To the potentiometer wire of L and `10Omega` resistance, a battery of emf 2.5 volts and a resistance R are connected in series. If a potential difference of 1 volt is balanced across L/2 length, the value of R is `Omega` will be

Options

A. 1
B. 1.5
C. 2
D. 2.5

Answer

4

Explanation

Total resistance `=(10+R)Omega`
`therefore` Total current through potentiometer wire
`I=(2.5)/((10+R))A`
Potential difference across AJ=1V

i.e. `I((10)/(2))=1VrArr(2.5)/((10+R))xx5=1`
10+R=12.5
`R=12.5-10=2.5Omega`.