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 is\[i-{{i}_{g}}.\] 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 \[=\frac{i-{{i}_{g}}}{i}=1-\frac{{{i}_{g}}}{i}\] \[=1-\frac{S}{S+G}=\frac{G}{S+G}\] \[=\frac{8}{2+8}=\frac{8}{10}\] \[=0.8\,A\]