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I have the following homework problem.

Consider a power supply with fixed emf $ε$ and internal resistance $r$ causing current in a load resistance $R$. In this problem, $R$ is fixed and $r$ is a variable. The efficiency is defined as the energy delivered to the load divided by the energy delivered by the emf.

When the internal resistance is adjusted for maximum power transfer, what is the efficiency?

The answer is 50% but I'm confused how this is calculated. Here is my thought process so far.

The power dissipated in the load is $I^2R$ and $\displaystyle I = \frac{ε}{r+R}$ so $\displaystyle P = \frac{ε^2R}{(r+R)^2} = \frac{ε^2}{\frac{r^2}{R} + 2r + R}$ and so power transferred to the load will be maximum when $\displaystyle \frac{r^2}{R} + 2r + R$ is a minimum. Taking the first derivative $\displaystyle \frac {d}{dr}(\frac {r^2}{R} + 2r + R) = (\frac{2r}{R} + 2)$ and this equals 0 when r = -R. The second derivative is $\displaystyle \frac{2}{R} > 0$ so this point would be a minimum and therefore max power transferred when r = -R.

But having $r = -R$ doesn't make sense to me. From what I read, for maximum power transfer $r$ should equal $+R$. So I've probably done something wrong in my calculations or assumptions in the previous paragraph. Also wouldn't max power be transferred when the internal resistance is as close to zero as possible?

Can someone please show me a proof that shows why $r = R$ for max power transfer and then how to calculate the efficiency.

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  • $\begingroup$ Your working seems right, the problem is that if you are free to chose $r$ then you want to make it as small as possible, even negative. Internal resistance of any form (in these circumstances) is bad as it wastes power. The value $r=R$ comes from the situation when you can vary $R$ and have a fixed internal resistance $r$. $\endgroup$ – Quantum spaghettification Mar 18 '15 at 20:25
  • $\begingroup$ Also you get the efficiency from power delivered to load/total power delivered. $\endgroup$ – Quantum spaghettification Mar 18 '15 at 20:26
  • $\begingroup$ @Joseph So if I choose a value of 0 for r wouldn't the efficiency be 100% instead of 50%? $\endgroup$ – Devin Crossman Mar 18 '15 at 20:30
  • $\begingroup$ Yes as no power would be lost over the internal resistance $r$. You will get the best efficiency when $r$ is as small as possible. $\endgroup$ – Quantum spaghettification Mar 18 '15 at 20:31
  • $\begingroup$ It shows a bit of respect to the reader to use good formatting. You could TeXify the variables in your leading paragraph. $\endgroup$ – DanielSank Mar 18 '15 at 21:12
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Are you sure you mean the internal resistance "r"? The internal resistance typically can not be adjusted, often this question is phrased in terms of the load resistance "R".

The power dissipated in the external load is: $$ P=\frac{V^2 R}{(r+R)^2}\;, $$ which you need to maximize. If you maximize with respect to R, you find R=r... If you maximize with respect to r, you find r=0; algebraically you found a zero at r=-R, but r can not equal -R! It has a physical boundary at r=0, which gives the maximum for fixed R (you are maximizing on a fixed interval, you need to check the endpoints!).

However, are you sure you want to maximize with respect to "r", or do you actually mean "R"?

Finally, divide the value of the maximize load power by the total power dissipated, which is $$ \frac{V^2}{(r+R)} $$ evaluated at the correct resistance to determine the efficiency.

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  • $\begingroup$ Thanks, every example I could find had variable load resistance but the problem I have states that the load resistance is fixed and the internal resistance is variable. $\endgroup$ – Devin Crossman Mar 18 '15 at 20:46
  • $\begingroup$ Then the answer is clearly r=0 as explained in my answer. And the efficiency is 1. $\endgroup$ – hft Mar 18 '15 at 20:47

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