(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...
Question 1 Report
(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. [\(\pi = \frac{22}{7}\)]
Answer Details
Theory. For a cell of e.m.f. E and internal resistance r driving current I through an external resistance R,
\[ E = I(R + r) \;\Rightarrow\; R = \frac{E}{I} - r = E\,I^{-1} - r \]Expected graph. A plot of R (vertical) against \(I^{-1}\) (horizontal) is a straight line of
\[ \text{slope } s = E \quad\text{and}\quad \text{intercept on the } R\text{-axis } C = -r \]So the slope gives the e.m.f. of the battery and the magnitude of the (negative) intercept gives its internal resistance.
Sample table (headings).
| R (Ω) | I (A) | \(I^{-1}\) (A\(^{-1}\)) |
|---|---|---|
| 1 | - | - |
| 2 | - | - |
| 3 | - | - |
| 4 | - | - |
| 5 | - | - |
| 6 | - | - |
Two precautions.
- The key was closed only while taking a reading, to prevent the battery running down and its e.m.f. drifting.
- Clean, tight connections were made and the eye placed squarely in front of the meter to avoid parallax when reading the ammeter.
(b)(i) Potential difference. The potential difference between two points in an electric field is the work done in moving one coulomb of positive charge from one point to the other.
(b)(ii) Resistivity calculation. Diameter \(=0.2\ \text{mm}\), so radius \(r = 0.1\ \text{mm} = 1\times10^{-4}\ \text{m}\); length \(L = 25\ \text{cm} = 0.25\ \text{m}\); resistance \(=7\ \Omega\).
\[ A = \pi r^{2} = \frac{22}{7}(1\times10^{-4})^{2} = 3.14\times10^{-8}\ \text{m}^{2} \] \[ \rho = \frac{RA}{L} = \frac{7\times3.14\times10^{-8}}{0.25} = 8.8\times10^{-7}\ \Omega\,\text{m} \]The resistivity of the wire is about \(8.8\times10^{-7}\ \Omega\,\text{m}\).
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