TEST OF PRACTICAL KNOWLEDGE QUESTION You are provided with a potentiometer x y; a jockey, J; a standard resistor, R, and other necessary apparatus. Connect ...

Answer Details

(a) Potentiometer-and-resistor experiment

With the jockey off the wire the current is \(I_0\). When J touches XY at C so that \(XC = l\), the current I is read for each length, and \(I^{-1}\) evaluated. A specimen table (values depend on your apparatus) is:

Graph: plot l (vertical) against \(I^{-1}\) (horizontal). Reading where the best-fit line meets the vertical axis (\(I^{-1}=0\)) gives the intercept value of l. Then evaluate \(I_0/I\) as required.

Two precautions:

• Make firm, brief contact of the jockey to avoid heating or scraping the wire.
• Open the key between readings to prevent the wire and cells from heating.

(b)(i) Potential difference

The potential difference between two points in a circuit is the work done (energy converted from electrical form) in moving one coulomb of charge from one point to the other: \[ V = \frac{W}{Q} \] Its S.I. unit is the volt (J C^-1).

(b)(ii) Length of the resistance wire

Taking the diameter as \(d = 0.2\ \text{mm} = 2\times10^{-4}\ \text{m}\), \(R = 7\ \Omega\), \(\rho = 8.8\times10^{-7}\ \Omega\text{m}\), \(\pi = \tfrac{22}{7}\).

Cross-sectional area: \[ A = \frac{\pi d^2}{4} = \frac{\tfrac{22}{7}\times(2\times10^{-4})^2}{4} = \frac{\tfrac{22}{7}\times4\times10^{-8}}{4} = 3.143\times10^{-8}\ \text{m}^2 \]

From \(R = \dfrac{\rho L}{A}\): \[ L = \frac{RA}{\rho} = \frac{7 \times 3.143\times10^{-8}}{8.8\times10^{-7}} = \frac{2.2\times10^{-7}}{8.8\times10^{-7}} = 0.25\ \text{m} \]

The length of the wire is 0.25 m (25 cm).