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Question
In an ammeter 0.5% of main current passes through galvanometer; If resistance of galvanometer is G, the resistance of ammeter will be.
Options
G/200
G/199
199 G
200 G
MCQ
Solution
G/199
Explanation:
Here Ig = 0.5% of I = 0.005 I
Is = I – Ig = 1 – 0.005 I = 0.995 I
S = `(IgG)/(Is) = (0.0051 xx G)/0.9951 = G/199`
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