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Resistance of a galvanometer coil is $100 Ω$ and permissible current through its coil is 10 mA. It can be converted into an ammeter of range 1A by connecting a resistance of

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1.01 $Ω$

1.9 $Ω$

2.1 $Ω$

0.9 $Ω$

answer is D.

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Detailed Solution

The resistance R is to be connected in parallel combination with the galvanometer coil.

$∴ R (1 - 10 \times 10^{- 3}) = 100 \times 10 \times 10^{- 3}$

$\Rightarrow R = \frac{1000 \times 10^{- 3}}{(1 - 0.01)} Ω = 1.01 Ω$

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