A resistance of `2 Omega` is connected in parallel to a galvanometer of resistance `48 Omega`. The fraction of the total current passing through the resistance of `2 Omega` is
A
`92%`
B
`94%`
C
`96%`
D
`98%`
Text Solution
Verified by Experts
The correct Answer is:
C
`S=((G)/(n-1))=((30)/(2-1))=30Omega` `I_g=((S)/(S+G))I=((30)/(30+30))I=0.51` Now , `S=((I_g)/(I-I_g))G=((0.5I)/(21-0.5I))30=10Omega`. Now , the effective resistance of galvanometer and its shunt is `10Omega`. Let R be the additional resistance connected across it. Thus , `(1)/(10)=(1)/(30+(1)/(R) therefore R=15 Omega`.
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