question_answer

The resistance of a galvanometer coil is R, then the shunt resistance required to convert it into a ammeter of range 4 times, will be

A)  4R            

B)  \[\frac{R}{3}\]

C)  \[\frac{R}{4}\]

D)  \[\frac{R}{5}\]

Correct Answer: B

Solution :

Key Idea: Shunt is connected in parallel with galvanometer. Ammeter is made by connecting a low resistance shunt 5 in parallel with galvanometer G. Since G and 5 are in parallel, the potential difference across them is same                 \[{{i}_{g}}\times G=(i={{i}_{g}})\times S\] Given,   \[G=R,\,\,i=4\,{{i}_{g}}\] \[\therefore \] \[S=\frac{{{i}_{g}}}{4\,{{i}_{g}}-{{i}_{g}}}\times R=\frac{{{i}_{g}}}{3\,{{i}_{g}}}\times R=\frac{R}{3}\] Note: Resistance of shunt is always low.