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Form 5 Physics – Chapter 7 – Lesson 4
7.4
Analysing Electromotive Force and Internal Resistance
7.4.1 Electromotive Force
1. A light bulb will light up when it is connected in series with a cell as shown in
Figure 2.45.
2. The cell is the source of energy and the light bulb is the energy consuming device.
3. The light bulb converts electrical energy into heat and light energy.
4. The electric charges that flow round the circuit transfer energy from the source to
the device.
5. In a cell, chemical reaction converts chemical energy into electrical energy. This
energy pushes the free electrons to move them from the negative terminal to the
positive terminal of the cell.
6. Work is done by the source in driving the charges around a complete circuit. This
work done is known as electromotive force.
7. The electromotive force (e.m.f) is the work done by a source in driving a unit
charge around a complete circuit.
7.4.2 Electromotive Force and Potential Difference
1. The definition for electromotive force (e.m.f) is similar to that of potential
difference (p.d.). However there is a distinction between e.m.f. and p.d.
2. The e.m.f of a cell is the energy supplied to a unit of charge within the cell.
3. The p.d. across a component in a circuit is the conversion of electrical energy into
other forms of energy when a unit of charge passes through the component.
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Form 5 Physics – Chapter 7 – Lesson 4
7.4.3 Internal Resistance
1. In an open circuit when there is no current flow, the potential difference, V across
the cell is the electromotive force, E of the cell.
2. In a closed circuit when there is a current flow, the potential difference, V across
the cell is smaller than the electromotive force, E of the cell.
3. The drop in potential difference across the cell is caused by the internal resistance
of the cell.
4. The internal resistance of a source or a cell is the resistance against the moving
charge due to electrolyte in the source or the cell.
5. Work is needed to drive a charge against the internal resistance.
6. This causes a drop in potential difference across the cell as the charge flows
through it.
7.4.4 Electromotive Force and Internal Resistance
1. A cell can be modeled as an e.m.f., E connected in series with an internal resistor,
r as shown in Figure 2.51.
2. When a high resistance voltmeter is connected across the terminals of the cell as
shown in Figure 2.52, the reading of the voltmeter gives the e.m.f., E of the cell.
3. If a resistor, R is then connected to the terminals of the cell as shown in Figure
2.53, the voltmeter reading is the potential difference, V across the resistor, R. It is
also the potential difference, V across the terminals of the cell.
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Form 5 Physics – Chapter 7 – Lesson 4
4. The value of the potential difference, V is less than the e.m.f., E of the cell. The
difference between E and V is due to the potential difference needed to drive the
current, I through the interval resistor, r of the cell. Hence,
E  V  Ir
Where,
e.m. f =

E  V  Ir
Potential
differnce
+
5. The internal resistance, r is given by: r 
Drop in potential
difference due to
internal resistance
E V
I
6. In the rheostat in Figure 2.56 is varied for a set of values for current, I and
potential difference, V, a graph of V against I can be plotted to get the values of
e.m.f., E and internal resistance, r.
7. The graph of V against I in Figure 2.57 is a straight line graph. The straight line
can be represented by the equation:
V  rI  E
[from E  V  Ir ]
8. If the straight line is extrapolated until it cuts the vertical axis V, the values of
I  0 and V  E are obtained. This shows that when no charges flow, the
potential difference across the cell is the electromotive force.
9. The gradient of the graph is negative showing that V always less than E by some
quantity Ir. The value of Ir is sometimes called the ‘lost volt’ due to the internal
resistance, r.
10. The internal resistance, r can be determined from the gradient of the graph.
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Form 5 Physics – Chapter 7 – Lesson 4
Example 7.4.1:
Figure 2.58 shows a 10  resistor connected in series to a cell. The voltmeter gives a
reading of 2.5 V across the 10  resistor.
From the e.m.f., E of the cell if the internal resistance, r is 2  .
Solution:
V = 2.5 V ; R = 10 
Current, I 
V 2. 5

 0.25 A
R 10
e.m.f., E = I R  r   0.2510  2  3.0 V
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