A galvanometer having a coil resistance of 60 | vedclass

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(A) To convert a galvanometer into an ammeter,a shunt resistance $S$ must be connected in parallel with the galvanometer.The formula for the shunt res

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putting in parallel a resistance of $15\,\Omega$

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(A) To convert a galvanometer into an ammeter,a shunt resistance $S$ must be connected in parallel with the galvanometer.The formula for the shunt resistance is given by $I_{g} \cdot G = (I - I_{g}) \cdot S$,where $I_{g}$ is the full-scale deflection current,$G$ is the galvanometer resistance,$I$ is the maximum current to be measured,and $S$ is the shunt resistance.Rearranging for $S$,we get $S = \frac{I_{g} \cdot G}{I - I_{g}}$.Given values are $I_{g} = 1.0\,A$,$G = 60\,\Omega$,and $I = 5.0\,A$.Substituting these values: $S = \frac{1.0 	imes 60}{5.0 - 1.0} = \frac{60}{4} = 15\,\Omega$.Thus,a resistance of $15\,\Omega$ must be connected in parallel.

$A$ galvanometer having a coil resistance of $60\,\Omega$ shows full-scale deflection when a current of $1.0\,A$ passes through it. It can be converted into an ammeter to read currents up to $5.0\,A$ by:

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