You are provided with a potentiometer XY, a voltmeter, V, a standard resistor R, an accumulator, E a plug key, K, a jockey, and connecting wires. Connect a ...

(a) Potentiometer experiment

Circuit: The accumulator \(E\), key \(K\) and standard resistor \(R\) are joined in series with the potentiometer wire \(XY\). The voltmeter \(V\) is connected between \(X\) and the jockey, so it reads the potential difference across the length \(l = XN\).

With the key closed, the jockey is pressed at \(N\) so that \(l = XN = 15.0\,\text{cm}\) and the voltmeter reading \(V\) is recorded. The procedure is repeated for \(l = 25.0, 35.0, 45.0, 55.0\) and \(65.0\,\text{cm}\). For each length \(l^{-1}\) and \(V^{-1}\) are evaluated.

Table of readings

S/N	\(l\,/\,\text{cm}\)	\(V\,/\,\text{volt}\)	\(l^{-1}\,/\,\text{cm}^{-1}\)	\(V^{-1}\,/\,\text{volt}^{-1}\)
1	15.0	0.60	0.0667	1.667
2	25.0	0.80	0.0400	1.250
3	35.0	1.10	0.0286	0.909
4	45.0	1.30	0.0222	0.769
5	55.0	1.60	0.0182	0.625
6	65.0	1.80	0.0154	0.556

Worked evaluations (row 1): \(l^{-1}=\dfrac{1}{15.0}=0.0667\,\text{cm}^{-1}\), \(V^{-1}=\dfrac{1}{0.60}=1.667\,\text{volt}^{-1}\); (row 6): \(l^{-1}=\dfrac{1}{65.0}=0.0154\,\text{cm}^{-1}\), \(V^{-1}=\dfrac{1}{1.80}=0.556\,\text{volt}^{-1}\).

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

Slope of the graph

Using two points on the line of best fit, \((l^{-1}=0.0154,\ V^{-1}=0.603)\) and \((l^{-1}=0.0667,\ V^{-1}=1.724)\):

\[ s=\frac{\Delta V^{-1}}{\Delta l^{-1}}=\frac{1.724-0.603}{0.0667-0.0154}=\frac{1.121}{0.0513}=21.85\ \text{cm}\,\text{volt}^{-1}. \]

Evaluate \(k=\tfrac{1}{s}\)

\[ k=\frac{1}{s}=\frac{1}{21.85}=0.046\ \text{volt}\,\text{cm}^{-1}. \]

Two precautions

1. I ensured that all connecting wires and terminals were clean and firmly tightened to avoid loose contacts.
2. I read the voltmeter with my eye directly in line with the pointer to avoid parallax error.

(b)(i) Four factors on which the resistance of a wire depends

1. The length of the wire \((R\propto l)\).
2. The cross-sectional area of the wire \((R\propto \tfrac{1}{A})\).
3. The nature (material / resistivity) of the wire.
4. The temperature of the wire.

(b)(ii) Heat produced in the wire

Resistance of the wire:

\[ R = (\text{resistance per unit length})\times(\text{length}) = 0.02\times100 = 2\,\Omega. \]

Given \(V=1.5\,\text{V}\) and \(t=1\,\text{minute}=60\,\text{s}\), the heat produced is

\[ H=\frac{V^{2}t}{R}=\frac{(1.5)^{2}\times60}{2}=\frac{2.25\times60}{2}=\frac{135}{2}=67.5\,\text{J}. \]

Check (using current): \(I=\dfrac{V}{R}=\dfrac{1.5}{2}=0.75\,\text{A}\), so \(H=I^{2}Rt=(0.75)^{2}\times2\times60=0.5625\times120=67.5\,\text{J}.\)