A potentiometer wire, 4 m long and resistance `10 Omega`, is connected in series with a resistance of 1988 `Omega` and a cell of emf 2V and internal resistance `2Omega`. What is the potential gradient along the wire?

Explanation

Data: L=4m, `r=4 Omega, E=2V, r_( c)=2Omega,V//L = 0.1 V//m`
Let R be the series resistance for which the desired potential drop is obtained. The current in the circuit is
`I=E/(r+r_(c) + R) = 2/(4+2+R)`
The potential difference across the wire is
`V=Ir = (2/(6+R)) xx 4 = 8/(6+R)`
`therefore` The potential drop per unit length of the wire `=V/L = 8/(4(6+R)) = 2/(6+R)`
`therefore 0.1 = 2/(6+R) therefore 6+R=20, therefore R=14 Omega`