Three identical cells each having an emf $1.5 \mathrm{V}$ and a constant internal resistance $2.0 \mathrm{\Omega }$, are connected in series with a $4.0 \mathrm{\Omega }$ resistor $R$, first as in the circuit $\left(\mathrm{i}\right)$ and second as in the circuit $\left(\mathrm{ii}\right)$.

Then $\frac{\text{Power in} R \text{in circui}\mathrm{t} \left(\mathrm{i}\right)}{\text{Power in }R \text{in circuit} \left(\mathrm{ii}\right)}=$

The circuit shown in the figure contains a battery, a rheostat and two identical lamps. What will happen to the brightness of the lamps if the resistance of the rheostat is increased?

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 an arrangement is found from the expression, $\frac{\mathrm{E}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{y} \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{s}\mathrm{i}\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{d} \mathrm{i}\mathrm{n} \mathrm{th}\mathrm{e} \mathrm{l}\mathrm{o}\mathrm{a}\mathrm{d}}{\mathrm{E}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{y} \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{s}\mathrm{i}\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{d} \mathrm{i}\mathrm{n} \mathrm{t}\mathrm{h}\mathrm{e} \mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{t}\mathrm{e} \mathrm{c}\mathrm{i}\mathrm{r}\mathrm{c}\mathrm{u}\mathrm{i}\mathrm{t}}$. Which of the following gives the efficiency in this case?