You are provided with a constantan wire, a 2\(\Omega\) standard resistor, an accumulator E, an ammeter A, a key K, and other necessary apparatus. Measure an...

Answer Details

Test of Practical Knowledge – Electricity

Circuit diagram

Measurements recorded

• E.m.f. of the accumulator, \(E = 2.0\ \text{V}\)
• Ammeter reading with the clip not in contact with the wire, \(I_{o} = 0.65\ \text{A}\)

Table of readings

\(l\) (cm)	\(l^{-1}\) (cm\(^{-1}\))	\(I\) (A)
90	0.0111	0.70
80	0.0125	0.75
70	0.0143	0.80
60	0.0167	0.85
50	0.0200	0.90

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

Slope of the graph

Taking two widely spaced points on the line of best fit, \((l^{-1}_{1},\,I_{1}) = (0.0100,\ 0.69)\) and \((l^{-1}_{2},\,I_{2}) = (0.0200,\ 0.91)\):

\[ s = \frac{\Delta I}{\Delta (l^{-1})} = \frac{0.91 - 0.69}{0.0200 - 0.0100} = \frac{0.22}{0.0100} = 22.0\ \text{A cm} \]

Intercept on the vertical axis

Extending the line of best fit back to \(l^{-1} = 0\), it cuts the \(I\)-axis at

\[ c = 0.47\ \text{A} \]

Evaluate \(k = \dfrac{c}{s}\)

\[ k = \frac{c}{s} = \frac{0.47}{22.0} = 0.021 \]

Current \(I\) when \(l = 55\ \text{cm}\)

\[ l^{-1} = \frac{1}{55} = 0.0182\ \text{cm}^{-1} \]

Reading this value on the horizontal axis and going up to the line of best fit:

\[ I = c + s\,(l^{-1}) = 0.47 + 22.0 \times 0.0182 = 0.87\ \text{A} \]

Two precautions

1. The key was opened (removed) immediately after each reading was taken, to prevent heating of the constantan wire and to conserve the accumulator.
2. The terminals and crocodile-clip contacts were kept clean and the connections firmly tightened to avoid contact resistance, and the ammeter was read with the eye directly above the pointer to avoid parallax.

(b)(i) Potential difference between two points

The potential difference between two points in an electric circuit is the work done in moving one coulomb of positive charge from one point to the other. It is measured in volts, where \(1\ \text{V} = 1\ \text{J C}^{-1}\).

(b)(ii) Two factors on which the resistance of a resistance wire depends

1. The length of the wire – resistance increases with length \((R \propto l)\).
2. The cross-sectional area (thickness) of the wire – resistance decreases as the area increases \(\left(R \propto \dfrac{1}{A}\right)\).

(It also depends on the resistivity/nature of the material and on the temperature of the wire.)