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A galvanometer of resistance 40 Ω and current passing through it is 100 µA per division. The full scale has 50 divisions. If it is converted into an ammeter of range 2A by using a shunt, then the resistance of ammeter is

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$\frac{40}{399} Ω$

0.1 Ω

$\frac{4}{399} Ω$

0.4Ω

answer is C.

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Detailed Solution

$S = \frac{G}{n - 1}; S = \frac{40}{\frac{2}{5 \times 10^{- 3}} - 1} = \frac{40}{399}$

$Resistance of the ammeter is \frac{1}{R_{a}} = \frac{1}{G} + \frac{1}{S}$

$\frac{1}{R_{a}} = \frac{399}{40} + \frac{1}{40} = \frac{400}{40} = 10$

$R_{a} = 0.1 Ω$

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