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The maximum current in a galvanometer ca...

The maximum current in a galvanometer can be 10 mA. Its resistance is `10 Omega`. To convert it into an ammeter of 1 A, what resistance should be connected in parallel with galvanometer (in `10^(-1)Omega)`?

Text Solution

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The correct Answer is:
A
`I_(G) = 10mA, G = 10Omega`
`S(I-I_(G)) = I_(G)G` where S is hunt in parallel , solve to get `S = 0.1 Omega`.
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