when a 12 ohm resistor is connected in series with a moving coil galvanometer than it subtraction reduces form 50 division to 10 division find the resistance of the galvanometer Share with your friends Share 0 Anshu Agrawal answered this Dear student, Here let the value of current from one division be i A. Thus the current through ammeter I=50i and through Galvanometer Ig=10i S=IgGI−Ig⇒G=(I−Ig)SIg=50i−10i10i×12Ω =4×12Ω =48Ω Regards 0 View Full Answer Tanmayee answered this Answer We know that for a galvanometer the deflection is proportional to the current passing through it. Thus initially , [V=RG∗50] now , when a 12 ohm resistor is connected the equation will become, [V=(RG+12)∗10] [(RG+12)∗10=50∗RG] [RG=3] ohm 1 Lijin Edward answered this We know that for a galvanometer the deflection is proportional to the current passing through it. Thus initially , [V=RG∗50] now , when a 12 ohm resistor is connected the equation will become, [V=(RG+12)∗10] [(RG+12)∗10=50∗RG] [RG=3] ohm 1