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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)`? |
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Answer» Correct Answer - 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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