The current flowing in the galvanometer G when key `K_2` is kept open is I. On closing the key `K_2`, the current in the galvanometer becomes `I//n`, where n is an integer. Figure

Obtain an expression for resistance G of the galvanometer in terms of R, S and n. To what form does this expression reduce when the value of R is very large as compared to S?

Explanation

When key `k_2` is open, current in the galvanometer is
`I=(epsilon)/(R+G)`
When key `K_2` is closed, the equivalent resistance of the parallel combination of G and S is
`R^'=(GS)/(G+S)`
Total current in the circuit,
`I_1=(epsilon)/(R+GS//(G+S))`
This current will divide between S and G in the inverse ratio of S and G.
Hence current through the galvanometer will be
`I_G=(S)/(S+G)=(S)/((S+G))xx(epsilon(G+S))/(R(G+S)+GS)`
`=(Sepsilon)/(RG+RS+GS)`
As per question, `I_G=I/n=1/n(epsilon)/((R+G))`
`:. (epsilon)/(n(R+G))=(Sepsilon)/(RG+RS+GS)`
On solving, `G=((n-1)RS)/(R-(n-1)S)`
When `Rgt gt S`, `G=((n-1)RS)/(R)=(n-1)S`