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Ohm's law states that  electric current is directly proportional to voltage provided that physical conditions like temperature remain constant i.e.
$$V = IR$$

On the other hand,

$$\text{Power = Voltage} \times \text{Current}$$

So here it seems that the higher the voltage, the lower the current, provided that the power remains constant (i.e. current is inversely proportional to the voltage here which is against Ohm's Law.).

Now my question is how do physicists explain this apparent contradiction? Or maybe this not a contradiction because I am analysing things incorrectly?

P.S.: I am a tenth grade student so please refrain from the usage of highly complicated terminologies in your answers.

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    $\begingroup$ This is not an answer to your question, but you know that Ohm's law it's not a fundamental law. While this law describes many (or most) cases, some materials/systems may not follow this relation $\endgroup$ – cinico Dec 2 '16 at 18:40
  • $\begingroup$ @cinico i know non ohmic resistors do not follow this relation but hows this related to the question. $\endgroup$ – user116688 Dec 2 '16 at 18:44
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    $\begingroup$ As I said, it's not the answer to your question but a side note that I thought it would be interesting to add because there are materials/systems with a "negative resistance". To be more specific, this doesn't mean that you will have a negative electrical current, but only that the current can decrease with increasing voltage. $\endgroup$ – cinico Dec 2 '16 at 19:33
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    $\begingroup$ So what you've discovered might be described as "current needs to go inversely with the voltage if you want to keep the same power output" as well as "current needs to go proportional to voltage if you want to keep the same effective resistance." Therefore what you have discovered is that there is no way to simultaneously modify both current and voltage in a way that keeps both the same: that is what the contradiction tells you. So if you raise the voltage and keep the resistance the same, you're going to be forced to increase both current and power dissipated. $\endgroup$ – CR Drost Dec 2 '16 at 20:53
  • $\begingroup$ I don't see any contradiction in what you describe. You changed the variables between the two equations hence anything can happen since you have some "free variable" that can make things turn out how you want. In this case: when you increase voltage you can just increase the power and tada, the current can increase too! $\endgroup$ – Bakuriu Dec 3 '16 at 11:03
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When you are going from an equation to a proportionality statement you need to be mindful of what is being kept constant.

$V=IR$ means that $I$ varies directly with $V$ if $R$ is constant.

$P=IV$ means that $I$ varies inversely with $V$ if $P$ is constant.

The only time you could get a contradiction is if you are comparing situations where the power is constant and also the resistance is constant. But if that's the case you'll find there is only one solution for $I$ and $V$, that is to say, with those restrictions $I$ and $V$ can't vary - directly or inversely.

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    $\begingroup$ "But if that's the case you'll find there is only one solution for $I$ and $V$" Technically for given values of $P$ and $R$, you will get a system of two equations in two variables. This system of equations might have a single solution. $\endgroup$ – Code-Guru Dec 3 '16 at 1:33
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    $\begingroup$ Specifically, there is a solution for the system of equations if $PR \ge 0$. That is, both $P$ and $R$ have the same sign. In most physical situations, they are both positive, so this requirement will be met. $\endgroup$ – Code-Guru Dec 3 '16 at 1:37
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There is no contradiction here. In fact, the power equation is often represented in different ways:

\begin{eqnarray}P & = & IV\\ P & = & I^2R \\ P & = & V^2/R \end{eqnarray}

The equations can be manipulated depending on which variables you are controlling.

Most commonly, your circuit has a particular resistance. Your power source has a particular voltage. And so you can figure out the current, by working out $I=V/R$. Bigger voltage is bigger current. Bigger resistance is smaller current. In this same situation, you can work out the power that's going into the circuit. You know all three, so the power will be the same whichever equation you use. But, since you know the voltage and the resistance, you might say that $P=V^2/R$. So doubling the voltage will quadruple the power.

To be specific, $V$,$I$, and $R$ are going to be related by Ohms law for a given circuit. If you adjust one, one of the others will change. The power law can then be used to figure out how much power is being used in the circuit.

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The thing is that proportionality means a constant in the formula. $R$ is a constant, but $P$ isn't.

  • Someone (Ohm) found out that $V$ is proportional to $I$: $$V \propto I$$ He found out by testing and experimenting. Everytime $V$ was doubled, $I$ doubled as well. This is what is called (direct) proportionality. We can write it with a proportionality constant, for example called $R$: $$V=RI$$ $R$ is constant, which causes the doubling of $V$ in our circuit to double $I$ as well.

  • Someone then also found the relationship: $$P=VI$$ by testing and experimenting. And yes, we could write this as: $$V=P\frac1I$$ which at first sight looks like inverse proportionality. Double the $V$ should then half the $I$.
    But $P$ is not constant. If we change $V$ in our circuit, both $I$ and $P$ change as well. Doubling $V$ does not half the $I$.

Conclusion is that $R$ is a proportionality constant, while $P$ certainly isn't (because it isn't constant). It looks like inverse proportionality at first sight, but isn't.

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Um ... my two cents....

I don't see where does Ohm's law even talk about power? All it talks about is that V=IR irrespective of the power consumed.

If voltage is high then current is low, but to maintain that low current with the high voltage you need high resistance. E.g. for a deltaV of 10V and 1 A current, u need a 10 Ohm resistor or rating higher than 10W (10Vx1A) of higher.

If you happen to put a 5 Ohm resistor across 10V then you will get 2A current and your power (i.e. heat generated) in the resistor will be 20W. Conversely if you want to retain the power of 10W with 2A then you need a 2.5 Ohm resistor which will give you 2.5Ohmx2A = 5V voltage drop and hence 5Vx2A = 10W power.

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    $\begingroup$ Note that this is not a forum in a traditional sense, but a Q&A site. I did not downvote, but I understand that people would downvote if you start with "Um ... my two cents....". Either you know the answer or you do not - any number of cents is generally not appreciated. $\endgroup$ – Sanchises Dec 3 '16 at 15:51

protected by Qmechanic Dec 3 '16 at 17:28

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