Question 1 Report
You are provided with an ammeter, resistor, key, metre bridge, and other necessary apparatus.
i. Connect a circuit as shown in the diagram above.
ii. Close the key and use the jockey to make contact with AB at N such that AN = d = 25cm
iii. Read and record the ammeter reading.
iv. Evaluate 1\(^{-1}\).
v. Repeat the procedures for values of d = 35cmm, 50cm, 65cm and 80cm. In each case, record I and determine 1\(^{-1}\)
vi. Tabulate your results.
vii. Plot a graph with log on the vertical axis and d on the horizontal axis.
viii. Determine the slope, s of the graph.
ix. State two precautions taken to obtain accurate results.
(b)i. Use your graph to determine the value of d = l = 1.5A.
ii. State two factors that affect the resistance of a wire.
The apparatus is connected so that the cell, key, ammeter and resistor \(R\) are joined in series to end \(A\) of the metre bridge wire, while the jockey is used to make contact with the wire at \(N\). Only the length \(AN = d\) of the bridge wire carries the current, so as \(d\) increases the resistance in the circuit rises and the ammeter reading \(I\) falls.
(ii) - (vi) Readings and table. For each contact length \(d = AN\) the key is closed, the ammeter reading \(I\) is taken, its reciprocal \(I^{-1}\) is evaluated and \(\log I^{-1}\) obtained. The results are tabulated below.
| S/N | d (cm) | I (A) | I-1 (A-1) | log I-1 |
|---|---|---|---|---|
| 1 | 25 | 0.65 | 1.54 | 0.19 |
| 2 | 35 | 0.55 | 1.82 | 0.26 |
| 3 | 50 | 0.44 | 2.50 | 0.40 |
| 4 | 65 | 0.30 | 3.33 | 0.52 |
| 5 | 80 | 0.23 | 4.35 | 0.64 |
(vii) Graph. \(\log I^{-1}\) is plotted on the vertical axis against \(d\) on the horizontal axis and the best straight line is drawn through the points.
(viii) Slope. Taking two widely separated points on the line, \((25,\;0.20)\) and \((80,\;0.64)\):
\[ s = \frac{\Delta(\log I^{-1})}{\Delta d} = \frac{0.64 - 0.20}{80 - 25} = \frac{0.44}{55} = 8.0\times10^{-3}\ \text{A}^{-1}\,\text{cm}^{-1}. \](ix) Two precautions.
(b)(i) Value of d when I = 1.5 A.
\[ I^{-1} = \frac{1}{1.5} = 0.67\ \text{A}^{-1}, \qquad \log I^{-1} = \log(0.67) = -0.17. \]The best-fit line is \(\log I^{-1} = 8.0\times10^{-3}\,d - 0.01\). Producing this line downwards until it meets \(\log I^{-1} = -0.17\) gives
\[ d = \frac{-0.17 + 0.01}{8.0\times10^{-3}} \approx -20\ \text{cm}. \]Because this value falls below \(d = 0\) (off the lower end of the scale), a current as large as \(1.5\ \text{A}\) cannot be obtained with this circuit: even the shortest usable length of wire keeps the current below about \(0.65\ \text{A}\).
(b)(ii) Two factors that affect the resistance of a wire.
(The resistivity or nature of the material and the temperature of the wire also affect its resistance.)
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