You have been provided with a resistance box, a voltmeter, a key, a battery, and other necessary materials. Connect the circuit as shown in the diagram abov...
Question 1 Report
You have been provided with a resistance box, a voltmeter, a key, a battery, and other necessary materials.
- Connect the circuit as shown in the diagram above.
- With the key K closed, read and record the voltmeter reading V\(_{0}\).
- Set the resistance R in the resistance box equal to I\(\Omega\).
- Close the key, read and record the potential difference V on the voltmeter.
- EvaluateR\(^{-1}\) and V\(^{-1}\) .
- Repeat the procedure for five other values or R = 2\(\Omega\), 3\(\Omega\), 4\(\Omega\), 5\(\Omega\) and 6\(\Omega\).
- Tabulate your readings.
- Plot a graph with V\(^{-1}\) on the vertical axis and R on the horizontal axis.
- Determine the slope, s, of the graph and the intercept, C, on the vertical axis.
- Evaluate c\(^{-1}\)
- State two precautions taken to obtain accurate results.
(b)i. Define the potential difference between two points in an electric circuit.
ii. Explain why the emf of a cell is greater than the p.d. across the call when it is supplying Current through an external resistance.
Answer Details
(a) E.m.f. and internal resistance experiment
The circuit is connected as shown below: the battery (e.m.f. \(E\), internal resistance \(r\)) is joined in series with the key \(K\) and the resistance box \(R\), while the voltmeter is connected across \(R\) to read the terminal potential difference \(V\).
When the key is closed and the box is set to a resistance \(R\), the same current flows through \(R\) and through the internal resistance \(r\), so the terminal p.d. is
\[ V=\frac{ER}{R+r}. \]Taking reciprocals of both sides,
\[ \frac{1}{V}=\frac{R+r}{ER}=\frac{1}{E}+\frac{r}{E}\cdot\frac{1}{R}. \]Hence a graph of \(V^{-1}\) against \(R^{-1}\) is a straight line of slope \(s=\dfrac{r}{E}\) and vertical intercept \(C=\dfrac{1}{E}\).
Table of readings
| S/N | R / \(\Omega\) | V / V | R\(^{-1}\) / \(\Omega^{-1}\) | V\(^{-1}\) / V\(^{-1}\) |
|---|---|---|---|---|
| 1 | 1.0 | 1.80 | 1.00 | 0.56 |
| 2 | 2.0 | 2.10 | 0.50 | 0.48 |
| 3 | 3.0 | 2.30 | 0.33 | 0.43 |
| 4 | 4.0 | 2.40 | 0.25 | 0.42 |
| 5 | 5.0 | 2.50 | 0.20 | 0.40 |
| 6 | 6.0 | 2.60 | 0.17 | 0.38 |
Graph
Plotting \(V^{-1}\) (vertical axis) against \(R^{-1}\) (horizontal axis) gives the straight line below.
Slope and intercept
Reading two points on the line of best fit, \((R^{-1}=1.00,\;V^{-1}=0.57)\) and \((R^{-1}=0.20,\;V^{-1}=0.40)\):
\[ s=\frac{0.57-0.40}{1.00-0.20}=\frac{0.17}{0.80}=0.21\ \Omega. \]The line cuts the vertical axis (at \(R^{-1}=0\)) at
\[ C=0.36\ \text{V}^{-1}. \]Evaluating \(C^{-1}\)
Since \(C=\dfrac{1}{E}\), the e.m.f. of the cell is
\[ C^{-1}=E=\frac{1}{0.36}=2.78\ \text{V}. \]The internal resistance follows from \(s=\dfrac{r}{E}\):
\[ r=sE=0.21\times2.78=0.58\ \Omega. \]Two precautions
- I read the voltmeter with my eye directly in line with the pointer and the scale to avoid parallax error, and checked that the pointer read zero (no zero error) before starting.
- I ensured all connecting wires were pressed tightly onto the terminals and opened the key immediately after each reading so that the battery would not run down and cause the readings to drift.
(b)(i) Potential difference
The potential difference between two points in an electric circuit is the work done in moving one coulomb of positive charge from one point to the other. It is measured in volts (V).
(b)(ii) Why the e.m.f. exceeds the terminal p.d.
The e.m.f. \(E\) is the total energy supplied by the cell to each coulomb of charge it drives round the whole circuit. When the cell delivers a current \(I\) through an external resistance, part of this energy is used up in driving the current through the cell's own internal resistance \(r\); this wasted "lost volt" is \(Ir\). The p.d. available at the terminals is therefore
\[ V=E-Ir, \]which is less than \(E\). The two are equal only when no current flows (open circuit), where \(Ir=0\).
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