Question

A cell of internal resistance $r$ 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 an 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 $E_l$ and $E_c$ be the energies dissipated in the load and in the complete circuit respectively.
Let $I$ be the current flowing through the circuit.


$E_l=I^{2}R$
$⟹E_l=\frac{E^2}{(R+r{)}^2}\times R$......(i)

Similarly, $E_c=\frac{E^2}{(R+r{)}^2}\times (R+r)$...(ii)
$\therefore$ Efficiency = $\frac{R}{(R+r)}$