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                \[{{i}_{g}}(G+S)=iS\] or                   \[\frac{{{i}_{g}}}{i}=\frac{S}{S+G}\] The fraction of current passing through shunt \[=\frac{i-{{i}_{g}}}{i}=1-\frac{{{i}_{g}}}{i}\] \[=1-\frac{S}{S+G}=\frac{G}{S+G}\] \[=\frac{8}{2+8}=\frac{8}{10}\] \[\text{= 0}\text{.8 A}\]=