It is required to measure the resistance of a circuit operating at `120 V`. There is only one galvanometer of current sensitivity `10^(-6)` A per division. How should the galvanometer be connected in the circuit to operate an ammeter? Why minimum resistance can be measured with such a galvanometer if its full-scale has `40` divisions?

Explanation

In circuit 1, `i` (main current)
`=(epsilon)/((R_(V)(X+R_(A)))/(R_(V)(X+R_(A)))xx(R_(V)(X+R_(A)))/(R_(T)+X+R_(A)))=epsilon`
`i` (ammeter reading)
`=` current in the branch
`=(epsilon(R_(V)+X+R_(A)))/(R_(V)(X+R_(A))(R_(V)+X+R_(A)))xxR_(V)=(epsilon)/(X+R_(A))`
X (apparent value) `=(X^(')-X)/(X)xx100=(R_(A))/(X)xx100`
When `Xto0,etatoinfty` and hence this circuit is unsuitable for low resistance
When `Xto0,etatoinfty` and hence this circuit is suitable for high resistance. in circuit 2.
`i^(')=(epsilon)/((R_(V)X)/(R_(V)+X)+R_(A))=(epsilon(R_(V)+X))/(X(R_(V)+R_(A))+R_(A)R_(V))`
`V^(')=(R_(V)X)/(R_(V)+X)xx(epsilon(R_(V)+X))/(X(R_(V)+R_(A))+R_(A)R_(V))`
`=(epsilon(R_(V)X)/((R_(V)+R_(A))+R_(A)R_(V))`
`X^(')=(V^('))/(i^('))=(epsilonR_(V)X)/(X(R_(V)+R_(A))+R_(A)R_(V))xx(X(R_(A)+R_(V))+R_(A)R_(V))/(epsilon(R_(V)+X))=(R_(V)X)/(R_(V)+X)`
`eta,%` error
`=(X^(')-X)/(X)100=(X)/(R_(V)+X)100`
(numerical value)
`=(100)/(1+R_(V))/(X)`
When `Xto0,etato0` and when `Xtoinfty,etato1`