13.5.3 Maximum Power Theorem

If you have a battery with internal resistance r, what should the external resistance R be, in order to draw the maximum output power P_out from your battery?

By P_out, we mean the power dissipated in the external resistance. So

$\displaystyle \begin{aligned}{{P}_{{out}}}&={{V}_{t}}I\\&=(\frac{R}{{R+r}}E)(\frac{E}{{R+r}})\end{aligned}$

Perhaps you feel that having a large external resistance R would result in a large P_out, since a large R brings V_t closer to the open circuit voltage of E. But note that increasing R also brings I towards zero. So in fact, P_out approaches zero as R approaches ∞.

So should we go for a small external resistance then? Having a small R does bring I closer to the short circuit current of $\displaystyle \frac{E}{r}$ . But it would also bring V_t towards zero. So in fact P_out approaches zero as R approaches 0.

So what is just-nice value of R that maximizes P_out? For that, we have to turn to math.

$\displaystyle {{P}_{{out}}}=\frac{R}{{{{{(R+r)}}^{2}}}}{{E}^{2}}$

By differentiating the P_out equation with respect to R and solving for $\displaystyle \frac{{dP}}{{dR}}=0$ or otherwise, it can be shown that maximum P_out is achieved when the external resistance R matches the internal resistance r.

So the maximum P_out turns out to be

$\displaystyle \begin{aligned}{{P}_{{\max }}}&=\frac{r}{{{{{(r+r)}}^{2}}}}{{E}^{2}}\\&=\frac{{{{E}^{2}}}}{{4r}}\end{aligned}$

Let’s remind ourselves that in theory, if your battery has zero internal resistance, infinite output power is attained by an external resistance of zero. In practice, however, your battery always has some internal resistance r. So infinite output power is not possible. Instead, maximum power of $\displaystyle \frac{{{{E}^{2}}}}{{4r}}$ is attained when the external resistance matches the internal resistance of your battery. This is called the maximum power theorem.

Recall that when $R=r$ , the efficiency is only 50% (because half the power is wasted in the internal resistance). That’s the price to pay if high power output is what you’re after. For example, if you want your headphone to be playing at the loudest possible volume, the electrical resistance of your headphone should be designed to match the internal resistance of the circuitry driving your headphone.