Circuit C withG=S
Circuit D withG=RSR−S
Circuit B withG=S
Circuit A withG=RS(R−S)
Here, in figure D, current will flow through the circuit when key K1 is closed and K2 is open. The current flowing through the galvanometer is proportional to the deflection in it.
I1=E/(R+G) = kθ
where, E - emf of the cell
R - resistance from the resistance box
G - galvanometer resistance
for current I
θ - galvanometer deflection for current I
k - proportionality constant.
When K2 is closed and by adjusting the shunt resistance S, we can make galvanometer deflection as θ/2. Then the current in the circuit is ;
I2=E/(R+(GS/(G+S)))
Now, a fraction, (S/(G+S)) of the current in the circuit is flows through the galvanometer, which is given by:
I′=I2(S/(G+S))=kθ/2
Now, from the above relations, we can get the resistance of the given galvanometer as:
G=RS/(R−S)