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A voltmeter of 250 mV range having a resistance of $9 Ω$ is converted into an ammeter of 250mA range. The value of necessary shunt is

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$2 Ω$

$10 Ω$

$0.1 Ω$

$1 Ω$

answer is C.

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Detailed Solution

$i = \frac{V}{R} = \frac{250}{10} = 25 m A$

We know $i_{g} = [\frac{S}{G + S}] i$

$\Rightarrow 25 = [\frac{S}{9 + S}] 250$

$\Rightarrow 9 S = 9 \Rightarrow S = 1 Ω$

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