The figure shows a potentiometer with a cell of $2.0 \; V$ and internal resistance $0.40 \; \Omega$ 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 are drawn from the standard cell,a very high resistance of $600 \; k \Omega$ 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 $emf$ $\varepsilon$ and the balance point found similarly,turns out to be at $82.3 \; cm$ length of the wire.
$(a)$ What is the value of $\varepsilon ?$
$(b)$ What purpose does the high resistance of $600 \; k \Omega$ have?
$(c)$ Is the balance point affected by this high resistance?
$(d)$ Would the method work in the above situation if the driver cell of the potentiometer had an $emf$ of $1.0 \; V$ instead of $2.0 \; V ?$
$(e)$ Would the circuit work well for determining an 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-D) The constant $emf$ of the given standard cell is $E_1 = 1.02 \; V$.
The balance point on the wire is $l_1 = 67.3 \; cm$.
When a cell of unknown $emf$ $\varepsilon$ replaces the standard cell,the new balance point on the wire is $l = 82.3 \; cm$.
The relation connecting $emf$ and balance point is $\frac{E_1}{l_1} = \frac{\varepsilon}{l}$.
Therefore,$\varepsilon = \frac{l}{l_1} \times E_1 = \frac{82.3}{67.3} \times 1.02 \approx 1.247 \; V$.
The value of the unknown $emf$ is $1.247 \; V$.
$(b)$ The purpose of using the high resistance of $600 \; k \Omega$ is to protect the galvanometer from high currents when the movable contact is far from the balance point.
$(c)$ No,the balance point is not affected by the presence of this high resistance because at the balance point,the current through the galvanometer is zero,so there is no potential drop across the high resistance.
$(d)$ The method would not work if the driver cell had an $emf$ of $1.0 \; V$ because the $emf$ of the cell to be measured ($1.02 \; V$ or $1.247 \; V$) would be greater than the potential drop across the potentiometer wire,making it impossible to find a balance point.
$(e)$ The circuit would not work well for determining an extremely small $emf$ because the balance point would be very close to end $A$,leading to a large percentage error. To modify the circuit,a series resistor should be connected with the potentiometer wire $AB$ to reduce the potential drop across $AB$ so that it is only slightly greater than the $emf$ being measured.