1. 6.1 Alternating current

2. CRO: note , 筆記
CRO: answer , 答案

3. Patterns formed by light emitting diodes
A and B are two 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? (Answer on p.256)

4. Experiment 6a
Drive an a.c. dynamo to light up a lamp.
Connect the dynamo to centre-zero milliammeter. Turn the dynamo and watch the deflection of the pointer.

5. 3. Connect the dynamo to a CRO
3. Connect the dynamo to a CRO. Turn it steadily and observe the waveform displayed.
video

6. From Experiment 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. Cycle: 1 complete alternation Frequency: number of cycles per second
e.g. a sinusoidal a.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. a Sinusoidal a.c.
The CRO shows a sinusoidal alternating voltage output from a low voltage power supply operated with the a.c. mains.
Period = 0.02 s,
i.e. frequency = 50 Hz
= frequency of electricity supply in HK

9. b Varying d.c.
When an a.c. flows through a diode, since a diode permits current to flow only in one direction, half cycles instead of full cycles are left.
When an LED is connected to a.c. power supply with current sensor, the sensor measures a varying d.c.

10. c Constant d.c.
The type of d.c. that we commonly use, e.g. in torches and mobile phones, is a steady or constant d.c., which gives a horizontal line on a CRO.

11. Example 1
Explain why the two LEDs in Let’s begin(p.254) give different patterns when a photo is taken with a moving camera?
Given the LED lights up only when the p.d. across it is above 2.2 V in the forward direction.

12. Battery (constant e.m.f.):
LED lights up continuously
∴ shows a line pattern in the photo.

13. 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. Which are a.c. currents/voltages? B C A
Check-point 1 – Q1
Which are a.c. currents/voltages? B C A D E F

15. cancelled Experiment 6b
The effective values of alternating current and voltage
Connect the circuit. Connect a CRO across the two ends of light bulb.
cancelled

16. cancelled Experiment 6b
The effective values of alternating current and voltage
Switch to the a.c. power supply. Apply a sinusoidal alternating voltage to light bulb.
Note the peak-to-peak value of the voltage.
cancelled
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 Idc from the ammeter.

17. 2 Effective value of an a.c.
For an a.c., its e.m.f. varies with time.
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.

18. 3 Root-mean-square value of an a.c.
It can be derived mathematically that
The effective value of an a.c. is the root-mean-square(r.m.s.) value of the current.
Irms = I 2 = mean value of I 2

19. Similarly, effective value of the alternating voltage is the r. m. s
Similarly, effective value of the alternating voltage is the r.m.s. value of the voltage, denoted by Vrms.
P = Irms2R =
Vrms2 R
= Vrms Irms (5)

20. An a.c. flowing through a resistor varies with time.
Example 2
An a.c. flowing through a resistor varies with time.
The current is fixed at 6 A in one half cycle and 3 A in another (in opposite direction).

21. (a) R.m.s. value of the a.c. = ?
1st half cycle: I 2 = 62
2nd half cycle: I 2 = (–3)2
62 + (–3)2 2 =
Irms = mean value of I 2
= 4.74 A
(b) Equivalent steady current of the a.c. = ?
Idc = Irms
= 4.74 A

22. The current of an a.c. source varies with time.
Check-point 2 – Q1
The current of an a.c. source varies with time.
(a) Express Irms in terms of I0.
Irms =
(2I0)2 + (–I0)2 2
= I0 5 2
(b) If 0.5 A of d.c. gives the same heating effect as this a.c., value of I0 = ?
= 0.5 I0 5 2
 I0 = A

23. 4 Root-mean-square and peak values
It can be derived mathematically that for sinusoidal a.c.
Irms = I0 2
Vrms = V0 2
&
The voltage of the mains in Hong Kong, 220 V, is the r.m.s. value.

24. The r.m.s. value of the mains supply in Hong Kong = 220 V.
Example 3
by mulimeter
The r.m.s. value of the mains supply in Hong Kong = 220 V.
Peak value of alternating voltage = ?
By Vrms = , V0 2
V0 = Vrms  2
= 220  2
= 311 V
by C.R.O.

25. A sinusoidal a.c. of 60 Hz has a peak value of 15 A.
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.
Irms = I0 2 = 15 2
= 10.6 A

26. Find the max. voltage from an a.c. mains of 110 V.
Check-point 3 – Q2
Find the max. voltage from an a.c. mains of 110 V.
Max. voltage
= Vrms 2 =
= 156 V

27. Experiment:
Bicycle generator , 自行車發電機

28. Do Practice 6.1
Handed on 09/04/2013

29. E n d