question_answer

In an ammeter, 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, then the resistance of ammeter will be:

A) \[\frac{G}{499}\]                      

B) \[\frac{499G}{500}\]

C) \[\frac{G}{500}\]                                  

D) \[\frac{500G}{499}\]

Correct Answer: C

Solution :

[c] Parallel combination \[{{V}_{g}}={{V}_{S}}\] \[{{I}_{g}}G=IS-{{I}_{g}}S\] \[IS={{I}_{g}}[G+S]\] \[IS=\frac{2}{1000}I\,[G+S]\] \[500S=G+S\] \[S=\frac{G}{499}\] \[{{R}_{ammeter}}=\frac{SG}{S+G}=\frac{\frac{G}{400}.G}{\frac{G}{400}+G}=\frac{\frac{{{G}^{2}}}{400}}{\frac{500G}{400}}\] \[{{R}_{ammeter}}=\frac{G}{500}\]