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Consider the following elementary derivation of the formula $P = VI$ for Joule heating of a resistor $R$ for the dissipated power $P$ where the voltage over the resistor is $V$ and the current $I$.

Assume the voltage is defined as $V = \dfrac{W}{q}$ where $W$ is the work done by moving (in thought) a positive test charge $q$ against the field lines form one terminal to the other in the resistor. This is the same (absolute value of) work as that done by the electric field on a charge which moves as part of the current thought the resistor.

Since we assume as steady current, the charge will not become accelerated so the energy (all work done by the electric field as it passes the resistor) must dissipated as heat.

Now consider the following picture which represents a perfect conductor (white) and a resistor (gray) for three instances of times. The moving charges consituting the current $I$ are illustrated by dots. Consider now the red dots, say $N$ times the charge $q$. Those $N$ charges pass the resistor in the time $t$. But in the same time $2N$ charges pass the cross section $A$, i.e. $I = \dfrac{2qN}{t}$.

enter image description here

Then you get:

$$ \begin{align} W &= NqU \\[6pt] &= \frac{1}{2} \frac{2Nq}{t} U t \\[6pt] &= \frac{1}{2} I U t \end{align} $$

So

$$ P = \frac{W}{t} = \frac{1}{2}UI $$

The factor $\frac{1}{2}$ shows that something must be wrong with this argument. You can see this if you consider a much larger time, then you would get even another factor.

My question is how can transform this idea into a correct argument with a correct result. The more elementary the better.

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Your error is to assume that only your red charges generate the heat, ie the red charges go through area $A$ and they are not replaced by any other charges.
If that were the case then the factor of $\frac 12$ would be correct.
However as the red charges move through the resistor black charges to the left of the red charges would move into the resistor and dissipate heat as well.
So for every red charge which moves a certain distance through the resistor and then out through area $A$ there will be a black charge which moves through the resistor such that the combined distance travelled by the red and black charges is the length of the resistor.

So overall the black and red charges do an amount of work $W=NqU+NqU = 2NqU$

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  • $\begingroup$ Thanks, yes that was the error. You have to consider the black charges two times. The first time when going out of the resistor (from the first to the second picture). Some of them do 1/3 of the possible work others 2/3... and then you have to consider the black charges going in from the second to the third picture. Some of them do 2/3 of the work, some of them 1/3 adding those two "black contributions" together gives $Nq$ which gives with the red $Nq$ contribution the $2Nq$. Is there any way to make the argument simpler? $\endgroup$ – Julia Jun 22 '16 at 11:09
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Since you equate $W$ with $NqU$, that means $W$ represents the amount of energy dissipated as heat during the time interval $N$ charges passed through the cross section. That time interval is $t / 2$, so the resulting power is $P = {{UIt / 2} \over {t / 2}} = UI$.

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