A galvanometer of resistance $40\,\Omega $ gives a deflection of $5\, divisions$ per $mA$. There are $50\, divisions$ on the scale. The maximum current that can pass through it when a shunt resistance of $2\,\Omega $ is connected is ................ $mA$
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$I_{G}=\frac{50}{5}=10 \mathrm{\,mA} ; \mathrm{R}_{\mathrm{G}}=40\, \Omega, \mathrm{R}_{\mathrm{s}}=2\, \Omega$
Maximum current,
$\mathrm{I}=\frac{\mathrm{R}_{\mathrm{G}}+\mathrm{R}_{\mathrm{S}}}{\mathrm{R}_{\mathrm{G}}} \times \mathrm{I}_{\mathrm{G}}=\frac{(40+2) \times 10}{2}=210 \mathrm{\,mA}$
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