Question

A cell of internal resistances $r$ is connected to a load of resistance $R$. Energy is dissipated in the load, but some thermal energy is also wasted in the cell. The efficiency of such arrangement is found from the expression

$\frac{\text{Energy dissipated in the load}}{\text{Energy dissipated in the complete circuit}}$

Which of the following gives the efficiency in this case?

• A. $\frac{r}{R}$
• B. $\frac{R}{r}$
• C. $\frac{r}{R+r}$
• D. $\frac{R}{R+r}$

Answer 1

The correct option is D $\frac{R}{R+r}$

Let current $I$ flow through the circuit.


Energy dissipated in load $=I^{2}R$,

And energy dissipated in the complete circuit $=I^2(r+R)$

Therefore, the efficiency of the given arrangement will be

$\frac{I^{2}R}{I^2(R+r)}=\frac{R}{R+r}$

Hence, option $\left(d\right)$ is correct.

$$\begin{array}{c}\text{Why this question?}\\ \text{Note: We can also find the efficiency when}\\ \text{power through R is maximum i.e when R=r.}\\ \text{At this condition Efficiency}=\frac{R}{R+R}=0.5\\ \text{MAXIMUM POWER TRANSFER THEOREM}\\ \text{For a given real battery the load resistance maximizes the power}\\ \text{if it is equal to the internal resistance of the battery.}\end{array}$$