1.

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.

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`


Discussion

No Comment Found

Related InterviewSolutions