You are provided with a voltmeter V, a chemical cell/ battery E; two standard resistors R, and R a potentiometer a key K a jockey, and other necessary mater...
Question 1 Report
You are provided with a voltmeter V, a chemical cell/ battery E; two standard resistors R, and R a potentiometer a key K a jockey, and other necessary materials.
- Set up a circuit as shown in the diagram above.
- Close the key K.
- Make contact with the potentiometer wire AB using the jockey at a point C such that AC = x = 20cm
- Read and record the voltmeter reading
- Evaluate x\(^{-1}\) and v\(^{-1}\)
- Repeat the procedure for other values of x= 30, 40, 50, 60 and 80cm.
- Tabulate your readings.
- Plot a graph with V\(^{-1}\) on the vertical axis and x\(^{-1}\) on the horizontal axis, starting both axes from the origin (0, 0).
- Determine the:
(a) slope, s, of the graph;
(b) intercept, c, on the vertical axis. - State two precautions taken to ensure accurate results.
(b)i. State the two devices in which ohm's law does not apply.
ii. A current of 1 A is supplied to two resistors of resistance 2\(\Omega\) and 3\(\Omega\) connected in parallel. Calculate the current in each resistor.
Answer Details
Circuit
The circuit is connected as shown, with the cell E and key K driving the potentiometer wire AB, and the voltmeter V connected across the length AC = x.
Table of readings
With the key K closed, the jockey is placed on the wire at C so that AC = x, and the voltmeter reading V is taken for each length. The reciprocals \(x^{-1}\) and \(V^{-1}\) are then evaluated.
| \(x\,/\,\text{cm}\) | \(V\,/\,\text{V}\) | \(x^{-1}\,/\,\text{cm}^{-1}\) | \(V^{-1}\,/\,\text{V}^{-1}\) |
|---|---|---|---|
| 20 | 0.15 | 0.050 | 6.667 |
| 30 | 0.20 | 0.033 | 5.000 |
| 40 | 0.30 | 0.025 | 3.333 |
| 50 | 0.40 | 0.020 | 2.500 |
| 60 | 0.50 | 0.017 | 2.000 |
| 80 | 0.60 | 0.013 | 1.667 |
Graph of \(V^{-1}\) against \(x^{-1}\)
The graph of \(V^{-1}\) (vertical axis) against \(x^{-1}\) (horizontal axis) is a straight line passing through the origin.
(a) Slope of the graph
Taking two widely separated points on the line of best fit, \((0.050,\;6.667)\) and \((0.013,\;1.667)\):
\[ s = \frac{V_2^{-1}-V_1^{-1}}{x_2^{-1}-x_1^{-1}} = \frac{6.667-1.667}{0.050-0.013} = \frac{5.000}{0.037} = 135.1\ \text{cm}\,\text{V}^{-1} \](b) Intercept on the vertical axis
The line of best fit passes through the origin, so the intercept on the vertical axis is
\[ c = 0 \]Two precautions
- Tight, clean connections at all terminals were ensured before taking readings.
- The key K was opened between successive readings so that the wire did not heat up.
(b)(i) Two devices in which Ohm's law does not apply
- A semiconductor diode (rectifier).
- A thermionic (vacuum) diode/valve.
(b)(ii) Current in each parallel resistor
A total current of \(1\ \text{A}\) is supplied to \(2\ \Omega\) and \(3\ \Omega\) in parallel. The current divides in the inverse ratio of the resistances:
\[ I_{2\Omega} = 1 \times \frac{3}{2+3} = \frac{3}{5} = 0.6\ \text{A} \] \[ I_{3\Omega} = 1 \times \frac{2}{2+3} = \frac{2}{5} = 0.4\ \text{A} \]Therefore \(0.6\ \text{A}\) flows through the \(2\ \Omega\) resistor and \(0.4\ \text{A}\) through the \(3\ \Omega\) resistor.
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