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The resistance of a galvanometer is $50 Ω$ and the current required to give full scale deflection is $100 μ A .$ In order to convert it into an ammeter, reading upto 10 A, it is necessary to put a resistance of

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$5 \times 10^{- 3} Ω in parallel$

$5 \times 10^{- 4} Ω in parallel$

$5 \times 10^{- 5} Ω in s e r i e s$

$5 \times 10^{- 3} Ω in s e r i e s$

answer is B.

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Detailed Solution

Resistance in parallel, $S = \frac{{Gi}_{g}}{i - i_{g}} = \frac{50 \times 100 \times 10^{- 6}}{(10 - 100 \times 10^{- 6})}$

$\Rightarrow S = 5 \times 10^{- 4} Ω$

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