question_answer

A galvanometer having a resistance of \[8\Omega \] is shunted by a wire of resistance\[2\Omega \]. If the total current is \[1\,\,A\], the part of it passing through the shunt will be

A) \[0.25\,\,A\]                      

B) \[0.8\,\,A\]

C)  \[0.2\,\,A\]                       

D)  \[0.5\,\,A\]

Correct Answer: B

Solution :

Key Idea: Potential difference across galvanometer should be equal to potential difference across shunt. The shunt and galvanometer are connected as shown in figure. Let total current through the parallel combination is \[i\], the current through the galvanometer is \[{{i}_{g}}\] and the current through the shunt\[i-{{i}_{g}}\]. The potential difference \[{{V}_{ab}}(={{V}_{a}}-{{V}_{b}})\] is the same for both paths, so                           \[{{i}_{g}}G=(i-{{i}_{g}})S\] or            \[{{i}_{g}}(G+S)=i\,\,S\] or            \[\frac{{{i}_{g}}}{i}=\frac{S}{S+G}\] The fraction of current passing through shunt                      \[=\frac{i-{{i}_{g}}}{i}=1-\frac{{{i}_{g}}}{i}\]                     \[=1-\frac{S}{S+G}=\frac{G}{S+G}\]                    \[=\frac{8}{2+8}=\frac{8}{10}\]                    \[=0.8\,\,A\]