Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(a) What is the value of ε?
(b) What purpose does the high resistance of 600 kΩ have?
(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell?
(e) Would the method work in the above situation, if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?
(f) Would the circuit work well for determining extremely small emf, say of the order of a few mV (such as the typical emf of a thermocouple)? If not, how will you modify the circuit ?

(a) Given,
Emf of the standard cell, E₁ =1.02 V
Distance where balance point is obtained, l₁= 67.3 cm
Resistance, R= 600 kΩ 
When the standard cell is replaced by another cell, balance point is obtained at, l₂= 82.3 cm
Emf of the new cell, E₂ = ?

Using the formula for comparison of emf,
                   $\boxed{\frac{{\mathrm{E}}_2}{{\mathrm{E}}_1}= \frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}}$  
$\therefore$   ${\mathrm{E}}_2 = \frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}\times {\mathrm{E}}_1 = \frac{82.3}{67.3}\times 1.02 = 1.247 \mathrm{V}.$

(b) The purpose of using high resistance of 600 kΩ is to reduce the amount of current passing through the galvanometer when the movable contact is far from the balance point. 

(c) No, the balance point is not affected by the presence of this resistance. 

(d) No, the balance point is not affected by the internal resistance of the driver cell. 

(e) No, it is necessary that the emf of the driver cell is more than the emf of the cells and hence, the balance point will not be obtained if the emf of the driver cell is 1 V instead of 2 V.

(f) For measurement of small emf, this circuit will not work well. 
Modification of circuit: 
In order to measure small emf, we can connect a high resistance in series with the cell of 2V. This will decrease the amount of current flowing in the potentiometre wire. Thereby, the potential difference will decrease and small emf can be measured. 
                              OR
The number of potentiometer wires is increased to 11 or 15 to get a potential gradient of 0.1 Vm^–1. The purpose discussed above will be served by implementing a single 1 metre long wire, with high series resistance.