2. 1© Manhattan Press (H.K.) Ltd. 15.11 Potentiometer Comparing resistances Measuring the e.m.f. of a cell Measuring the internal resistance of a cell

3. 2 © Manhattan Press (H.K.) Ltd. 15.11 Potentiometer (SB p. 170) Potentiometer Potentiometer – measure p.d. and e.m.f. without drawing voltage V  V = k 1. Measuring p.d. - as an ideal voltmeter - X and Y are connected to +ve and –ve terminals of a voltmeter

4. 3 © Manhattan Press (H.K.) Ltd. 15.11 Potentiometer (SB p. 170) Potentiometer 2. Measuring e.m.f. - X and Y are connected across the cell - e.m.f. of the cell (E) = p.d. across the length

5. 4 © Manhattan Press (H.K.) Ltd. 15.11 Potentiometer (SB p. 171) Comparing resistances Compare resistances of two resistors by comparing p.d. across them (connected in series) Go to Example 18 Example 18

6. 5 © Manhattan Press (H.K.) Ltd. 15.11 Potentiometer (SB p. 173) Measuring the e.m.f. of a cell potentiometer – be calibrated before uses V = E (e.m.f. of the cell)

7. 6 © Manhattan Press (H.K.) Ltd. 15.11 Potentiometer (SB p. 173) Measuring the e.m.f. of a cell Notes: When using a voltmeter to measure the e.m.f. of a cell, a current flows from the cell through the voltmeter to produce a deflection. Hence, the p.d. across the terminals of the cell as measured by the voltmeter is: V = E – Ir < E (I  0) This means that the voltmeter reading is less than the actual e.m.f. of the cell.

8. 7 © Manhattan Press (H.K.) Ltd. 15.11 Potentiometer (SB p. 174) Measuring the internal resistance of a cell

9. 8 © Manhattan Press (H.K.) Ltd. 15.11 Potentiometer (SB p. 175) Measuring the internal resistance of a cell intercept on 1/R- axis Go to Example 19 Example 19 Go to Example 20 Example 20

10. 9 © Manhattan Press (H.K.) Ltd. 1. The flow of electrons is the cause of an electric current. 2. Current is the rate of flow of charge: 3. One coulomb is the quantity of electric charge for a steady current of 1 ampere flowing through a given point for 1 second. 15.1 Electric current

11. 10 © Manhattan Press (H.K.) Ltd. 4. The steady average velocity of the free electrons in the direction of drift (or in the direction opposite to that of the electric field) in the metal, is known as the drift velocity of the free electrons. 5. The current can also be defined as: I = nAvQ where n is the number of free electrons per unit volume of the wire, A is the cross- sectional area of the wire, v is the drift velocity, Q is the charge carried by an electron (i.e. 1.6 ×10 –19 C). 15.1 Electric current

12. 11 © Manhattan Press (H.K.) Ltd. 6. The potential difference between two points in a circuit is the energy converted from electrical potential energy to other forms, when one coulomb of charge passing between the points outside the source. 7. We have 15.2 Potential difference (p.d.)

13. 12 © Manhattan Press (H.K.) Ltd. 8. The electromotive force (e.m.f.) of a source is defined as the energy supplied to each coulomb of charge by the source to drive the charge around a complete circuit. 15.3 Electromotive force (e.m.f.)

14. 13 © Manhattan Press (H.K.) Ltd. 9. Resistance is defined as: 10. The ohm (1 Ω) is the resistance of a conductor when a current of one ampere flows through the conductor, the potential difference across it is 1 volt. 15.4 Resistance and Ohm’s Law

15. 14 © Manhattan Press (H.K.) Ltd. 11. Ohm’s Law states that the steady current through a metallic conductor is directly proportional to the potential difference across it, provided that the temperature and other physical conditions remain constant. 12. The internal resistance (r) of a cell is due to the resistance of the chemicals in the cell. E = I (R + r) 15.4 Resistance and Ohm’s Law

16. 15 © Manhattan Press (H.K.) Ltd. 13. The resistance R of a uniform wire is directly proportional to its length (l) and inversely proportional to its cross-sectional area (A). That is, where ρ is a constant known as the resistivity of the material of the wire, provided that other physical conditions like temperature remain unchanged. 15.5 Factors affecting resistance of a conductor

17. 16 © Manhattan Press (H.K.) Ltd. 14. For the resistors connected in series, the equivalent resistance of the resistors (R) is: R = R 1 + R 2 + R 3 +... 15. For the resistors connected in parallel, the equivalent resistance of the resistors (R) is: 15.6 Resistor networks

18. 17 © Manhattan Press (H.K.) Ltd. 16. Kirchhoff’s First Law refers to any point in a network. It states that: The algebraic sum of the currents at a junction of a circuit is zero. Therefore, the current arriving a junction equals the current leaving the junction. 15.7 Kirchhoff’s Laws

19. 18 © Manhattan Press (H.K.) Ltd. 17. The power dissipated in the external resistor of resistance R is maximum when R = r, the value of the internal resistance of the cell. 15.8 Maximum power and efficiency

20. 19 © Manhattan Press (H.K.) Ltd. 18. A potential divider is an arrangement of resistors which is used to obtain a fraction of the potential difference provided by a voltage supply. 15.9 Potential divider

21. 20 © Manhattan Press (H.K.) Ltd. 19. The Wheatstone bridge is a device used to measure the resistance accurately. A simple form of Wheatstone bridge is the metre bridge. 15.10 Wheatstone bridge and metre bridge

22. 21 © Manhattan Press (H.K.) Ltd. 20. A potentiometer is used to measure the potential differences and e.m.f. to a high degree of accuracy. 21. A potentiometer can also be used to: (a) Compare resistances (b) Measure the e.m.f. of a cell (c) Measure the internal resistance of a cell 15.11 Potentiometer

23. 22 © Manhattan Press (H.K.) Ltd.

24. 23 © Manhattan Press (H.K.) Ltd. End

25. 24 © Manhattan Press (H.K.) Ltd. Q: Q: The circuit shown is used to compare the resistance R of an unknown resistor with a standard 100 Ω resistor. When X is connected to Y and then to Z, the distances from one end of the slide wire of the potentiometer to the balance point are 400 mm and 588 mm, respectively. If the length of the slide wire is 1.00 m, what is the value of R ? 15.11 Potentiometer (SB p. 172) Solution

26. 25 © Manhattan Press (H.K.) Ltd. Solution: Let I = current passing through the 100 Ω resistor and R. When X is connected to Y,  = 400 mm, then p.d. across 100 Ω = p.d. across 400 mm I x 100 = k x 400................... (1) where k = constant. When X is connected to Z, = 588 mm, then p.d. across (100 + R) = p.d. across 588 mm I x (100 + R) = k x 588........... (2) Return to Text 15.11 Potentiometer (SB p. 172)

27. 26 © Manhattan Press (H.K.) Ltd. Q: Q: Two cells A and B and a centre-zero galvanometer G are connected to a uniform slide wire OS by two sliding contacts X and Y as shown in the figure. The length of the slide wire is 1.0 m and has a resistance of 12 Ω. With OY of length 75 cm, the galvanometer shows no deflection when OX is 50 cm. When Y is moved to be in contact with the end of the wire at S, the length OX is 62.5 cm when the galvanometer is balanced. The e.m.f. of cell B is 1.0 V. Calculate (a) the potential difference across OY when Y is 75 cm from O and the galvanometer is balanced, (b) the potential difference across OS when Y is in contact with S and the galvanometer is balanced, (c) the internal resistance of cell A, and (d) the e.m.f. of cell A. 15.11 Potentiometer (SB p. 175) Solution

28. 27 © Manhattan Press (H.K.) Ltd. Solution: (a) OX = 50 cm when the galvanometer is balanced. Hence, p.d. across 50 cm = e.m.f. of cell B = 1.0 V ∴ p.d. across OY (75 cm) = = 1.5 V (b) When the slider Y is at S, OX = 62.5 cm ∴ p.d. across OX (62.5 cm) = e.m.f. of cell B = 1.0 V ∴ p.d. across OS (100 cm) = = 1.6 V (c) When OY = 75 cm and OX = 50 cm, the galvanometer is balanced and no current flows from cell B. Hence, the current flowing along the slide wire is from cell A. From (a), p.d. across 75 cm (OY) = 1.5 V 15.11 Potentiometer (SB p. 175)

29. 28 © Manhattan Press (H.K.) Ltd. Solution (Cont’d): Return to Text 15.11 Potentiometer (SB p. 176)

30. 29 © Manhattan Press (H.K.) Ltd. Q: Q: In the circuit, cell A has an e.m.f. of 2.0 V and negligible internal resistance. XY is a uniform wire of length 100 cm and resistance 5.0 Ω. Cell B has an e.m.f. of 1.5 V and internal resistance of 0.80 Ω. Calculate the length XP required to produce zero deflection in the galvanometer G (a) in the circuit shown, (b) when a 1.0 Ω resistor is placed in series with A, (c) when this 1.0 Ω resistor is removed from A and placed in series with B, and (d) when this 1.0 Ω resistor is placed in parallel with B. 15.11 Potentiometer (SB p. 176) Solution

31. 30 © Manhattan Press (H.K.) Ltd. Solution: (a) In the circuit shown, the potentiometer is used to measure the e.m.f. of cell B. When the galvanometer is balanced, p.d. across XP = e.m.f. of cell B But p.d. across XY (100 cm) = 2.0 V (b) When a 1.0 Ω resistor is placed in series with A, the current through wire XY: 15.11 Potentiometer (SB p. 177)

32. 31 © Manhattan Press (H.K.) Ltd. Solution (Cont’d): Return to Text (c) When the 1.0 Ω resistor is connected in series with cell B and the galvanometer is balanced, no current flows through the cell B or the 1.0 Ω resistor. (d) When the 1.0 Ω resistor is placed in parallel with cell B, the p.d. V across the 1.0 Ω resistor is measured by the potentiometer, and V is given by: 15.11 Potentiometer (SB p. 177)