In balanced meter bridge, the resistance of bridge wire is `0.1Omega cm` . Unknown resistance X is connected in left gap and `6Omega` in right gap, null point divides the wire in the ratio 2:3 . Find the current drawn the battery of 5V having negligible resistance
A. 1A
B. 1.5A
C. 2A
D. 5A

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1 Answers

Correct Answer - A
`(X)/(R)=(l_(x))/(l_(r))`
`(X)/(6)=(2)/(3)" ":.x=4Omega,R=6Omega`
Now, `R_(p)=(R_(1).R_(2))/(R_(1)+R_(2))=(10xx10)/(10+0)=5Omega`
`I=(E)/(R_(p)+r)=(5)/(5+0)=1A`.

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