The resistance of a galvanometer is $50\,\Omega $ and current required to give full scale deflection is $100\,μA$ in order to convert it into an ammeter for reading upto $10\,A$. It is necessary to put an resistance of
AIIMS 2010, Medium
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$\mathrm{G}=50\, \Omega$, $\mathrm{I}_{\mathrm{G}}=100\, \mu \mathrm{A}$, $\mathrm{I}=10\, \mathrm{A}$
Shunt, $\mathrm{S}=\left(\frac{\mathrm{I}_{\mathrm{G}}}{\mathrm{I}-\mathrm{I}_{\mathrm{G}}}\right) \mathrm{G}$
$\Rightarrow $ $\mathrm{S}=\left(\frac{100 \times 10^{-6}}{10-100 \times 10^{-6}}\right) \times 50$
$=\frac{10^{-4}}{10} \times 50=5 \times 10^{-4}\,\Omega$
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