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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

 $$  I_{G}=10 \mathrm{mA}, R_G=10 \Omega $$

$$ R\left(I-I_{G}\right)=I_{G} \times R_G $$

$$R(1-0.001)=0.001\times 10$$

$$R=\dfrac{0.001}{0.999}$$

$$R=0.1\Omega$$

Physics
NCERT
Standard XII

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