TEST OF PRACTICAL KNOWLEDGE QUESTION Measure and record the length XY of the resistance wire provided. Connect the circuit shown in the diagram. With R= O\(...
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. [\(\pi\) =\(\frac{22}{7}\)]
(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:
Plot of voltage V against current I. The slope is obtained from two widely separated points on the best-fit line.
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, \(d=0.2\,\text{mm}=2.0\times10^{-4}\,\text{m}\).
Plot of voltage V against current I. The slope is obtained from two widely separated points on the best-fit line.
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, \(d=0.2\,\text{mm}=2.0\times10^{-4}\,\text{m}\).