A printer-friendly version of this document is available as a PDF file (40 Kbytes).

Battery & resistor games

These notes are a wrapped-up version of the problem solved in the tutorial. It is a well known problem in Physics and Engineering (although I don't have a precise reference for it).

The problem: Assume we have a DC circuit as the one sketched in Figure 1. The battery has the EMF $\mathcal{E}$ and internal resistance $r$. The external resistor has resistance $R$ which can be varied (from 0 to $\infty$). The task: find the value of $R$ for which the power $P_R$ delivered to/dissipated on $R$ reaches its maximum.

Figure 1: Circuit diagram for one battery + resistor

To solve the problem we apply Kirchhoff's law for the loop. We have

$$\mathcal{E} = I r + I R,$$ (1)

where stands for the current through the circuit. The power delivered to is given by

$$P_R = P(R) = I^2 R = \frac{\mathcal{E}^{2} R}{(R+r)^{2}}.$$ (2)

At this point we can just simply take the derivative of with respect to , set it to zero and solve for . The function $P(R)$ at the solution will have an extremum; to check that we have a maximum we would need to take the second derivative, $P^{\prime\prime}(R)$ and see if it is negative when evaluate at the ``suspect'' point.

Figure 2: A sketch of the $P(R)$ behaviour

It would be a bit more educating to proceed at a slower pace, with a bit less mathematics (at least at the beginning). Let us first stare a bit at equation (2), which is now rewritten in the form

$$P(R) = \frac{ \mathcal{E}^2 }{r^2} \times \frac{R}{ \left(\displaystyle 1+\frac{R}{r} \right)^2 }.$$ (3)

Now it becomes obvious that for $R\rightarrow 0$ the expression for $P_R$ also goes to 0, and almost linearly (indeed, in the denominator we have $(1+ R/r)^2 \approx 1$, hence $P_R \propto R$ in this regime). For $R\rightarrow \infty$ the picture is different, but $P_R\rightarrow 0$ as well. In this case the denominator may be approximated by $(R/r)^2$; therefore, the $P(R)$ behaves like $R/ (R/r)^2 \propto 1/R$. The behaviour of $P(R)$ in both extreme cases is sketched in Fig. 2. Clearly, there must be a value(s) of $R$ for which $P(R)$ reaches maximum. To find the value of $R$ for which $P_R$ reaches the extremum we still need to do the math. The first derivative on ``the top of the hill'' has to vanish:

$$\frac{\mathrm{d}P(R)}{\mathrm{d}R} = \mathcal{E}^2 \frac{r-R}{(r+R)^3} =0.$$ (4)

The unique solution of the above equation is $R =r$. To convince yourself that this value provides a maximum for $P_R$ you have to take the second derivative of $P(R)$ and check it is negative when evaluated at $R =r$. This task is left to the reader. For completeness, let us calculate the value of $P_{max}$. We have

$$P_R\Bigl\vert _{R=r} = \frac{1}{4} \frac{\mathcal{E}^2}{r} = \frac{1}{4}I_{short}^2 r.$$ (5)

Here $I_{short}$ stands for the shortcircuit current - this is the maximum current the battery can deliver. One may complicate the problem by considering (see Fig. 3) two batteries connected in parallel and delivering power to a (variable) resistor $R$. The reader is again advised to perform the same task: find the condition $R$ should obey for the power dissipated on it to reach its maximum. As an incentive here is the answer for the power:

$$P_R = \frac{ \left( \displaystyle \frac{\mathcal{E}_1}{r_1... ...+ R \left(\frac{1}{r_1} + \frac{1}{r_2} \right)\right)^2} R.$$ (6)

The power is maximum when

$$R = r_{e} = \frac{r_1 r_2}{r_1 + r_2}$$ (7)

and the expression for the maximum power is similar to (5), with $r$ being replaced by $r_e$ and $I_{short}$ being now the sum of the shortcircuit currents.

Figure 3: The circuit for the 2 batteries + resistor case

Last revised: 02/07/05 Written by Sorin Codoban