional to resistance?

Looking for more of a conceptual answer than a mathematical one.

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    \$\begingroup\$ There's no conceptual answer. It's a very simple and totally basic mathematical equation, no magic here. There will be no difference even if You change electrical values P, V, R and I to abstract ones like a, b, c and d. \$\endgroup\$ – Jakub Rakus Oct 6 '15 at 17:40
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I've made a couple of plots showing the two different cases. The X and Y axes are current and voltage, and I've also overlaid constant power curves (V=P/I) and constant resistance curves (V=R*I).

enter image description here enter image description here

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These are two totally different situations:

  1. If you keep the voltage constant and change the resistance, the power dissipated is \$P = V^2 / R\$. Note that this also must change the current.

  2. If you keep the current constant and change the resistance, the power dissipated is \$P = I^2 R\$. Note that this also must change the voltage.

Since it is impossible to change the resistance and keep both the voltage and current constant, there's no conflict.

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    \$\begingroup\$ This helps thank you! I think this is the very important part that I was overlooking: "it is impossible to change the resistance and keep both the voltage and current constant" \$\endgroup\$ – Sean McDonnell Oct 6 '15 at 17:55
  • \$\begingroup\$ Yes, that's the key. The generic P = IV definition is a bit easier -- you can picture how the power changes when the current goes up and the voltage goes down or vice-versa. \$\endgroup\$ – Greg d'Eon Oct 6 '15 at 17:57
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Power is \$P = IV\$. You can only relate it to resistance if you decided to fix one and replace the other by Ohm's law. Ohm's law states that voltage across a resistor is linearly proportional to the current flowing through it, or \$V = IR\$. This can be arranged to say that current is inversely proportional to voltage, or \$I = \frac{V}{R}\$.

If you assume \$I\$ is constant and replace \$V\$ with \$IR\$, you get that power is proportional to resistance (\$P = I \cdot (I R)\$). If you assume \$V\$ is constant and replace \$I\$ with \$\frac{V}{R}\$, you get that power is inversely proportional to resistance (\$P = V \cdot \frac{V}{R}\$).

A way of visualize this is think of a constant voltage source (i.e. a battery). When there's a large resistance connected, very little current can flow so very little power is being outputted by the battery, and the resistor won't get too warm because there's less power. If you reduce the resistance more current will flow and the resistor will get warmer because you've increased the power.

Current sources are a bit difficult to visualize, but one way you can think of it is a variable voltage source which increases or decreases the voltage until the current flowing from it is the desired value. If you attach a small resistance, the source doesn't need to work very hard to get an amount of current flowing so little power is expended. If you attach a large resistance the source needs to work much harder to get the same current flowing, thus a lot of power is expended.

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  • \$\begingroup\$ This is very helpful thank you! Especially the last two paragraphs. One follow up question: There are two assumptions in your second paragraph. Does that mean that each version of the equation is only valid when those assumptions are true? \$\endgroup\$ – Sean McDonnell Oct 6 '15 at 17:53
  • \$\begingroup\$ Yes, the two equations are mutually exclusive. \$\endgroup\$ – helloworld922 Oct 6 '15 at 18:02
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For \$P = I^2R\$ and on a voltage fed circuit (i.e. a battery and a resistor) if you doubled the resistance then current halves so no, it isn't so simple to just say power is proportional to resistance unless you are talking about a constant current circuit.

And, if you applied the \$P=\dfrac{V^2}{R}\$ formula to a circuit fed with a constant current you'd find that as you increased resistance, power would also increase.

You have to choose the most appropriate formula but, understanding what really happens means you can use either formula.

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see there is no concept that R is directly proportion to P. according to equations P--I2R and P--V/R R seem to be directly proportional but that is not so R stands as an substitution and is constant it does not increase or decrease power it is just constant

but in second equation it stands as a proportional quantity thus affect power indirectly

hope it helped you

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$$ P=I^2R \tag{1} \label{1} $$ $$ P=\frac{V^2}{R} \tag{2} \label{2} $$

Comparing \$\eqref{1}\$ and \$\eqref{2}\$, we get

$$\begin{align} I^2R & = \frac{V^2}{R} \\ I^2R^2 & = V^2 \end{align}$$

square-rooting on both sides

$$ \sqrt{I^2R^2} = \sqrt{V^2} $$ $$ IR = V $$

and removing the constant of resistivity

$$R \sim V$$

Hence, when the p.d. is directly proportional to the resistance (by Ohm's law), power can be directly as well as inversely proportional to resistance at the same time.

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I think you're confusing power with work. Work is the amount of energy converted, for example resistance creates heat from pressure or voltage. This is a quantity of heat. Power is the speed at which this heat is created, or how fast.

For example, walking a mile burns 350 calories, but it takes 30 minutes. Sprinting a mile also burns 350 calories, but only takes 5 minutes. Sprinting requires 6 times more power, even though the same amount of work has been done. So, power consists of two things heat created or energy spent, and time.

Resistance of an object is neither energy spent or a period of time. So in and of itself, resistance has no relationship with work or with time interval. None of these units are compatible by themselves. It's like comparing the compressive strength of steel with the boiling point of water. They measure two completely different things. In and of themselves they have no relationship. However, you add a conditional component that both can share, and the comparative changes for each can can create a connection. For instance, add a variable component in the mix, like adding an electrical current to both the boiling water and the steel, then measuring the steel strength and the boiling point to see if that changes either one or both of their measurements. Now you have a comparison you can make, not to one another directly, but to how they both react to this new component.

Let's say, adding an electrical current to the water lowers its boiling point, and adding the same electrical current, lowers the strength of the steel. You can say in regards to the electrical current, that both the boiling point of water, and the strength of the steel are directly proportional because they both go down. Now, this is not real but it shows you how the relationship between two units of measurement can change.

Same is true for resistance of a medium and the rate of heat created by it. Resistance is a static measurement based on the characteristics of a material component. Power is a dynamic measurement based on the conditions or multiple components, (amount of electrical current per second) amps, and (differential charge of the conductor) voltage.

Hope that this is a better conceptualization than just bouncing formulas around.

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I can simply say that in the first equation P = V^2/R power is inversely proportional to R because it's in an parallel circuit remember that in a parallel circuit voltage is constant throughout. (And we have V in the equation) And in the second equation P=I^2R. P is directly proportional to R because it's in series circuit remember taht in a series circuit curret is constant..... (And we have I in the equation also) Hope it helps... :)

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  • \$\begingroup\$ Why do you assume a parallel circuit for calculating power by voltage and a series circuit for calculating it by current? \$\endgroup\$ – jusaca Sep 19 at 5:36

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