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 S in parallel with galvanometer G. Since G and S 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.