Physics - Current Electricity Question with Solution | TestHub

In Fig. call I , the ammeter current directed down, ${\mathrm{I}}_1$ the current up through the left-hand emf, and ${\mathrm{I}}_2$ the current up through the right-hand emf. Loop and junction equations give $\mathrm{E}-{\mathrm{I}}_{\mathrm{r}}-{\mathrm{I}}_1{\mathrm{fR}}_0=0,\mathrm{E}-\mathrm{Ir}-{\mathrm{I}}_2(1-\mathrm{f}){\mathrm{R}}_0=0$, and $\mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2$. Solving, $\mathrm{I}=\mathrm{E}/\left(\mathrm{r}+{\mathrm{R}}_0\mathrm{f}-{\mathrm{R}}_0{\mathrm{f}}^2\right)$. The denominator is smallest when $\mathrm{f}=0$ or 1 , making I a maximum. Since $\mathrm{f}-{\mathrm{f}}^2=\frac{1}{4}-{\left(\frac{1}{2}-\mathrm{f}\right)}^2,\mathrm{I}$ is smallest for $\mathrm{f}=\frac{1}{2}$.