Question

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. $\frac{1}{499}G$
• B. $\frac{499}{500}G$
  
• C. $\frac{1}{500}G$
• D. $\frac{500}{499}G$

Answer 1

The correct option is C $\frac{1}{500}G$


From circuit diagram, potential difference is same across both resistances, then
$\left(\frac{2I}{1000}\right)G=\left(\frac{998I}{1000}\right)S\Rightarrow S=\frac{G}{499}$
$\therefore$Total resistance of ammeter,
$R=\frac{SG}{S+G}=\frac{\left(\frac{G}{499}\right)G}{\left(\frac{G}{499}\right)+G}=\frac{G}{500}$