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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. |
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Answer» SOLUTION :`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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