qu[![question 24][1]][1]2nd pagederivation for maximum currentSuppose in a circuit the battery has emf 6V and internal resistance 3ohm. It is connected to an external resistance = R ohm. According to my book the maximum power dissipated is when R = 3 ( I.e. internal resistance). They have derived the result and it seems fine. Now power = I^2 ( R+3) and I = 6/(R + 3) so power should be 36/(R+3). But if we put R=3 power is 6 w and if we put R = 0 power is 12 w. So shouldn't power be maximum when external resistance is 0?

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    $\begingroup$ I suspect the book means that the maximum power dissipated in the load $R$ is when $R=3$. If you set $R=0$ then the power dissipated in $R$ is obviously zero, and if you set $R=\infty$ the power dissipated in $R$ is obviously zero. So the maximum of the power dissipated in $R$ is somewhere in between. With a bit of algebra you can easily show the maximum is at $R=3$. $\endgroup$ – John Rennie Feb 4 '16 at 15:39
  • $\begingroup$ I made a mistake actually. According to my book current is maximum when external resistance = internal resistance. But since power =I^2R, power will also be maximum when current is maximum. The question says power dissipated in circuit and not in R @john Rennie $\endgroup$ – rishabh gupta Feb 4 '16 at 16:14
  • $\begingroup$ The book considers only components "external" to the power supply to be the circuit... or it is poorly phrased. $\endgroup$ – Floris Feb 4 '16 at 16:20
  • $\begingroup$ Current is a maximum when $R_\text{ext}=0$ not when $R_\text{ext}=R_\text{int}$. $\endgroup$ – John Rennie Feb 4 '16 at 16:27
  • $\begingroup$ I am posting the proof given in my book. $\endgroup$ – rishabh gupta Feb 4 '16 at 16:29

Your confusion is between two related concepts.

  1. Power dissipated in total = internal power + external power. If that is the power you are talking about, then an external short circuit will maximize the current and therefore maximize total power, $V\cdot I$.
  2. Power delivered to the load. That is the thing addressed by the maximum power transfer theorem, and it requires internal resistance = external resistance.

The proof follows simply. If we have internal resistance $R_i$ and external resistance $R_o$, then the total resistance is $R_i+R_o$. The current is $\frac{V}{R_i+R_o}$ and the voltage across the external resistor is current times resistance. It follows that power in the external resistor is $\frac{V}{R_i+R_o}\frac{V\cdot R_o}{R_i+R_o}$

To find the maximum of that power, we take the derivative w.r.t $R_o$ and set it to zero:

$$\begin{align}\frac{dP}{d(R_o)}\propto \frac{-2R_o}{\left(R_i+R_o\right)^3}+\frac{1}{\left(R_i+R_o\right)^2}&=0\implies\\ -2R_o+R_i+R_o&=0\implies\\ R_o&=R_i\end{align}$$

  • $\begingroup$ Thanks, now I understand the concept of power dissipation. But I don't understand in a circuit why is the current maximum when external resistance = internal resistance? $\endgroup$ – rishabh gupta Feb 4 '16 at 16:26
  • $\begingroup$ The current is not maximum; the power dissipation in the external load is. $\endgroup$ – Floris Feb 4 '16 at 16:29
  • $\begingroup$ I am posting the proof given in my book $\endgroup$ – rishabh gupta Feb 4 '16 at 16:30
  • $\begingroup$ Moreover if we just say that external resistance = internal resistance -1, and for a moment we don't consider the prove, then the power would be more than the power when external resistance = internal resistance. Isn't it? @Floris $\endgroup$ – rishabh gupta Feb 4 '16 at 17:05

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