In a galvanometer there is a deflection of 10 divisions per mA. The internal resistance of the galvanometer is 60 `Omega`. If a shunt of `2*5 Omega` is connected to the galvanometer, calculate the maximum current in which the galvanometer can be dead.

Explanation

Resistance of galvanometer `= G = 60 Omega`
Shunt ` = S+ 2*5 Omega`
In a galvanometer, there is a deflection of 10 divisions per mA
As there are total 50 divisions on scale of galvanometer
`therefore I_(G) = (1)/(10) xx 50 = 5 mA = 5 xx 10^(-3)A`
`S = (I_(G))/(I-I_(G))G`
`I-I_(G) = (I_(G)G)/(S)`
`I = (I_(G) + (I_(G))/(S)G) = I_(G) (1 + (G)/(S))`
`I = 5 xx 10^(-3) (1 + (60)/(2*5))`
`I = 5 xx 10^(-3) (1 + 24)`
`I = 125 xx 10^(-3)A`
`I = 0*125 A`