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Explain how a moving-coil galvanometer is converted into an ammeter. Derive the necessary formula.

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A moving-coil galvanometer is converted into an ammeter by reducing its effective resistance by connecting a low resistance S across the coil. Such a parallel low resistance is called a shunt since it shunts a part of the current around the coil, shown in below figure. That makes it possible to increase the range of currents over which the meter is useful.

An ammeter is a modified galvanometer

Let I be the maximum current to be measured and Ig the current for which the galvanometer of resistance G shows a full-scale deflection. Then, the shunt resistance S should be such that the remaining current I – Ig = Is is shunted through it.

In the parallel combination, the potential difference across the galvanometer = the potential difference across the shunt

∴ IgG = Is

= (I – Ig)S

∴ S = \(\left(\cfrac{I_g}{I-I_g}\right)C\)

This is the required resistance of the shunt. The scale of the galvanometer is then calibrated so as to read the current in ampere or its submultiples (mA. µA) directly.

[Notes : (1) Thick bars of manganin are used for shunts because manganin has a very small temperature coefficient of resistivity. (2) The fraction of the current passing through the galvanometer and shunt are, respectively,

\(\cfrac{I_g}I\) = \(\cfrac{S}{S+G}\) and \(\cfrac{I_g}I\) = \(\cfrac{S}{S+G}\)

(3) On the right hand side of Eq. (1), dividing both the numerator and denominator by I , we get,

S = \(\cfrac{1}{(I/I_g)-1}\). G = \(\cfrac{G}{p-1}\)

where p = I/Ig is the range-multiplying factor, i.e., the current range of the galvanometer can be increased by a factor p by connecting a shunt whose resistance is smaller than the galvanometer resistance by a factor p – 1.

∴ p = \(\cfrac{G+S}{S}\)

If RA is the resistance of the ammeter,

RA = \(\cfrac{GS}{G+S}\) = \(\cfrac GP\)]

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