A potentiometer wire of length `100 cm` having a resistance of `10 Omega` is connected in series with a resistance `R` and a cell of emf `2V` of negligible internal resistance. A source of emf

of `10 mV` is balanced against a length of `40 cm` of the potentiometer wire. What is the value of resistance `R` ?

Answer

` 790 Omega `

Explanation

Here length of potentiometer wire `L = 100 cm ` resistance of potentiometer wire `r = 10 Omega `
`EMF ` of cell `epsilon = 2 V`
Current through the potentiometer wire
`1 = (epsilon)/(R + r) = (2) (R + 10)`
The resistance of `40 cm ` length of potentiometer wire
`r' = (10)/(100) xx 40 = 4 Omega `
since emf of the `10 mV ( = 10 xx 10^(-3)V)` is balanced againest a length `40 cm` of potentiometer wire so
`1 r' = 10 xx 10^(-3) or (2)/(R + 10) xx 4 = 10 ^(-2)`
On solving `R = 790 Omega `