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A galvanometer of resistance $20 Ω$ gives full scale deflection with a current of 0.004 A. To convert it into an ammeter of range 1 A, the required shunt resistance should be

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$0.38 Ω$

$0.21 Ω$

$0.08 Ω$

$0.05 Ω$

answer is C.

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Detailed Solution

Maximum current which can pass through galvanometer $I_{g} = 0.004 A$

Let $R$ be the resistance of shunt.

We know potential drop across $A B$

= Potential drop across CD

$\begin{array}{l} R (I - I_{g}) = I_{g} (20) \\ \Rightarrow R (1 - 0.004) = 0.004 \times 20 \Rightarrow R = 0.08 Ω \end{array}$

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