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A potentiometer wire of length 100 cm ha...

A potentiometer wire of length 100 cm has a resistance of `100Omega` it is connected in series with a resistance and a battery of emf 2 V and of negligible internal resistance. A source of emf 10 mV is balanced against a length of 40 cm of the potentiometer wire. what is the value of the external resistance?

A

`790Omega`

B

`890Omega`

C

`990Omega`

D

`1090Omega`

Text Solution

Verified by Experts

The correct Answer is:
A
The current in the poteniometer iwre AC is `I=(2)/(10+R)`
The potential difference across the potentiometer wire is `V="current"xx"resistance"=(2)/(10+R)xx10`
The length of the wire is l=100cm
So, the potential gradient along the wire is `k=(V)/(l)=((2)/(10+R))xx(10)/(100)`
The source of emf 10mV is balanced against a length of 40cm of the potentiomer wire
`i.e. 10xx10^(-3)=kxx40`
or `10xx10^(-3)=(2)/((10+R))xx(40)/(10)("Using(i)")`
`or R=790Omega`
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