The resistance of a galvanometer is $50\,\Omega $ and it requires $2\,\mu A$ per two division deflection. The value of the shunt required in order to convert this galvanometer into ammeter of range $5\,A$ is (The number of divisions on the galvanometer scale on one side is $30$)
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The shunt $\mathrm{S}$ to be connected is given by
$\mathrm{S}=\frac{\mathrm{I}_{\mathrm{g}} \mathrm{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}$
$\mathrm{I}_{\mathrm{g}}=\left(\frac{\mathrm{I}}{\theta}\right) \times \mathrm{N}=\left(\frac{2 \times 10^{-6}}{2}\right) \times 30$
$\Rightarrow \mathrm{I}_{\mathrm{g}}=30 \times 10^{-6} \mathrm{\,A}$
$\therefore \mathrm{S}=\frac{30 \times 10^{-6} \times 50}{5-30 \times 10^{-6}} \cong 3 \times 10^{-4}\, \Omega$
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