The driver cell of a potentiometer has an emf of 2 V and negligible internal resistance. The potentiometer wire has a resistance of 3 Ω. Calculate the resistance needed in series with a wire if a p.d. 5.0 mV is required across the whole wire. The wire is 100 cm long and a balanced length of 60 cm is obtained for a thermocouple of emf E. What is the value of E?

The driver cell of a potentiometer has an emf of 2 V and negligible internal resistance. The potentiometer wire has a resistance of 3 Ω. Calculate the resistance needed in series with a wire if a p.d. 5.0 mV is required across the whole wire. The wire is 100 cm long and a balanced length of 60 cm is obtained for a thermocouple of emf E. What is the value of E?
Solution:

emf of driver cell, $V_0$ = 2 V
resistance of potentiometer wire, $R_{AB}$ = 3 Ω
resistance to be connected in series, $R$ = ?
p.d. across AB, $V$ = 5 mV = 5 × 10 ^-3
length of potentiometer wire, $L$ = 100 cm
balancing length for thermoemf , $l$ = 60 cm
thermoemf, $E$ = ?
Note that! For the first case just look at the upper section of the figure and for the next part you have to consider the thermocouple (indicted by V in the diagram) that generates thermoemf E.
For first part,
Since $R$ and $R_{AB}$ are in series, the total resistance in the circuit is, $$R_T=R+3$$ Now, $$\begin{array}{rl}{V}_0& =I(R+3)\\ I& =\frac{V_0}{R+3}\end{array}$$ Again, $$\begin{array}{rl}V& =I\times R_{AB}\\ 5\times {10}^{-3}& =\frac{V_0}{R+3}\times 3\\ 5\times {10}^{-3}R+15\times {10}^{-3}& =6\\ R& =1197\mathrm{\Omega }\end{array}$$
For second part,
$$\begin{array}{rl}\text{p.d. across l}& =kl\\ E& =\frac{V_0}{L}\times l\because \text{p.d. across wire = E (for balancing length)}\\ E& =\frac{5\times {10}^{-3}}{100}\times 60\\ \therefore E& =3\times {10}^{-3}V\end{array}$$ Hence, the value of E is 3mV .