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 r

Question

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 r

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%`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - 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`.