In a galvanometer there is a deflection of 10 divisions per `mA`. The internal resistance of the galvanometer is `60Omega`. If a shunt of `2*5Omega` is connected to the galvanometer and there are 50 divisions in all on the scale of galvanometer what maximum current can this galvanometer read?

Answer

125

Explanation

Since the galvanometer has 50 divisions, so current for full scale deflection is
`I_g =1/10 xx 50 mA= 5mA= 5 xx10^(-3)A`
`G = 60 Omega,S=2.5 Omega`
Let I be the maximum current which a galvanometer can read when shunted with resistance S, then
`I_g = (IS)/(G+S) ` or `I=(I_(g)(G+S))/S`
`I=((5xx10^(-3))(6+2.5))/(2.5)=125 xx10^(-3)A`
`I = 125 mA`