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The resistance of galvanometer is 999Ome...

The resistance of galvanometer is `999Omega`. A shunt of `1 Omega` is connected to it.If the main current current is `10^(-2)A`, what is the current flowing through the galvanometer .

Text Solution

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`G=999Omega,S=1Omegai=10^(-2)A,i_(g)=?`
`i_(g)=i((S)/(G+S))=10^(-2)xx((1)/(999+1))=10^(-5)A`
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NARAYNA-MOVING CHARGES AND MAGNETISM-EXERCISE - 4
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