6.1Alternating current. CRO: answer, 答案answer 答案 CRO: note, 筆記note 筆記

6.1 Alternating current

CRO: answer , 答案

CRO: note , 筆記

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 themwith a moving camera:

A line pattern B dot patternWhy are different patterns formed by LEDs? (Answer on p.256)

Experiment 6a

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.

video

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

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.

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. = 20

e.g. a sinusoidal a.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.

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.

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.

Example 1Explain 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.

Battery (constant e.m.f.):

LED lights up continuously∴ shows a line pattern in the photo.

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.

Check-point 1 – Q1

A B C

D

Which are a.c. currents/voltages?

E F

Experiment 6bThe effective values of alternating current and voltage

1. Connect the circuit. Connect a CRO across the two ends of light bulb.

Experiment 6bThe 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 Idc from the ammeter.

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.

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.

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

3 Root-mean-square value of an a.c.

It can be derived mathematically that

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)

Example 2

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.

(a) R.m.s. value of the a.c. = ?

1st half cycle: I 2 = 62

2nd half cycle: I 2 = (–3)2

= 4.74 A

(b) Equivalent steady current of the a.c. = ?

Idc = Irms = 4.74 A

Irms = mean value of I 262 + (–3)2

2=

Check-point 2 – Q1

The current of an a.c. source varies with time. (a) Express Irms in terms of I0.

(b) If 0.5 A of d.c. gives the same heating effect as this a.c., value of I0 = ?

I0 = 0.316 A

Irms =(2I0)2 + (–I0)2

2

= 0.5I0 5

2

= I0 5

2

4 Root-mean-square and peak values

The voltage of the mains in Hong Kong, 220 V, is the r.m.s. value.

& Vrms = V0

2 Irms =

I0

2

It can be derived mathematically that for sinusoidal a.c.

Example 3

The r.m.s. value of the mains supply in Hong Kong = 220 V.

Peak value of alternating voltage = ?

= 311 V

By Vrms = , V0

2V0 = Vrms 2

= 220 2

by mulimeter

by C.R.O.

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

Irms = I0

2

= 15

2

Check-point 3 – Q2

Find the max. voltage from an a.c. mains of 110 V.

= 156 V

Max. voltage = Vrms 2

= 110

2

Experiment:

Bicycle generator , 自行車發電機

Do Practice 6.1

Handed on 09/04/2013

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