Question

# The maximum current in a galvanometer can be $$10 \mathrm{mA}$$ Its resistance is $$10 \Omega .$$ To convert it into an ammeter of 1 A, what resistance should be connected in parallel with galvanometer $$\left(\text { in } 10^{-1} \Omega\right) ?$$

Solution

## $$I_{G}=10 \mathrm{mA}, R_G=10 \Omega$$$$R\left(I-I_{G}\right)=I_{G} \times R_G$$$$R(1-0.001)=0.001\times 10$$$$R=\dfrac{0.001}{0.999}$$$$R=0.1\Omega$$PhysicsNCERTStandard XII

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