(a) Briefly explain how a galvanometer is converted into an ammeter.(b) A galvanometer coil has a resistance of $15\Omega$ and it shows full-scale deflection for a current of 4 mA. Convert it into an ammeter of range 0 to 6 A.

(a) By connecting a small resistance called Shunt(s) in parallel to the coil of a galvanometer. a galvanometer can be converted into an ammeter.

According to formula,

$S=\frac{I_{g}G}{I-I_{g}}$

Where, $G=$ resistance

$I_{g}=$ Current in the galvanometer.

(b) The formula for the shunt is given as :

$S=\frac{I_{g}G}{I-I_{g}}$

We have given ;

$G=15\Omega ,\; \; \; Ig=4\times 10^{-3}A$

$I=6A ,\; \; \; IgG=(I-I_{g})S$

$S=\frac{4\times 10^{-3}\times 15}{6.4\times 10^{-3}}$

$S=0.01\Omega$

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