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?

Answer

790

Explanation

Current through the potentiometer wire, `I = 2/(R + 10)`
Resistance of 40 cm length of wire.
`=40 cm xx (10 Omega)/(100 cm) = 4 Omega`
Since parallel difference across 40 cm length of wire.
i.e. `I xx 4 = 10 mV`
`R_(ex) = (2/(R + 10)) xx 4 = 10 xx 10^(-3)` or `R = 790 Omega`