When a resistance of 5Ω is shunted with a moving coil galvanometer,...

The formula to calculate the maximum current through the galvanometer, when a shunt resistance ${r}_{1}$ is connected is given by{\left({i}_{G}\right)}_\max = \dfrac{{r}_{1}}{{r}_{1} + {R}_{G}}{i}_\max . . . \left(1\right)

When a series resistance ${r}_{2}$ is connected, the potential difference across the galvanometer is given by

V = {\left({i}_{G}\right)}_\max\left[{R}_{G} + {r}_{2}\right] . . . \left(2\right)

From equations (1) and (2), it can be written that

\begin{cases}V & = & \dfrac{{r}_{1}}{{r}_{1} + {R}_{G}}{i}_\max\left({R}_{G} + {r}_{2}\right) \\ & ⇒ & V{r}_{1} + V{R}_{G} = {i}_\max{r}_{1}{R}_{G} + {i}_\max{r}_{1}{r}_{2} \\ & ⇒ & \left(V - {i}_\max{r}_{1}\right){R}_{G} = {i}_\max{r}_{1}{r}_{2} - V{r}_{1} \\ & ⇒ & {R}_{G} = \dfrac{{i}_\max{r}_{1}{r}_{2} - V{r}_{1}}{V - {i}_\max{r}_{1}} . . . \left(3\right)\end{cases}

Substitute the values of the known parameters into equation (3) to calculate the required galvanometer resistance.

$\begin{cases}{R}_{G} & = & \dfrac{0 . 250A \times 5Ω \times 1050Ω - 25V \times 5Ω}{25V - 0 . 250A \times 5Ω} \\ & = & 50Ω\end{cases}$