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` ?

Explanation

Resistance of 40 cm length of potentimoeter wire `= 10/100 xx 40 = 4 Omega`
Current through potentiometer wire,
`I= ("pot.diff.")/("resistance") = (10 mV)/(4Omega) = (10 xx 10^(-3)V)/(4Omega)`
`=2.5 xx 10^(-3)A`
The same current flows through the potentiometer wire and the external circuit having resistance R.
Total resistance of potnetiometer circuit
`(R + 10)Omega`
`:. 2. 5 xx 10^(-3)A = 2V/((R+10)Omega)`
or `R+ 10 = 2/(2.5 xx 10^(-3))=800`
or `R = 800 - 10 = 790 Omega`