Measure and record the length XY of the resistance wire provided. Connect the circuit shown in the diagram. With R= O\(\Omega\), close the key, K. Read and ...
Question 1 Report
- Measure and record the length XY of the resistance wire provided.
- Connect the circuit shown in the diagram.
- With R= O\(\Omega\), close the key, K. Read and record the current 1\(_{o}\) and the voltage drop V\(_{o}\) across the resistance wire.
- Setting R = 1\(\Omega\). close the key. Read and record the current, I, and the corresponding voltage drop, V across the wire.
- Repeat the procedure for five other values of R= 5, 10, 20, 40, and 60\(\Omega\). Tabulate your readings.
- Plot a graph of V on the vertical axis against 1 on the horizontal axis.
- Determine the slope of the graph
- State two precautions taken to ensure accurate results.
(b)i. Mention and state the law on which the experiment in (a) is based.
ii. A piece of resistance vire of diameter 0.2 mm and resistance m has a resistivity of 8.8 x 10\(^{-7}\)\(\Omega\)m, calculate the length of the Wire. [\(\pi\) =\(\frac{22}{7}\)]
Answer Details
(a) Readings and graph
Length of resistance wire, XY = 100 cm.
| Resistance, R (Ω) | Current, I (A) | Potential difference, V (V) |
|---|---|---|
| 0 | 0.15 | 2.65 |
| 1 | 0.15 | 2.65 |
| 5 | 0.15 | 2.65 |
| 10 | 0.10 | 1.85 |
| 20 | 0.08 | 1.40 |
| 40 | 0.06 | 0.90 |
| 60 | 0.04 | 0.70 |
Graph of V against I:
Using two widely separated points on the line of best fit, \[ (I_1,V_1)=(0.025\,\text{A},0.50\,\text{V}), \qquad (I_2,V_2)=(0.100\,\text{A},1.75\,\text{V}) \] \[ \text{Slope}=\frac{\Delta V}{\Delta I}=\frac{1.75-0.50}{0.100-0.025}=\frac{1.25}{0.075}=16.7\ \Omega. \] Thus, the resistance of the wire is approximately 16.7 Ω.
Precautions:
- Connections were clean, tight and correctly made before taking readings.
- The key was opened immediately after each reading to prevent heating of the resistance wire.
(b)(i)
The experiment is based on Ohm's law. Ohm's law states that the current through a metallic conductor is directly proportional to the potential difference across its ends, provided that temperature and other physical conditions remain constant. Thus,
\[\frac{V}{I}=\text{constant}.\](b)(ii)
Diameter of wire, \(d=0.2\,\text{mm}=2.0\times10^{-4}\,\text{m}\).
\[A=\frac{\pi d^2}{4}=\frac{\frac{22}{7}(2.0\times10^{-4})^2}{4}=3.14\times10^{-8}\,\text{m}^2.\]Using \(R=7\,\Omega\), \(\rho=8.8\times10^{-7}\,\Omega\text{m}\), and \(R=\frac{\rho L}{A}\),
\[L=\frac{RA}{\rho}=\frac{7\times3.14\times10^{-8}}{8.8\times10^{-7}}=0.25\,\text{m}.\]Length of the wire = \(0.25\,\text{m}\).
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