A galvanometer with its coil resistance $25\,\Omega $ requires a current of $1\,mA$ for its full deflection. In order to construct an ammeter to read up to a current of $2\,A,$ the approximate value of the shunt resistance should be
JEE MAIN 2018, Diffcult
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A ccording to question, current through galvanometer, $\mathrm{I}_{\mathrm{g}}=1\, \mathrm{mA}$
Current through shunt $\left(I-I_{g}\right)=2\, A$
Galvanometer resistance $\mathrm{R}_{\mathrm{g}}=25\, \Omega$
Resistance of shunt, $\mathrm{S}=?$
$\mathrm{I}_{\mathrm{g}} \mathrm{R}_{\mathrm{g}}=\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right) \mathrm{S}$
$ \Rightarrow S = \frac{{{{10}^{ - 3}} \times 25}}{2}$
$S \simeq 1.25 \times {10^{ - 2}}\,\Omega $
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