Topic 4: Oscillations and Waves

1. Topic 4: Oscillations and Waves TEST Friday 12th March

2. Topic 5: Electric currents Can you look through the contents and definitions? Definition TEST Friday 19th March

3. Stand up!

4. ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺

5. ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ cell energy electron lamp

6. Electrons Hi, I’m Eleanor the electron. ☺

7. Coulomb of charge (electrons) Think of it as a “bag of electrons” (containing 6000000000000000000 electrons!) ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺

8. A ☺ ☺ ☺ ☺ ☺ I’m counting how many coulombs of electrons go past me every second Current ☺ ☺ The rate of flow of electric charge (number of Coulombs flowing past a point in the circuit every second). I = Q/t ☺ ☺ ☺ ☺ 1 Amp = 1 coulomb per second

9. Let’s build some circuits! ☺

10. In a series circuit Current is the same at any point in the circuit 2.5 A 2.5 A 2.5 A 2.5 A

11. In a parallel circuit The current splits (total current stays the same) 2.5 A 2.5 A 1.25 A 1.25 A

12. ☺ ☺ ☺ V ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ Voltage(emf) I’m checking the difference in energy (per coulomb) between the 2 red arrows 1 Volt = 1 Joule per coulomb

13. ☺ ☺ ☺ V ☺ ☺ ☺ ☺ ☺ ☺ ☺ ☺ Voltage (p.d.) I’m checking the difference in energy (per coulomb) before and after the lamp 1 Volt = 1 Joule per coulomb

14. p.d. and e.m.f Electric potential difference between two points is the work done per unit charge to move a small positive charge between two points. Electromotive force is the total energy difference per unit charge around the circuit (it is the potential difference when no current flows in a circuit).

15. Let’s build some circuits! ☺

16. In a series circuit The sum of thep.d.s across the lamps equals the emf across the cells 9 V 3 V 3 V 3 V

17. In a parallel circuit In a simple parallel circuit, p.d. across each lamp equals the e.m.f. across the cells 5 V 5 V 5 V

18. Stand up!

19. V A Resistance Measures how difficult it is for current to flow. Measured in Ohms (Ω) Resistance = voltage/current R = V/I

20. V I R X Ohm’s Law • V = IR

21. Let’s do a practical!

22. Resistance • R is proportional to the length of wire – WHY? R α L • R is inversely proportional to the cross sectional area of wire – WHY? R α 1/A • R depends on the type of material – WHY?

23. Resistivity R = ρL A where R = resistance in Ohms L = Length of conductor in metres A = cross sectional area of conductor in m2 ρ = resistivity of the material in Ohms.meters

24. Example The resistivity of copper is 1.7 x 10-8Ωm. What is the resistance of a piece of copper wire 1 m in length with a diameter of 0.1mm?

25. Example The resistivity of copper is 1.7 x 10-8Ωm. What is the resistance of a piece of copper wire 1 m in length with a diameter of 0.1mm? radius = 0.05mm = 5 x 10-5m cross sectional area = πr2 = 3.14x(5 x 10-5)2 = 7 x 10-9 m2 R = ρL/A = (1.7 x 10-8 x 1)/ 7 x 10-9 = 2.42 Ω

26. Let’s do another practical!

27. Homework • BOTH electricity practicals (resistance of different thicknesses of wire AND voltage-current characteristics of filament lamps) to be handed in Wednesday 14th April.

28. A V Resistance of a lamp Vary the voltage and current using a variable resistor (rheostat). Plot a graph of resistance against current Resistance = voltage/current R = V/I

29. Resistance of a lamp • As the current in a lamp increases, its resistanceincreases. Why?

30. Ohmic behaviour • p.d. is proportional to the current Metal wires at constant temperature

31. Non-Ohmic behaviour • p.d. is not proportional to the current

32. V A Power The amount of energy used by a device per second, measured in Watts (Joules per second) Power = voltage x current P = VI

33. Power dissipated in a resistor/lamp • P = VI • From Ohm’s law, V = IR • So P = VI = I2R • From Ohm’s law also, I = V/R • So P = VI = V2/R

34. Total energy So the total energy transformed by a lamp is the power (J/s) times the time the lamp is on for in seconds, E = VIt E = energy transformed (J) V = Voltage (also called p.d.) I = current (A) t = time (s)

35. Electronvolt • Electronvolt – the energy gained by an electron when it moves through a potential difference of one volt.

36. Questions! • Page 316 and 317 questions 2, 5, 8, 9, 10, 12, 13, 14, 15, 17, 18.

37. Internal resistance • Connecting a voltmeter (VERY high resistance) across the terminals of a cell measures the EMF of the cell (no current flowing) V

38. Internal resistance • We have assumed so far that the power source has no resistance…….not a good assumption!

39. Internal resistance • In actuality the p.d. across a cell is less than the EMF due to energy lost in the INTERNAL RESISTANCE

40. Internal resistance • To help us visualize this, a cell is represented as a “perfect” cell attached in series to the internal resistance, given the symbol r.

41. Internal resistance • The p.d. across a cell (V) is then equal to the EMF (ε)minus the voltage lost across the internal resistance (=Ir) V = ε - Ir

42. Example • A cell of emf 12V and internal resistance 1.5 Ω produces a current of 3A. What is the p.d. across the cell terminals? • V = ε - Ir • V = 12 – 3x1.5 • V = 7.5 V

43. Another practical!

44. Adding resistances

45. In series and parallel

46. Reading and taking notes • Pages 320 to 328 • Read and make your OWN NOTES. • I will collect these in the first lesson back after the holiday.

47. Ideal meters • Voltmeters – infinite resistance! • Ammeters – Zero resistance!

48. Potential divider

49. LDRs, Thermistors in potential divider circuits

50. Simple!