If we look at $P=VI$, we see that if the current doubles then the potential difference is halved
but this doesn't seem to make sense according to $V=IR$. If we look at that equation, since the resistance remains constant, doubling the current should increase the potential difference. Can someone explain what I'm misunderstanding in simple terminology?
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2 Answers
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When you say "we see that if the current doubles then the potential difference is halved," you're assuming that $P$ is fixed, whereas when you say "doubling the current should increase the potential difference" you're assuming $R$ is fixed. But in fact, it isn't possible to change the current while keeping both of these things constant.
Let us assume that we're actually keeping $R$ fixed, i.e. we're working with a normal resistor. Let's try to work out what happens to the power if we increase the current. Then we have $V=IR$, and we also have $P=IV$. Substituting the first into the second, we have that $P=I^2R$. So if you double the current, the power will increase by a factor of four.
So for a fixed resistor, the power doesn't stay constant, and that's why the potential difference isn't halved when you double the current.
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Your confusion probably stems from the fact that you are not looking at $P$ being fixed.
If you double the current, but want power to remain constant, then voltage must be halved.
Sometimes the mechanics of the math alone can cause confusion, but if you try to relate it to what is physically going on, it helps to clear things up.
