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A galvanometer of resistance $4 Ω$ and current passing through it is $100 μ A$ 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

\frac{4}{399} Ω

0.01 Ω

0.4 Ω

answer is A.

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Detailed Solution

G = 40 Ω

$i_{g} = 100 μ A perdiuision$

$i = 2 A$

(Values are missing in question)

For 50 divisions

$i_{g} = 100 \times 10^{- 6} \times 50 = 5 \times 10^{- 3} A$

$i = 2 A$

$n = \frac{2}{5 \times 10^{- 3}} = 400$

$S = \frac{G}{n - 1} = \frac{40}{400 - 1} = \frac{40}{399} Ω$

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