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The resistance of galvanomet is 999 Omega. A shunt of 1 Omega 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 = 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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