Question: (a) In Fig. 27-4a, show that the rate at which energy is dissipated in Ras thermal energy is a maximum when R =r. (b) Show that this maximum power is $P={\epsilon }^2/4r$.

In the given circuit, the battery with an emf of $\epsilon$ and internal resistance r with an external resistance R is given.

The rate at which the energy is dissipated as thermal energy to an external resistor is called the output power of the battery. Thus, using the concept of Ohm's law, the maximum rate value can be calculated for the relation of external and internal resistances.

Formulae:

The rate at which energy is dissipated as thermal energy to a resistor,

$P=I^{2}R$ (i)

The voltage equation using Ohm’s law,

V =IR (ii)

Here l is the current, and R is the resistance.

From the given figure, the current flow in the circuit is given using equation (ii), as follows:

$I=\frac{\epsilon }{R+r}$

Now, using this current value in equation (i) and then equating the differentiation value of power with the resistance to zero will give the following maximum condition as follows:

$$\begin{gathered} \frac{dP}{dR}=0 \\ \frac{d\left[{\left({\displaystyle \frac{\epsilon }{R+r}}\right)}^{2}R\right]}{dR}=0 \\ \frac{d\left[{\displaystyle \frac{{\epsilon }^{2}R}{{\left(R+r\right)}^2}}\right]}{dR}=0 \\ solvingfurtheras, \\ \frac{{\epsilon }^2{\left(R+r\right)}^2-2\left(R+r\right){\epsilon }^{2}R}{{\left(R+r\right)}^4}=0 \\ {\epsilon }^2{R}^2+{\epsilon }^2{r}^2+2{\epsilon }^{2}Rr-2{\epsilon }^2{R}^2-2{\epsilon }^{2}Rr=0 \\ {\epsilon }^2{r}^2-{\epsilon }^2{R}^2=0 \\ R=r \end{gathered}$$

Hence, it is shown that the rate of dissipation is maximum at the condition R =r.

Now, using the current value from part (a) in equation (i), the rate of dissipation or the power output of the given circuit can be given as follows:

$P={\left(\frac{\epsilon }{R+r}\right)}^{2}R$

Thus, the maximum power of the circuit can be given using the above condition R =r as follows:
$$\begin{gathered} P={\left(\frac{\epsilon }{r+r}\right)}^{2}r \\ =\frac{{\epsilon }^{2}r}{4r^2} \\ =\frac{{\epsilon }^2}{4r} \end{gathered}$$

Hence, the maximum power is$\frac{{\epsilon }^2}{4r}$ .