Q.Explain the conversion of galvanometer into an ammeter.

A galvanometer is converted into an ammeter by shunting it with a small parallel resistance so that only a safe fraction of the total current passes through its coil.

1. Need for conversion. A galvanometer can safely carry only a very small current $I_g$ (its full-scale deflection current) and has a coil resistance $G$. To measure larger currents $I$ without damaging the coil, and without appreciably disturbing the circuit being measured (an ammeter must have very low resistance), a low-resistance shunt $S$ is connected in parallel with the galvanometer.

2. Working principle. When the combination carries a total current $I$, it divides between the galvanometer (carrying $I_g$) and the shunt (carrying $I-I_g$), the two branches having equal potential difference since they are in parallel:

$I_g G = (I-I_g)S$

3. Value of shunt. Solving for $S$:

$S = \dfrac{I_g\,G}{I-I_g}$

Choosing $S$ this way makes the meter read full-scale exactly when the external current is $I$, and any smaller current is shown proportionally.

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