JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Current Electricity

For the voltage range of a single galvanometer we want the largest total resistance in series with it, so that a given full-scale current corresponds to the largest terminal voltage. Placing all four components in one series loop makes the whole string carry the galvanometer's full-scale current $I_g$, giving a range $I_g(2R_C+2R)$. If instead the two resistors and one galvanometer are put in series while the second galvanometer is connected in parallel with the first, the two identical coils share the line current, so the series part carries $2I_g$ when each coil reaches full scale; the terminal voltage is then $2I_g(2R)+2I_g\cdot \frac{R_C}{2}=4I_{g}R+I_g{R}_C$. Since $R_C<R/2$, this exceeds the all-series value by $I_g(2R-R_C)>0$, so the parallel-coil arrangement actually gives the larger voltage range. For the current range we want the smallest shunt across the coil, so that the largest line current produces full-scale deflection; connecting every component in parallel provides the lowest shunt resistance and hence the maximum current range.