The length of a potentiometer wire is $L$. $A$ cell of e.m.f. $E$ is balanced at a length $\frac{L}{3}$ from the positive end of the wire. If the length of the wire is increased by $\frac{L}{2}$,at what distance will the same cell give a balance point?

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?

In a potentiometer, when the cell in the secondary circuit is shunted with a $4 \ \Omega$ resistance, the balance is obtained at a length of $120 \ cm$ of the wire. Now, when the same cell is shunted with a $12 \ \Omega$ resistance, the balance point shifts to a length of $180 \ cm$. The internal resistance of the cell is . . . . . . $\Omega$.

To determine the internal resistance of a cell by using a potentiometer,the null point is at $1 \ m$,when shunted by $3 \ \Omega$ resistance and at a length $1.5 \ m$,when cell is shunted by $6 \ \Omega$ resistance. The internal resistance of the cell is (in $\Omega$)

$A$ potentiometer circuit shown in the figure is set up to measure the $e.m.f.$ of a cell $E$. As the point $P$ moves from $X$ to $Y$, the galvanometer $G$ shows deflection always in one direction, but the deflection decreases continuously until $Y$ is reached. In order to obtain a balance point between $X$ and $Y$, it is necessary to

The figure shows a potentiometer wire $AB$ having a resistance of $5 \text{ } \Omega$ and a length of $10 \text{ m}$. The e.m.f. of the battery in the primary circuit is $5 \text{ V}$ and the external resistance is $45 \text{ } \Omega$. If the e.m.f. of the cell in the secondary circuit is $0.4 \text{ V}$, find the balancing length $AP$. (Internal resistance is negligible) (in $\text{ m}$)