Question

For a moving coil galvanometer, show that `S = G/(n - 1)`, where S is shunt resistance, G is galvanometer resistance and n is ratio of total current to the full scale deflection current.

Answer 1

To turn a Moving Coil Galvanometer (MCG) into an ammeter, connect a low value resistance in parallel with the galvanometer, as shown in the figure.

Let, G be the resistance of the galvanometer.

I_g be the current flowing through the galvanometer.

S be the shunt (low value resistance).

I_s be the current flowing through shunt.

I be the total current and it is given by

I = I_g + I_s   ...(i)

P.D. across S = P.D. across G  ....(As G and S are parallel to each other.)

∴ S.I_s = G.I_g

Putting, I_s = I − I_g    ....[From equation (i)]

∴ S(I − I_g) = G I_g

∴ S = `((I_g)/(I - I_g))G`

Putting, I = nI_g

∴ S = `(GI_g)/((nI_g - I_g))`

= `G/(n - 1)`     ...(ii)

Where n = `I/(I_g)` is the range-multiplying factor.

∴ The current range of the galvanometer can be increased by a factor of n by connecting a shunt whose resistance is smaller than the galvanometer resistance by a factor of n − 1.