In a galvanometer there is a deflection of 10 divisions per mA. The initial resistance of galvanometer is $60 Ω$ . If a shunt of $2.5 Ω$ is connected to the galvanometer and there are 50 divisions in all, on the scale of galvanometer , what maximum current this galvanometer can read?

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111 mA

170 mA

125 mA

150 mA

answer is B.

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Detailed Solution

As the galvanometer has 50 divisons current required to produce full scale deflection

$I_{g} = \frac{1}{10} \times 50 m A = 5 m A = 0.005 A$

$R_{g} = 60 Ω; R_{s} = 2.5 Ω$
Let i be the max current

$I_{g} = i (\frac{R}{R_{g} + R_{s}})$
$5 \times 10^{- 3} = i . (\frac{2.5}{60 + 2.5})$

$i = \frac{62.5 \times 5 \times 10^{- 3}}{2.5} = 125 m A$

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