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

$\eta=\frac{\text { energy dissipated in the load }}{\text { energy dissipated in the complete circuit }}$
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

Answer: (D)

Assuming current $I$ flows through the circuit.

Energy dissipated in $\operatorname{load}=I^{2} R$
Energy dissipated in the complete circuit $=I^{2}(r+R)$
$\therefore $ The efficiency $=\frac{I^{2} R}{I^{2}(R+r)}=\frac{R}{R+r}$