A galvanometer of resistance `95 Omega`, shunted resistance of `5 Omega`, gives a deflection of `50` divisions when joined in series with a resistance of `20 k Omega` and a `2 V` accumulator. What is the current sensitivity of the galvanomter (in div`//mu A`)?

Explanation

In accordance with given problem, the situation is depicted by the circuit diagram in figure. As here `20kOmega` is much greater greater than the resistance of shunted galvanometer `(lt5Omega)`, the current in the circuit will be `I=(2)/(20xx10^(3))=10^(-4)A=100muA` and as this current produces deflection of 50 divisiions in the galvanometer `CS=(theta)/(I)=(50di v)/(100muA)=(1 d iv)/(2muA)`