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The resistance of galvanometer is 999 Om...

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

Text Solution

Verified by Experts

`G = 999 Omega, S = 1 Omega = i 10^(-2) A , i_g = ?`
`i_g = i ((S)/(G+S)) = 10^(-2) xx ((1)/(999+1)) = 10^(-5) A.`
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