A) 19 G
B) 20 G
C) G/20
D) G/19
Correct Answer: D
Solution :
Key Idea: Potential difference across galvanometer and shunt is same. Let \[{{i}_{g}}\]current passes through galvanometer, and \[i-{{i}_{g}}\]through shunt S, then \[{{i}_{g}}G=(i-{{i}_{g}})S\] \[S=\left( \frac{{{i}_{g}}}{i-{{i}_{g}}} \right)G\] Given, \[{{i}_{g}}=5A,\,i=100A,\] \[\therefore \] \[S=\left( \frac{5}{100-5} \right)G=\frac{G}{19}\]You need to login to perform this action.
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