You are provided with a battery of e.m.f. E, a key... | WAEC SSCE - Physics - 2019 (Practical)

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

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.

Read lesson note on Capacitors (JAMB)

Read lesson note on Simple A.C Circuits (JAMB)

Answer Details