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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. SMJKKH/NPC/F5P/C7/L4 1 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. SMJKKH/NPC/F5P/C7/L4 2 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. SMJKKH/NPC/F5P/C7/L4 3 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 0.2510 2 3.0 V SMJKKH/NPC/F5P/C7/L4 4