You are provided with a potentiometer x y; a jockey, J; a standard resistor, R, and other necessary apparatus. Connect a circuit as shown in the diagram abo...

(a) Potentiometer (rheostat) current experiment

The circuit is connected as shown, with the standard resistor \(R\) and the ammeter in series with the length \(XC=l\) of the potentiometer wire. With the jockey off the wire the ammeter reads \(I_0\); when the jockey contacts XY at C the current \(I\) is read for each length \(l\), and \(l^{-1}\) is evaluated.

Table of readings   (\(I_0 = 0.8\ \text{A}\))

S/N	\(l\) (cm)	\(I\) (A)	\(l^{-1}\) (cm\(^{-1}\))
1	25.0	1.20	0.040
2	40.0	1.00	0.025
3	55.0	0.95	0.018
4	70.0	0.85	0.014
5	85.0	0.80	0.012

Graph of \(I\) (vertical) against \(l^{-1}\) (horizontal):

Deduction from the graph. The line of best fit cuts the vertical axis (where \(l^{-1}=0\)) at

\[ I = 0.66\ \text{A}. \]

Hence

\[ \frac{I_0}{I} = \frac{0.8}{0.66} = 1.21. \]

Two precautions:

1. The key was left open whenever readings were not being taken, to prevent heating of the wire.
2. The jockey was pressed firmly but was not rolled along or allowed to scratch the potentiometer wire.

(b)(i) Potential difference

The potential difference between two points in an electric circuit is the work done (energy converted from electrical form to other forms) in moving one coulomb of charge from one point to the other:

\[ V = \frac{W}{Q}. \]

Its S.I. unit is the volt (\(\text{J C}^{-1}\)).

(b)(ii) Length of the resistance wire

Given: diameter \(d = 0.2\ \text{m}\), resistance \(R = 7\ \Omega\), resistivity \(\rho = 8.8\times10^{-7}\ \Omega\text{m}\), \(\pi = \tfrac{22}{7}\).

Cross-sectional area of the wire:

\[ A = \frac{\pi d^{2}}{4} = \frac{\tfrac{22}{7}\times(0.2)^{2}}{4} = \frac{\tfrac{22}{7}\times 0.04}{4} = 0.03143\ \text{m}^{2}. \]

From \(R = \dfrac{\rho L}{A}\):

\[ L = \frac{RA}{\rho} = \frac{7 \times 0.03143}{8.8\times10^{-7}} = \frac{0.22}{8.8\times10^{-7}} = 0.025\times10^{7}\ \text{m}. \] \[ \boxed{L = 2.5\times10^{5}\ \text{m}.} \]