question_answer

In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, the resistance of ammeter will be [AIPMT 2014]

A)  \[\frac{1}{499}G\]         

B)  \[\frac{499}{500}G\]

C)  \[\frac{1}{500}G\]

D)  \[\frac{500}{499}G\]

Correct Answer: A

Solution :

For ammeter \[0.0002l\times G=0.998l\times {{r}_{g}}\] \[{{r}_{g}}=\frac{0.002}{0.998}G\] Þ \[{{r}_{g}}=0.002004G=\frac{1}{499}\times G\] Resistance of ammeter is approximately equal to resistance of shunts.