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

Options

A. `1 A`
B. `1.5 A`
C. `2 A`
D. `5A`

Answer

A

Explanation

The resistance of the bridge wire `=0.1xx100 =10 Omega`
The null point divides the wire of length 100 cm, in the ratio of 2 : 3.
`therefore l_(x)=40 cm and l_(R )=60 cm `

` therefore (X)/(6)=(40)/(60)=(2)/(3) therefore X= 4 Omega`
In the balanced condition, `X= 4 Omega and 6 Omega` are in series.
`therefore` Their effective resistance `= 10 Omega`
Similarly, the resistance of the wire `=10 Omega`
The resistances in both the branches are in parallel.
`therefore` The resistance of the Wheatstone's network
`(1)/(R )=(1)/(10)+(1)/(10) therefore R= 5 Omega`
`therefore " Current " = (E )/(R )=(5)/(5)= 1A`