Moving Charges and Magnetism - Result Question 75
77. In an ammeter $0.2 %$ of main current passes through the galvanometer. If resistance of galvanometer is $G$, the resistance of ammeter will be :
[2014]
(a) $\frac{1}{499} G$
(b) $\frac{499}{500} G$
(c) $\frac{1}{500} G$
(d) $\frac{500}{499} G$
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Answer:
Correct Answer: 77. (c)
Solution:
- (c) As $0.2 %$ of main current passes through the galvanometer hence $\frac{998}{1000} I$ current through the shunt.
$(\frac{2 I}{1000}) G=(\frac{998 I}{1000}) S \Rightarrow S=\frac{G}{499}$
Total resistance of Ammeter
$R=\frac{S G}{S+G}=\frac{(\frac{G}{499}) G}{(\frac{G}{499})+G}=\frac{G}{500}$
