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(A) The shunt resistance $S$ required to convert a galvanometer of resistance $G$ into an ammeter of range $I$ is given by $S = \frac{I_g G}{I - I_g}$

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(A) The shunt resistance $S$ required to convert a galvanometer of resistance $G$ into an ammeter of range $I$ is given by $S = \frac{I_g G}{I - I_g}$,where $I_g$ is the full-scale deflection current of the galvanometer.For an ammeter,the range $I$ is large,which makes the denominator $(I - I_g)$ large,resulting in a very small shunt resistance $S$.For a milliammeter,the range $I$ is smaller (in the milliampere range),which makes the denominator $(I - I_g)$ smaller,resulting in a comparatively larger shunt resistance $S$.Therefore,the shunt resistance of an ammeter is less than that of a milliammeter.

Two identical galvanometers are converted into an ammeter and a milliammeter. For the same current,the value of the shunt of the ammeter as compared to that of the milliammeter is

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