A galvanometer has a resistance of $20\,\Omega$ and reads full-scale when $0.2\, V$ is applied across it. To convert it into a $10\, A$ ammeter, the galvanometer coil should have a
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Given, $V=0.2 V, R_{g}=20 \Omega$ and $I=10 A$
Hence, $I_{g}=\frac{0.2}{2}=0.01 A$
To convert galvanometer into a $10 \mathrm{A}$ ammeter, the galvanometer coil should have to connect a shunt resistance in parallel, given by
$S=\frac{R_{g} I_{g}}{I-I_{g}}$
$S=\frac{(20)(0.01)}{10-0.01}$
$S=\frac{0.2}{9.99}=0.02 A$
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