1. P.1 Book 4 Section 6.1 Alternating current Patterns formed by light emitting diodes A.c. and d.c. Check-point 1 Effective value of an a.c. Root-mean-square value of an a.c. Check-point 2 Root-mean-square and peak values Check-point 3 6.1Alternating current

2. P.2 Book 4 Section 6.1 Alternating current Patterns formed by light emitting diodes A and B are identical light emitting diodes (LEDs). A is connected to a battery, while B an a.c. power supply. Take a photo of them with a moving camera: A  line pattern B  dot pattern Why are different patterns formed by LEDs?

3. P.3 Book 4 Section 6.1 Alternating current 1 A.c. and d.c. Current and e.m.f. produced by a small a.c. generator (dynamo) Expt 6a

4. P.4 Book 4 Section 6.1 Alternating current Experiment 6a Current and e.m.f. produced by a small a.c. generator (dynamo) 2.Connect the dynamo to centre-zero milliammeter. Turn the dynamo and watch the deflection of the pointer. 1.Drive an a.c. dynamo to light up a lamp.

5. P.5 Book 4 Section 6.1 Alternating current Experiment 6a Current and e.m.f. produced by a small a.c. generator (dynamo) 6.1 Expt 6a - Current and e.m.f. produced by a small a.c. generator (dynamo) 3.Connect the dynamo to a CRO. Turn it steadily and observe the waveform displayed. Video

6. P.6 Book 4 Section 6.1 Alternating current 1 A.c. and d.c. From Expt 6a: the current and the e.m.f. produced by a simple a.c. dynamo change direction periodically. An a.c. varies periodically with time in both magnitude and direction.

7. P.7 Book 4 Section 6.1 Alternating current 1 A.c. and d.c. Cycle: 1 complete alternation Frequency: number of cycles per second Period: duration of one cycle Peak value of e.m.f. =  0  peak-to-peak value of e.m.f. = 2  0

8. P.8 Book 4 Section 6.1 Alternating current 1 A.c. and d.c. a Sinusoidal a.c. Period = 0.02 s, i.e. frequency = 50 Hz = frequency of electricity supply in HK The CRO shows a sinusoidal alternating voltage output from a low voltage power supply operated with the a.c. mains.

9. P.9 Book 4 Section 6.1 Alternating current 1 A.c. and d.c. b Varying d.c. When a.c. flows through a diode, half cycles instead of full cycles are left ( ∵ diode permits current to flow only in one direction). When an LED is connected to a.c. power supply with current sensor, the sensor measures a varying d.c.

10. P.10 Book 4 Section 6.1 Alternating current 1 A.c. and d.c. c Constant d.c. The type of d.c. that we commonly use is a steady or constant d.c., which gives a horizontal line on a CRO. Different patterns formed by LEDs Example 1

11. P.11 Book 4 Section 6.1 Alternating current Example 1 Different patterns formed by LEDs Why do the two LEDs give different patterns when a photo is taken with a moving camera? (LED lights up only when the p.d. across it is above 2.2 V in the forward direction)

12. P.12 Book 4 Section 6.1 Alternating current Example 1 Different patterns formed by LEDs Battery (constant e.m.f.): LED lights up continuously ∴ shows a line pattern in the photo.

13. P.13 Book 4 Section 6.1 Alternating current Example 1 Different patterns formed by LEDs A.c. power supply: LED lights up only when  is above 2.2 V  turns off when  is below 2.2 V or –ve ∴ shows a dot pattern in the photo.

14. P.14 Book 4 Section 6.1 Alternating current Check-point 1 – Q1 A B C D Which are a.c. currents/voltages? E F

15. P.15 Book 4 Section 6.1 Alternating current 2 Effective value of an a.c. The effective values of alternating current and voltage Expt 6b

16. P.16 Book 4 Section 6.1 Alternating current Experiment 6b The effective values of alternating current and voltage 1.Connect the circuit. Connect a CRO across the two ends of light bulb.

17. P.17 Book 4 Section 6.1 Alternating current Experiment 6b The effective values of alternating current and voltage 6.2 Expt 6b - The effective values of alternating current and voltage 3.Switch to the d.c. power supply. Adjust the rheostat so that light bulb gives the same brightness in the a.c. power supply. Read the d.c. voltage from the CRO and the current I dc from the ammeter. Video 2.Switch to the a.c. power supply. Apply a sinusoidal alternating voltage to light bulb. Note the peak-to-peak value of the voltage.

18. P.18 Book 4 Section 6.1 Alternating current 2 Effective value of an a.c. From Expt 6b: Same brightness  Equal average power developed by the a.c. and the chosen d.c.  size of steady d.c. = effective value of a.c. The effective value of an a.c. is equal to the steady current which produces the same heating effect in the same resistance in the same time.

19. P.19 Book 4 Section 6.1 Alternating current Mean or the average power P = I 2 R = I 2 R.........................(2) A steady current I dc gives the same heating effect (in terms of brightness) in light bulb. Power of the bulb P dc = I dc 2 R....(3) 3 Root-mean-square value of an a.c. When a light bulb of resistance R is lit by an a.c. I, instantaneous power of the bulb P = I 2 R...................................(1)

20. P.20 Book 4 Section 6.1 Alternating current A.c & d.c circuits give the same heating effect: I 2 R = I dc 2 R  I dc = I 2.............(4) i.e. steady current I dc of value I 2 produces the same heating effect as the a.c. I in bulb. ∴ I 2 represents the effective value of the a.c. 3 Root-mean-square value of an a.c.

21. P.21 Book 4 Section 6.1 Alternating current I 2 can be obtained by taking square root of the mean/average value of the square of current.  root-mean-square value (r.m.s. value) of the current, denoted by I rms. 3 Root-mean-square value of an a.c. The effective value of an a.c. is the r.m.s. value of the current. I rms = I 2 = mean value of I 2

22. P.22 Book 4 Section 6.1 Alternating current 3 Root-mean-square value of an a.c. Similarly, effective value of the alternating voltage is the r.m.s. value of the voltage, denoted by V rms. R.m.s. value of an alternating current P = I rms 2 R = V rms 2 R = V rms I rms.............(5) Example 2

23. P.23 Book 4 Section 6.1 Alternating current Example 2 R.m.s. value of an alternating current The current is fixed at 6 A in one half cycle and 3 A in another (in opposite direction). An a.c. flowing through a resistor varies with time.

24. P.24 Book 4 Section 6.1 Alternating current Example 2 R.m.s. value of an alternating current (a) R.m.s. value of the a.c. = ? 1 st half cycle: I 2 = 6 2 2 nd half cycle: I 2 = (–3) 2 = 4.74 A (b) Equivalent steady current of the a.c. = ? I dc = I rms = 4.74 A I rms = mean value of I 2 6 2 + (–3) 2 2 =

25. P.25 Book 4 Section 6.1 Alternating current Check-point 2 – Q1 The current of an a.c. source varies with time. (a) Express I rms in terms of I 0. (b) If 0.5 A of d.c. gives the same heating effect as this a.c., value of I 0 = ?  I 0 = 0.316 A I rms = (2I 0 ) 2 + (–I 0 ) 2 2 = 0.5 I0I0 5252 = I0= I0 5252

26. P.26 Book 4 Section 6.1 Alternating current 4 Root-mean-square and peak values The relationship between V rms, I rms and peak values V 0, I 0 depends on the waveform of a.c. For a sinusoidal case, 2tT2tT sin V = V 0........(6) 2tT2tT sin I = I 0........(7)

27. P.27 Book 4 Section 6.1 Alternating current 4 Root-mean-square and peak values From (1) and (7), the instantaneous power developed by a sinusoidal a.c. in a pure resistor is P = I 2 R 2tT2tT = I 0 sin R 2 2tT2tT sin = I 0 2 R 2 2tT2tT sin P = P 0...................(8) 2

28. P.28 Book 4 Section 6.1 Alternating current 4 Root-mean-square and peak values How the power P for a pure resistor varies with time t : The value of power is non-negative and fluctuating between 0 and max. power P 0.

29. P.29 Book 4 Section 6.1 Alternating current 4 Root-mean-square and peak values Areas above the dotted line fit into the empty regions over completed cycles. (passing through P 0 ) 1212 P 0...............(9) 1212 In this case, the mean power P =

30. P.30 Book 4 Section 6.1 Alternating current 4 Root-mean-square and peak values  The voltage of the mains in Hong Kong, 220 V, is the r.m.s. value. & Peak value of mains supply in Hong Kong Example 3  & P0P0 1212 P = I rms 2 R = I02RI02R 1212 = V rms 2 R V02RV02R 1212 Using (5), (9) and the max. power P 0 = I 0 2 R =, V02RV02R V rms = V02V02 (for sinusoidal a.c.) I rms = I02I02

31. P.31 Book 4 Section 6.1 Alternating current Example 3 Peak value of mains supply in Hong Kong The r.m.s. value of the mains supply in Hong Kong = 220 V. Peak value of alternating voltage = ? = 311 V By V rms =, V02V02 V 0 = V rms  2 = 220  2

32. P.32 Book 4 Section 6.1 Alternating current Check-point 3 – Q1 A sinusoidal a.c. of 60 Hz has a peak value of 15 A. Find the r.m.s. value of the a.c. = 10.6 A I rms = I02I02 = 15 2

33. P.33 Book 4 Section 6.1 Alternating current Check-point 3 – Q2 Find the max. voltage from an a.c. mains of 110 V. = 156 V Max. voltage = V rms 2 = 110 2

34. P.34 Book 4 Section 6.1 Alternating current The End