TEST OF PRACTICAL KNOWLEDGE QUESTION (a) You are provided with a battery, an ammeter, a voltmeter, a resistance box, a key and connection wires. (i) Set up ...
(a) You are provided with a battery, an ammeter, a voltmeter, a resistance box, a key and connection wires.
(i) Set up circuit as shown in the diagram above.
(ii) With the key opened, measure and record the e.m.f. \(E_o\) of the battery
(iii) With the key closed, select the resistance \(R=1\) on the resistance box. Read and record the current, 1Ω.
(iv) Evaluate \(I^{-1}\).
(v) Repeat the procedure for five other values of \(R=2\Omega\), \(3\Omega\), \(4\Omega\), \(5\Omega\), and \(6\Omega\).
In each case, record \(I\) and evaluate \(I^{-1}\) results.
(vii) Plot a graph with \(R\) on the vertical axis and \(I^{-1}\) on the horizontal axis.
(Viii) Determine the slope, \(s\), of the graph.
(ix) Determine the intercept \(C\), On the vertical axis.
(x) State two precautions taken to ensure accurate results.
(b)(i) Define potential difference in an electric field.
(ii) A piece of resistance wire of diameter 0.2 mm and length 25 cm has a resistance of 7Ω. Calculate the resistivity of the wire of the battery. [\(\pi = \frac{22}{7}\)]
(a) Test of practical knowledge: e.m.f. and internal resistance of a battery
The components are connected in a single series loop: battery, ammeter, resistance box and key, with the voltmeter connected across the battery terminals as shown.
Circuit for measuring the e.m.f. and internal resistance of the battery: battery (E₀, r), ammeter A and resistance box R in series with the key K, and voltmeter V connected across the battery terminals.
(ii) e.m.f. with key open: With the key open no current flows, so the voltmeter reads the full e.m.f. of the battery:
\[ E_o = 2.0\ \text{V} \]
(iii)–(vi) Readings: With the key closed, each value of \(R\) is selected on the resistance box, the steady current \(I\) is read on the ammeter, and \(I^{-1}\) is evaluated. The results are tabulated below.
\(R\ (\Omega)\)
\(I\ (\text{A})\)
\(I^{-1}\ (\text{A}^{-1})\)
1
1.33
0.75
2
0.80
1.25
3
0.571
1.75
4
0.444
2.25
5
0.364
2.75
6
0.308
3.25
(vii) Graph of \(R\) (vertical axis) against \(I^{-1}\) (horizontal axis):
Straight-line plot of R (vertical) against I^{-1} (horizontal). Slope = 2.0 V = e.m.f.; intercept on the R-axis = -0.5 Ω = -r.
Theory used. For a cell of e.m.f. \(E\) and internal resistance \(r\) driving current \(I\) through an external resistance \(R\):
\[ E = I(R + r) \]
Dividing through by \(I\) and rearranging:
\[ R = E\left(\frac{1}{I}\right) - r = E\,I^{-1} - r \]
Comparing with \(y = mx + c\), a plot of \(R\) against \(I^{-1}\) is a straight line of slope \(E\) and intercept \(-r\) on the \(R\)-axis.
(viii) Slope of the graph. Taking two widely separated points on the line of best fit, \((I^{-1}_1, R_1) = (0.75,\ 1.0)\) and \((I^{-1}_2, R_2) = (3.25,\ 6.0)\):
The slope equals the e.m.f. of the battery, \(E = 2.0\ \text{V}\), in agreement with the open-key reading \(E_o\).
(ix) Intercept on the vertical axis. Extending the line to \(I^{-1} = 0\):
\[ C = R - s\,I^{-1} = 1.0 - (2.0 \times 0.75) = 1.0 - 1.5 = -0.5\ \Omega \]
So \(C = -0.5\ \Omega\). Since the intercept equals \(-r\), the internal resistance of the battery is
\[ r = 0.5\ \Omega \]
(x) Two precautions:
Parallax error was avoided by looking perpendicularly at the pointer and scale of both the ammeter and the voltmeter when taking readings.
The key was opened immediately after each reading to prevent the battery from running down and to avoid heating of the resistance box, which would change its resistance.
(b)(i) Definition of potential difference
The potential difference between two points in an electric field is the work done in moving a unit positive charge from one point to the other.
\[ V = \frac{W}{Q} \]
It is measured in volts (V), where one volt is one joule per coulomb.
(a) Test of practical knowledge: e.m.f. and internal resistance of a battery
The components are connected in a single series loop: battery, ammeter, resistance box and key, with the voltmeter connected across the battery terminals as shown.
Circuit for measuring the e.m.f. and internal resistance of the battery: battery (E₀, r), ammeter A and resistance box R in series with the key K, and voltmeter V connected across the battery terminals.
(ii) e.m.f. with key open: With the key open no current flows, so the voltmeter reads the full e.m.f. of the battery:
\[ E_o = 2.0\ \text{V} \]
(iii)–(vi) Readings: With the key closed, each value of \(R\) is selected on the resistance box, the steady current \(I\) is read on the ammeter, and \(I^{-1}\) is evaluated. The results are tabulated below.
\(R\ (\Omega)\)
\(I\ (\text{A})\)
\(I^{-1}\ (\text{A}^{-1})\)
1
1.33
0.75
2
0.80
1.25
3
0.571
1.75
4
0.444
2.25
5
0.364
2.75
6
0.308
3.25
(vii) Graph of \(R\) (vertical axis) against \(I^{-1}\) (horizontal axis):
Straight-line plot of R (vertical) against I^{-1} (horizontal). Slope = 2.0 V = e.m.f.; intercept on the R-axis = -0.5 Ω = -r.
Theory used. For a cell of e.m.f. \(E\) and internal resistance \(r\) driving current \(I\) through an external resistance \(R\):
\[ E = I(R + r) \]
Dividing through by \(I\) and rearranging:
\[ R = E\left(\frac{1}{I}\right) - r = E\,I^{-1} - r \]
Comparing with \(y = mx + c\), a plot of \(R\) against \(I^{-1}\) is a straight line of slope \(E\) and intercept \(-r\) on the \(R\)-axis.
(viii) Slope of the graph. Taking two widely separated points on the line of best fit, \((I^{-1}_1, R_1) = (0.75,\ 1.0)\) and \((I^{-1}_2, R_2) = (3.25,\ 6.0)\):
The slope equals the e.m.f. of the battery, \(E = 2.0\ \text{V}\), in agreement with the open-key reading \(E_o\).
(ix) Intercept on the vertical axis. Extending the line to \(I^{-1} = 0\):
\[ C = R - s\,I^{-1} = 1.0 - (2.0 \times 0.75) = 1.0 - 1.5 = -0.5\ \Omega \]
So \(C = -0.5\ \Omega\). Since the intercept equals \(-r\), the internal resistance of the battery is
\[ r = 0.5\ \Omega \]
(x) Two precautions:
Parallax error was avoided by looking perpendicularly at the pointer and scale of both the ammeter and the voltmeter when taking readings.
The key was opened immediately after each reading to prevent the battery from running down and to avoid heating of the resistance box, which would change its resistance.
(b)(i) Definition of potential difference
The potential difference between two points in an electric field is the work done in moving a unit positive charge from one point to the other.
\[ V = \frac{W}{Q} \]
It is measured in volts (V), where one volt is one joule per coulomb.
