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

$\frac{Energydissipatedintheload}{Energydissipatedinthecompletecircuit}$

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)

$\frac{R}{R+r}$

Let current I flows through the circuit.

Energy dissipated in load of resistance R is $I^{2}R$ and

Energy dissipated in the complete circuit is $I^2(r+R)$

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