You are provided with a battery of e.m.f. E, a key K, a voltmeter, a standard resistor R0 = \(2\Omega\), a resistance box R, and some connecting wires. i. M...
Question 1 Report
You are provided with a battery of e.m.f. E, a key K, a voltmeter, a standard resistor R0 = \(2\Omega\), a resistance box R, and some connecting wires.
i. Measure and record the e.m.f. E of the battery.
i. Set up a circuit as shown in the diagram above with the key open.
iii. Set the resistance on the resistance box to R 22.
iv. Close the key, read and record the potential difference V on the voltmeter.
v. Evaluate \(V^{-1}\)
vi. Repeat the procedures for five other values of R = \(5\Omega\) \(10\Omega\), \(12\Omega\), \(15\Omega\) and \(20\Omega\). In each case, record V and evaluate V\(^{-1}\)
vii. Tabulate the results.
viii. Plot a graph with R on the vertical axis and \(V^{-1}\) l on the horizontal axis, starting both axes from the origin (0,0).
ix. Determine the slope, s, of the graph and the intercept c on the vertical axis.
x. Calculate \(\propto\) and \(\beta\) from the equations s = R0 \(\propto\) and c= - (R0+B).
xi. State two precautions taken to obtain accurate results.
Answer Details
Determination of the e.m.f. and internal resistance of a battery
The e.m.f. of the battery, measured with the key open, is:
\(E=2.00\ \text{V}\)
With \(R_0=2\ \Omega\), the resistance box was set successively to \(2, 5, 10, 12, 15\) and \(20\ \Omega\). The voltmeter was connected across the standard resistor \(R_0\).
Table of readings
| S/N | \(R\ (\Omega)\) | \(V\ (\text{V})\) | \(V^{-1}\ (\text{V}^{-1})\) |
|---|---|---|---|
| 1 | 2 | 0.800 | 1.250 |
| 2 | 5 | 0.500 | 2.000 |
| 3 | 10 | 0.308 | 3.250 |
| 4 | 12 | 0.267 | 3.750 |
| 5 | 15 | 0.222 | 4.500 |
| 6 | 20 | 0.174 | 5.750 |
Graph of \(R\) against \(V^{-1}\)
The graph is a straight line. Using two well-separated points on the line, \((V^{-1}=1.25\ \text{V}^{-1},\ R=2.0\ \Omega)\) and \((V^{-1}=5.75\ \text{V}^{-1},\ R=20.0\ \Omega)\):
\[s=\frac{\Delta R}{\Delta(V^{-1})}=\frac{20.0-2.0}{5.75-1.25}=\frac{18.0}{4.50}=4.0\ \Omega\,\text{V}.\]
From \(R=sV^{-1}+c\),
\[c=R-sV^{-1}=2.0-(4.0\times1.25)=-3.0\ \Omega.\]
Hence,
\[\alpha=\frac{s}{R_0}=\frac{4.0}{2.0}=2.0\ \text{V},\]
\[\beta=-c-R_0=-(-3.0)-2.0=1.0\ \Omega.\]
Therefore, \(\alpha\), the e.m.f. of the battery, is \(2.0\ \text{V}\), while \(\beta\), the internal resistance of the battery, is \(1.0\ \Omega\).
Precautions
- The key was closed only while taking a reading, to prevent heating and polarization of the battery.
- All connections were clean and tight, and the voltmeter was read at eye level to avoid parallax error.
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