For Fig. 31-35, show that the average rate at which energy is dissipated in resistance R is a maximum when R is equal to the internal resistance r of the ac generator. (In the text discussion we tacitly assumed that r = 0.)

The internal resistance is r.

The oad resistance is R

The voltage is $\epsilon$.

Equating the derivative of power with respect to load resistance to zero, shows that the average rate at which energy is dissipated in resistance Ris maximum whenRis equal to the internal resistance rof the ac generator.

Formula:

The dissipation of power through load resistance is,

$P_L=I_{rms}^{2}R$

Here, $l_{rms}$is the rms current, R is the resistance, and $P_L$is the load power.

The dissipation of power through load resistance is,

$P_L=I_{rms}^{2}R$

The rms current is,

$I_{rms}=\frac{I}{\sqrt{2}}$

Hence, the power will be,

$P_L=\frac{1}{2}{I}^{2}R$ ….. (1)

According to Ohm’s law, current through the load is

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

Putting this value in equation1, you get

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

For maximum power,

$$\begin{gathered} \frac{d{P}_L}{dR}=0 \\ \frac{{\epsilon }^2}{2}\left(\frac{{\left(R+r\right)}^2-2R\left(R+r\right)}{{\left(R+r\right)}^4}\right)=0 \\ \left(\frac{1}{{\left(R+r\right)}^2}-\frac{2R}{{\left(R+r\right)}^3}\right)=0 \\ d \end{gathered}$$

Therefore,

$$\begin{gathered} \frac{1}{{\left(R+r\right)}^2}=\frac{2R}{{\left(R+r\right)}^3} \\ R+r=2Rl \\ R=r \end{gathered}$$

Hence, the average rate at which energy is dissipated in resistance R is maximum when R is equal to the internal resistance r of the ac generator.