A potentiometer wire of length 1 m has a resistance of 10 `Omega` . It is connected to a 6 V battery in series with a resistance of 5 `Omega` . Determine the emf of the primary cell which gives a balance point at 40 cm.

Explanation

Here emf of battery `epsi_0` =6V, length of potentiometer wire L = 1 m, length for balancing the cell of emf `epsi` l = 40 cm = 0.4 m, resistance of potentiometer `R = 10 Omega` and series resistance `R. =5 Omegs`
Potential gradient `k = (epsi_0 R)/((R + R.)L)`
and emf of cell `epsi = kl = (epsi_0 Rl)/((R + R.)L) = (6 xx 10 xx 0.4)/((10 + 5) xx 1) = 1.6 V`