MODUL • Fizik TINGKATAN 5
Gerakan magnet Gerakan magnet Motion of a magnet Motion of a magnet Gerakan magnet Gerakan magnet Motion of a magnet Motion of a magnet
change of magnetic flux or the rate of cutting of the magnetic flux
perubahan fluks magnet atau kadar pemotongan fluks magnet
HUKUM-HUKUM ARUHAN ELEKTROMAGNET (Klasifikasi) / LAWS OF ELECTROMAGNETIC INDUCTION (Classifying)
Hukum Faraday Faraday’s Law Menyatakan bahawa magnitud d.g.e teraruh berkadar langsung dengan kadar States that the magnitude of the induced e.m.f. is directly proportianal to the rate of meningkatkan kelajuan gerakan relatif antara magnet dan solenoid increasing the speed of relative motion between the magnet and solenoid
Gerakan magnet Gerakan magnet Motion of a magnet Motion of a magnet Gerakan magnet Gerakan magnet Motion of a magnet Motion of a magnet Arus aruhan atau d.g.e aruhan boleh ditingkatkan melalui: The induced current or induced e.m.f can be increased by: meningkatkan kekuatan medan magnet increasing the strength of the magnet field menambahkan bilangan lilitan pada solenoid increasing the number of turns of solenoid
Hukum Aruhan Elektromagnet Laws of Electromagnetic Induction (a) (b) (c) UNIT 3
UNIT 3
menentang perubahan fluks magnet yang menghasilkannya
yang menentang gerakan Magnet dimasukkan Magnet is put in Magnet dimasukkan Magnet is put in Magnet dikeluarkan Magnet is taken out Magnet dikeluarkan Magnet is taken out
Hukum Lenz Lenz’s Law Menyatakan bahawa arus aruhan terhasil sentiasa mengalir pada arah yang State that an induced current always flows in such a direction so as to oppose the magnet fluks magnet masuk ke dalam solenoid itu. Kutub selatan (S) dihasilkan pada hujung solenoid menentang kutub selatan (S) magnet yang bergerak masuk Induced current produces a magnetic flux that opposes the motion of the magnet entering the solenoid. South pole (S) is produced at the end of the
menghasilkan change of magnetic flux producing this induced current
Video
teraruh
Arus ke solenoid.
(a) (b)
Hukum Lenz
Lenz's Law
https://goo.gl/2oEQSM
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MODUL • Fizik TINGKATAN 5
Gerakan relatif untuk menghasilkan arus teraruh
Relative motion to produce induce currents
Mekanisme penghasilan arus teraruh atau d.g.e teraruh:
Mechanism of the production of induced current or induced e.m.f.:
Apabila gerakan relatif antara konduktor dan magnet dilakukan, konduktor akan memotong garis medan magnet dan mengubah
fluks magnet. / Whenever there is a relative motion between a conductor and a magnet, the conductor cuts across the magnetic lines of
force and changes the magnetic flux.
Rajah menunjukkan arus
Rajah Gerakan relatif
Diagram Relative motion teraruh dihasilkan
Diagram shows the induced current occurs
Solenoid Arus aruhan dihasilkan apabila:
Solenoid Induced current is produced when:
(a) Gerakan magnet ke dalam
Magnet bar solenoid yang pegun , atau
Bar magnet gerakan magnet melintasi wayar yang
pegun. / Moving the magnet into a
Galvanometer berpusat sifar stationary solenoid
Zero centred galvanometer or moving
the magnet over a stationary wire. Galvanometer akan terpesong.
The galvanometer will show a deflection.
(b) Gerakan wayar / solenoid
melintasi magnet pegun.
Moving the wire / solenoid over
Galvanometer a stationary magnet. UNIT 3
berpusat sifar Wayar kuprum
Zero centred Copper wire (c) Gerakan magnet dan wayar / solenoid
UNIT 3
galvanometer
dalam arah bertentangan.
Moving magnet and wire / solenoid in
Magnet the opposite directions.
magnadur
Magnadur (d) Gerakan magnet dan wayar / solenoid
magnet dalam arah yang sama tetapi pada laju
Dening berbeza
besi .
bentuk-U Moving the magnet and wire /
Iron yoke solenoid in the same direction but at
different speeds.
Arus teraruh dihasilkan apabila: Galvanometer akan terpesong.
Induced current is produced when: The galvanometer will show a deflection.
P Q (a) Menutup dan membuka suis S. P Q
Closing and opening switch S.
S (b) Melaraskan reostat R . S
Adjusting rheostat R .
R (c) Gerakan gegelung P mendekati atau R
menjauhi gegelung Q.
Moving the coil P nearer to or further
from the coil Q.
(d) Menggantikan bateri dengan bekalan
kuasa arus ulang-alik dan suis S
ditutup.
Replacing the battery with a.c. power
supply and close the switch.
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MODUL • Fizik TINGKATAN 5
Galvanometer Arus teraruh dihasilkan apabila: Galvanometer akan terpesong.
Magnet bar berpusat sifar Induced current is produced when: The galvanometer will show a deflection.
Bar magnet Zero centred
galvanometer
Ladung mengayunkan bandul di
Bob dalam medan magnet.
oscillating the pendulum bob in
a magnetic field.
Wayar kuprum
Copper wire
Magnet bar
Bar magnet
Konduktor
Conductor
Eksperimen Eksperimen untuk menentukan magnitud arus teraruh
Experiment An experiment to determine the magnitude of an induced current
Magnitud arus teraruh bergantung kepada laju gerakan relatif antara magnet dan konduktor.
Inferens The magnitude of an induced current depends on the speed of the relative motion between the conductor and
Inference
magnet. UNIT 3
Laju gerakan relatif antara magnet dan konduktor bertambah (diukur melalui ketinggian magnet di atas
UNIT 3
solenoid), maka magnitud arus teraruh bertambah.
Hipotesis
Hypothesis As the speed of the relative motion between the conductor and magnet increases (indicated by height of the magnet
above the solenoid), the magnitude of the induced current also increases.
Tujuan Mengkaji hubungan antara laju gerakan relatif antara magnet dan konduktor dengan magnitud arus teraruh.
Aim To investigate the relationship between the speed of the relative motion between the conductor and magnet and the
magnitude of an induced current.
Pemboleh ubah dimanipulasikan: / Manipulated variable:
Laju gerakan relatif antara konduktor dan magnet. Laju ini diwakili oleh ketinggian magnet di atas solenoid, H,
sebelum magnet dilepaskan. / The speed of the relative motion between the conductor and magnet. This speed is
Pemboleh ubah represented by the height, H, of the magnet above the solenoid, before it is released.
Variables Pemboleh ubah bergerak balas: / Responding variable:
Magnitud arus teraruh / The magnitude of the induced current
Pemboleh ubah dimalarkan: / Constant variable:
Bilangan lilitan solenoid dan kekuatan magnet / Number of turns of solenoid and the strength of the magnet
Senarai radas dan
bahan Galvanometer berpusat sifar yang sensitif, solenoid, magnet bar dan pembaris.
List of apparatus Sensitive zero-centre galvanometer, solenoid, bar magnet and ruler.
and materials
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MODUL • Fizik TINGKATAN 5
Magnet bar
Bar magnet
Susunan radas Pembaris Solenoid
Arrangement of the Ruler Solenoid
apparatus
Paip PVC (atau kadbod)
PVC (or cardboard) pipe
1 Ketinggian magnet di atas solenoid diukur dengan pembaris, H = 2.0 cm.
The height of the magnet above the solenoid is measured by a ruler, H = 2.0 cm.
2 Magnet dijatuhkan ke dalam solenoid dan bacaan galvanometer dicatatkan = I.
The magnet is dropped into the solenoid and the reading of the galvanometer is recorded = I.
Prosedur
Procedure 3 Eksperimen diulangi dengan ketinggian magnet di atas solenoid yang berbeza,
H = 4.0 cm, 6.0 cm, 8.0 cm dan 10.0 cm.
The experiment is repeated with different heights of the magnet above the solenoid, H = 4.0 cm, 6.0 cm,
8.0 cm and 10.0 cm.
Penjadualan data Ketinggian magnet / Height of magnet, H / cm 2.0 4.0 6.0 8.0 10.0 UNIT 3
Tabulation of the
data Arus teraruh / Induced current, I / μA
UNIT 3
Arus teraruh / Induced current
I / μA
Analisis data
Analysis of the data
Ketinggian magnet
Height of magnet
0
H / cm
Aplikasi Aruhan Elektromagnet
Applications of Electromagnetic Induction
Penjana Arus Terus / D.C. Generator Penjana Arus Ulang-alik / A.C. Generator
Magnet kekal Putaran Putaran
Permanent magnet Rotation Rotation Magnet kekal
Permanent magnet
C D C D
S S
N A N A
B B
Gegelung satah/angker Komutator Berus karbon Gegelung satah/angker
Commutator
Coil Current} Litar luar Carbon brush Coil Gelang } Litar luar
Berus karbon Q P Arus External Galvanometer gelincir External
Carbon brush circuit Galvanometer Slip rings circuit
Galvanometer
Galvanometer
A A B B A A 105 A B B © Nilam Publication Sdn. Bhd.
A
A
A
P
P
Q P Q Q P P Q A Q P A Q Q P P Q B Q P B Q P B B A A A A B B
B B A A
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B B A A B B B B A A B B
MODUL • Fizik TINGKATAN 5
A A A B B B A A A A A B B B A A
A
B
A
A
B
A
A
A
P A Q
P
Q P Q Q P P A Q Q P P Q Q P P B Q Q P B Q P P B A B A A B A B B
B Q Q
P P
P
P B Q
A Q Q
Q Q
P P
P A
P
Q
Q
B A
A B
A A
B
B B B A A A
A
B
B B B A A A B B B B B A A A B B
A
A
B
B
B
B
B
B
(a) (b) (c) (d) (e) (a) (b) (c) (d) (e)
d.g.e. = 0 d.g.e. = d.g.e. = 0 d.g.e. = d.g.e. = 0 d.g.e. = 0 d.g.e. = d.g.e. = 0 d.g.e. = d.g.e. = 0
e.m.f. = 0 maksimum e.m.f. = 0 maksimum e.m.f. = 0 e.m.f. = 0 maksimum e.m.f. = 0 maksimum e.m.f. = 0
e.m.f. = e.m.f. = e.m.f. = e.m.f. =
maximum maximum maximum maximum
d.g.e teraruh d.g.e teraruh
induced e.m.f. induced e.m.f.
Bilangan
putaran Bilangan
Number of putaran
0 rotations 0 Number of
rotations
(a) Apabila angker/gegelung satah dalam keadaan menegak, (a) Bermula dengan gegelung satah/angker dalam keadaan
sisi BC dan AD bergerak selari dengan garis medan magnet. menegak, sisi BC dan AD bergerak dalam keadaan selari
Maka, tiada pemotongan garis medan magnet. Maka, tiada dengan garis medan magnet. Maka, tiada pemotongan garis
arus aruhan dihasilkan. medan magnet. Maka, tiada arus aruhan dihasilkan.
When the plane of the coil is in a vertical position, the sides Starting with the plane of the coil in the vertical position, the
sides BC and AD move parallel with the magnetic field lines.
BC and AD move parallel with the magnetic field lines. There is no cutting of magnetic lines of force. Therefore, no
There is no cutting of magnetic lines of force. Therefore, no induced current is produced. UNIT 3
induced current is produced.
UNIT 3
(b) Dalam keadaan mendatar, BC bergerak ke atas dan
(b) Dalam keadaan mendatar, BC bergerak ke atas dan AD AD bergerak ke bawah. Sisi BC dan AD memotong
bergerak ke bawah. Sisi BC dan AD memotong garis garis medan magnet
pada sudut tegak. Oleh itu,
medan magnet pada sudut tepat. Oleh itu, arus teraruh yang arus teraruh dan d.g.e. teraruh yang maksimum dihasilkan.
maksimum dihasilkan serta d.g.e. maksimum In the horizontal position, BC moves upwards and AD moves
diaruhkan. / In the horizontal position, BC moves upwards and downwards. The sides BC and AD cut the magnetic field lines
AD moves downwards. The sides BC and AD cut the magnetic at right angles. Hence, the induced current produced is
field lines at right angles. Hence, the induced current produced maximum and the induced e.m.f. is maximum.
is maximum and the induced e.m.f. is maximum.
(c) Selepas keadaan ini, arus teraruh (d.g.e. teraruh) berkurangan
(c) Selepas keadaan ini, arus teraruh (d.g.e. teraruh) berkurangan sehingga ke sifar, apabila angker/gegelung satah dalam
sehingga ke sifar apabila angker/gegelung satah dalam keadaan menegak semula kerana tiada
keadaan menegak semula kerana tiada pemotongan garis pemotongan garis medan magnet.
medan magnet. After this position, the current starts to decrease until it is zero
After this position, the current starts to decrease until it is zero
when the plane of the coil is in a vertical position because the when the plane of the coil is in a vertical position
magnetic field lines are not cut. because the magnetic field lines are not cut.
(d) Arah arus di dalam litar luar tidak berubah. (d) Proses ini diulang.
The direction of the current in the external circuit does not The process is repeated.
change.
(e) Arah arus di dalam litar luar berubah-ubah setiap kali
(e) Proses ini diulang. gegelung/angker melepasi keadaan menegak.
The process is repeated. The direction of the current in the external circuit changes each
time the coil passes the vertical position.
(f) Arus dalam litar luar ialah arus terus (iaitu,
arus yang mengalir dalam satu arah). (f) Arus dalam litar luar ialah arus ulang-alik .
The current in the external circuit is a direct The current in the external circuit is an alternating
current (that is, the current flows in one direction). current.
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Teknik Menjawab [Format Kertas 2 : Perbandingan]
Answering Technique [Paper 2 Format : Comparison]
Rajah (a) menunjukkan seorang budak perempuan sedang mengayuh basikal di jalan rata pada waktu malam. Rajah (b) menunjukkan
basikal tersebut dikayuh mendaki bukit yang curam pada waktu malam. Rajah (c) menunjukkan reka bentuk dinamo basikal tersebut.
Diagram (a) shows a girl cycling on flat roads at night. Diagram (b) shows the bike cycled climbing a steep uphill road. Diagram (c) shows
design of the bicycle’s dynamo.
Magnet kekal Pencengkam roda
Permanent Wheel grip
Malap magnet
Dim Teras besi
Iron core
Gegelung dawai
Cerah
Bright Wire coil
Punca bertebat
Insulated source
Jalan rata Jalan mendaki bukit Dinamo (c)
Flat road Climb uphill road Dynamo (c)
(a) (b)
(a) Lengkapkan jadual di bawah dengan membandingkan aspek-aspek yang dinyatakan dalam jadual:
Complete the table below by comparing the aspects mentioned:
Aspek / Aspect Rajah (a) / Diagram (a) Rajah (b) / Diagram (b)
Laju kayuhan basikal Lebih laju Kurang laju UNIT 3
The cycling speed of the bike Higher speed Lower speed
UNIT 3
Kadar pemotongan fluks magnet oleh gegelung dawai dinamo Lebih besar Lebih kecil
The cutting rate of magnetic flux by the coil wire of dynamo Larger Smaller
Keadaan nyalaan mentol Lebih cerah Kurang cerah
Lighting condition of bulb Brighter Dimmer
Arus elektrik yang terhasil Lebih besar Lebih kecil
Current produced Larger Smaller
(b) Hubung kaitkan: / Relate:
(i) laju kayuhan basikal dan keadaan nyalaan mentol. / the cycling speed of bike and the lighting condition of bulb.
Semakin bertambah laju kayuhan basikal, semakin cerah nyalaan mentol.
As the cycling speed of bike increases, the bulb lights up brighter.
(ii) kadar pemotongan fluks magnet oleh gegelung dawai dinamo dan arus elektrik yang terhasil.
the cutting rate of magnetic flux by the coil wire of dynamo and current produced.
Apabila kadar pemotongan fluks magnet oleh gegelung dawai dinamo semakin bertambah, arus elektrik yang
terhasil semakin bertambah.
As the cutting rate of magnetic flux by the coil wire of dynamo increases, the current produced increases.
(c) Berdasarkan jawapan-jawapan anda di (b)(i) dan (b)(ii), deduksikan
Based on your answers in (b)(i) and (b)(ii), deduce
(i) konsep fizik / physics concept: Aruhan elektromagnet / Electromagnetic induction
(ii) hukum fizik / laws of physics: Hukum Lenz / Lenz’s law
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MODUL • Fizik TINGKATAN 5
Teknik Menjawab [Format Kertas 2 : Kefahaman / Esei Pendek]
Answering Technique [Paper 2 Format : Comprehension / Short Essay]
Dinamo basikal / Bicycle’s dynamo
Terangkan bagaimana dinamo menyalakan lampu basikal. • Magnet kekal diputarkan.
Explain how the dynamo lights up the bike’s bulb.
[4 markah / marks] The permanent magnet is rotated.
• Fluks magnet dipotong oleh gegelung dawai secara gerakan relatif.
Magnet kekal Pencengkam roda The magnetic flux is cut by the coil wire through the relative motion.
Permanent Wheel grip
magnet • Arus aruhan terhasil melalui aruhan elektromagnet.
Teras besi
Kotak dinamo Iron core The induced current produced through the electromagnetic induction.
Dynamo box Gegelung dawai
Wire coil • Tenaga kinetik → Tenaga elektrik → Tenaga cahaya
Punca bertebat Kinetic energy → Electric energy → Light energy
Insulated source
Teknik Menjawab [Format Kertas 2 : Bahagian B] UNIT 3
Answering Technique [Paper 2 Format : Part B]
UNIT 3
Cadang dan terangkan pengubahsuaian untuk menjadikan dinamo berfungsi secara lebih berkesan.
Suggest and explain modifications to make the dynamo work more efficiently.
Cadangan / Suggestion Sebab / Reason
Kekuatan magnet kekal: Menghasilkan arus yang lebih besar
Strength of permanent magnet:
Lebih kuat / Stronger [M1] Produce larger induced current [M2]
Bahan teras: Mudah dimagnet dan mudah dinyah-magnetkan
Material of core:
Besi lembut / Soft iron [M3] Easier to be magnetised and easier to be demagnetised [M4]
Bilangan lilitan gegelung dawai: Menambahkan kadar pemotongan fluks magnet
The number of turns of coil wire:
Lebih banyak / More [M5] To increase the cutting rate of magnetic flux [M6]
Luas keratan rentas / diameter dawai untuk membuat gegelung: Rintangan lebih kecil
Cross sectional area / diameter of wire of coil:
Lebih besar / Larger [M7] Smaller resistance [M8]
Bahan yang sesuai untuk gegelung dawai: Rintangan lebih kecil
Suitable material of coil wire:
Kuprum / Copper [M9] Smaller resistance [M10]
Saiz / diameter pencengkam roda: Menambahkan kadar pemotongan fluks magnet
Size / diameter of wheel gripper:
Lebih kecil / Smaller [M11] To increase the cutting rate of magnetic flux [M12]
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MODUL • Fizik TINGKATAN 5
Perbandingan antara Arus Terus dan Arus Ulang-alik
Comparison between Direct Current and an Alternating Current
Arus terus / Direct current Arus ulang-alik / Alternating current
Maksud: / Meaning: Maksud: / Meaning:
Arus yang mengalir dalam satu arah yang tetap sahaja. Arus yang mengalir pergi dan balik dalam arah berlawanan
secara berkala.
The current flowing in one fixed direction only. The current flowing to and from in opposite directions periodically.
Menghasilkan medan magnet yang tetap pada konduktor elektrik. Menghasilkan medan magnet yang berubah-ubah pada konduktor
Generates fixed magnetic field on the electrical conductor. elektrik. / Generates varying magnetic field on the electrical conductor.
Boleh mengalir melalui perintang tetapi tidak boleh mengalir Boleh mengalir melalui perintang dan kapasitor.
melalui kapasitor. Can flow through the resistor and capacitor.
Can flow through the resistor but cannot flow through the capacitor.
Sumber: bateri atau sel kering / Source: battery or dry cell Sumber: Bekalan kuasa TNB (Dijana dengan frekuensi 50 Hz)
Source: TNB power supply (Generated with a frequency of 50 Hz )
Simbol / Symbol :
Simbol / Symbol :
Sumber a.u.
Sumber a.t. / d.c. source a.c. source
I (A) UNIT 3
I (A)
I (A) I (A) t (s)
UNIT 3
t (s)
t (s) t (s)
Magnitud arus adalah tetap, Magnitud dan arah arus berubah,
Arus dari bateri Arus dari penjana a.t. arah berlawanan arus dari penjana a.u
Current from battery Current from d.c. generator The magnitude of current is Magnitude and direction of
fixed, opposite direction current are changing, current from
a.c. generator
Latihan / Exercise
1 Rajah menunjukkan arus ulang-alik di mana magnitud berubah dengan masa.
The diagram shows an alternating current with a magnitude that changes with time.
I / A
(a) Apakah arus puncak?
5.0 What is the peak current?
(b) Apakah tempoh arus ulang-alik?
What is the period of the alternating current?
0 t / s
0.02 0.04 0.06 0.08 0.10 (c) Apakah frekuensi arus ulang-alik?
What is the frequency of the alternating current?
–5.0
Penyelesaian / Solution
(a) 5 A (daripada graf / from the graph)
(b) 0.04 s (daripada graf / from the graph)
1
(c) f = ——— = 25 Hz
0.04 s
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MODUL • Fizik TINGKATAN 5
Transformer
3.4 Transformers
Alat yang digunakan untuk meningkatkan atau menurunkan beza Teras besi lembut berlapis
Transformer keupayaan bekalan arus ulang-alik berdasarkan prinsip aruhan Laminated soft iron core
Transformers increases
elektromagnet. / Electrical device which or decreases an
alternating current based on the principle of electromagnetic induction.
` Terdiri daripada dua gegelung wayar yang dililitkan berasingan pada
teras besi lembut berlapis. / Consists of two coils of wire wound round
separately on a laminated soft-iron core.
` Gegelung yang disambungkan kepada voltan input dinamakan
Struktur gegelung primer . / The coil connected to the input voltage is
Structure Gegelung
called the primary coil . Coil
` Gegelung yang disambungkan kepada voltan output dinamakan Nota / Note:
gegelung sekunder . • Frekuensi beza keupayaan
The coil connected to the output voltage is called the secondary coil . sekunder, V , adalah sama
s
dengan frekuensi beza
keupayaan primer, V .
p
The frequency of the secondary
voltage, V is the same as that of
s
the primary voltage, V .
p
Simbol 240 V 24 V • Magnitud beza keupayaan UNIT 3
Symbol sekunder, V bergantung kepada
UNIT 3
s
kadar bilangan lilitan gegelung
primer dan sekunder.
Gegelung primer Gegelung sekunder The magnitude of the secondary
Primary coil Secondary coil voltage, V depends on the ratio
s
of the number of turns of the
• Transformer bekerja berdasarkan prinsip aruhan elektromagnet. primary and secondary coils.
A transformer works on the principle of electromagnetic induction.
• Apabila beza keupayaan ulang-alik, V , dibekalkan ke gegelung primer,
p
arus ulang-alik mengalir melalui gegelung.
When an a.c. voltage, V , is applied to the primary coil of the transformer,
p
an alternating current flows through the coil.
• Teras besi lembut dimagnetkan. / The soft-iron core is magnetised.
• Arus ulang-alik mempunyai arah dan magnitud yang berubah. Oleh itu,
Prinsip terdapat perubahan arah dan magnitud medan magnet.
kerja An alternating current has varying directions and magnitudes.
Working So there is a change in the direction and magnitude of magnetic field.
principle
• Teras besi lembut membekalkan medan magnet yang berubah-ubah arah
dan magnitud dalam gegelung sekunder. / The soft-iron core provides a
varying magnetic field in the secondary coil.
• Medan magnet yang berubah-ubah juga wujud dalam gegelung
sekunder. / A varying magnetic field also occurs in the secondary coil.
• Medan magnet yang berubah-ubah akan menghasilkan
arus aruhan dan d.g.e ulang-alik teraruh, V , dalam
s
gegelung sekunder. / The varying magnetic field will produce an
induced current and alternating voltage, V in the secondary coil.
s
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MODUL • Fizik TINGKATAN 5
Kuasa / Power
Dalam transformer unggul, tiada tenaga yang hilang semasa proses Nota/Note:
pemindahan beza keupayaan. Mengapakah transformer tidak
In an ideal transformer, there is no energy loss during the process of boleh berfungsi dengan bekalan
transforming the voltage. kuasa arus terus?
Why the transformer does not work
Oleh itu, kuasa output = kuasa input with a d.c. power supply?
Hence, output power = input power • Jika beza keupayaan arus terus
I V = I V
s s p p digunakan pada gegelung
Di mana / where, primer, beza keupayan output
I = Arus sekunder / Secondary current pada gegelung sekunder akan
s
I = Arus primer / Primary current menjadi sifar.
p
Ciri-ciri Beza keupayaan sekunder / Secondary potential difference If d.c. voltage is applied to the
Characteristics V = primary coil of the transformer,
s
V = Beza keupayaan primer / Primary potential difference the output voltage at the
p
secondary coil will be zero.
Kecekapan / Efficiency
• Ini adalah kerana bekalan
Dalam transformer sebenar, sebahagian tenaga akan hilang akibat dari: kuasa arus terus membekalkan
In real transformers, some energy will be lost due to: arus mantap dalam gegelung
(a) kesan pemanasan dalam gegelung dawai primer di mana ia mempunyai
heating effect in the coils magnitud tetap dan arah yang
(b) kesan arus pusar yang teraruh dalam teras besi tetap.
eddy currents which are induced in the iron core This is because a d.c. power
(c) memagnetkan dan menyahmagnetkan teras besi supply gives a constant current in
magnetisation and demagnetisation of iron core primary coil which has constant
(d) kebocoran garis medan magnet (fluks magnet) magnitude and direction.
leakage of magnetic field lines (magnetic flux leakage) • Oleh itu, medan magnet yang
dihasilkan di dalam teras besi UNIT 3
Kuasa output lembut mempunyai magnitud
UNIT 3
Kecekapan = Kuasa Input × 100% dan arah yang tetap.
Kecekapan So the magnetic field produced
Efficiency in the soft iron core has constant
Efficiency = Output power × 100% magnitude and direction.
Input power
• Teras besi lembut tidak
membentuk garis medan
=
N s V s magnet berubah-ubah di dalam
N p V p gegelung sekunder.
Di mana / where The magnet does not create a
changing magnetic flux in the
N = Bilangan lilitan gegelung sekunder secondary coil.
s
Rumus = Number of turns in the secondary coil
Formula N = Bilangan lilitan gegelung primer
p
= Number of turns in the primary coil
V = Beza keupayaan sekunder
s
= Secondary potential difference
V = Beza keupayaan primer
p
= Primary potential difference
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MODUL • Fizik TINGKATAN 5
JENIS TRANSFORMER (Klasifikasi)
TYPES OF TRANSFORMERS (Classifying)
Gegelung sekunder
Rajah / Diagram disambung kepada
voltan output Rajah / Diagram
Secondary coil is
connected to output
voltage
Formula:
N V
S
—– = —–
S
N P V P
V > V V < V
S P S P
Transformer UNIT 3
Transformer Injak Turun
UNIT 3
Injak Naik Gegelung primer Step-down
Step-up disambung kepada transformer
transformer voltan input
Primary coil is
connected to input
voltage
N > N P N < N P
S
S
Dua gegelung wayar
dililit berasingan pada
teras besi lembut
berlamina / Two coils
of wire wound round
separately on a laminated
soft-iron core
Video
Transformer
Transformer
https://goo.gl/ycstGS
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MODUL • Fizik TINGKATAN 5
Faktor yang mempengaruhi kecekapan transformer dan cara untuk meningkatkan kecekapannya
Factors that affect the efficiency of a transformer and ways to improve the efficiency
Punca kehilangan tenaga Cara mengatasi Sebab
Cause of energy loss Way to overcome Reason
Rintangan gegelung : Gunakan dawai kuprum tebal • Rintangan lebih rendah.
The resistance of coil : sebagai gegelung. Lower resistance.
• kesan pemanasan / the effect of heating Use thick copper wire as
• menghasilkan haba / heat is produced coil. • Mengurangkan penghasilan
• E = Q = I Rt haba .
2
To reduce the heat produce.
Kesan histerisis / Hysterisis effect Gunakan teras besi lembut. Besi lembut lebih mudah dimagnetkan dan
• Kehilangan tenaga semasa proses Use soft iron core. lebih mudah dinyah-magnetkan.
pemagnetan dan penyah-magnetan. Soft iron is easy to be magnetised and easy to
Loss of energy during the process of be demagnetised.
magnetisation and demagnetisation.
• Tenaga hilang dalam bentuk haba.
Energy is lost in form of heat.
Kebocoran fluks magnet . (i) Reka bentuk teras : Sambungan fluks magnet lebih baik antara
Design of core:
The leakage of magnetic flux . gegelung primer dan gegelung sekunder.
• Tenaga elektrik hilang kerana kehilangan (ii) Melilitkan gegelung sekunder di Better linkage of magnetic flux between the
sebahagian fluks magnet yang terhasil di atas gegelung primer. primary coil and the secondary coil.
gegelung primer. Wind the secondary coil on the top of
Electrical energy is lost due to loss a part the primary coil. UNIT 3
of the magnetic flux in the primary coil. (iii) Mengurangkan jarak antara
UNIT 3
• Menyebabkan d.g.e teraruh di gegelung gegelung sekunder dan gegelung
sekunder menjadi lebih kecil. primer.
Causing induced e.m.f in the secondary Reduce the distance between the
coil becomes smaller.
primary coil and the secondary coil.
Arus pusar dalam teras. Teras berlamina Menghasilkan haba arus
Eddy currents in the core. digunakan. pusar.
• Pemutaran elektron dalam teras Laminated core is used. To heat the eddy currents.
disebabkan medan magnet yang berubah-
ubah. / Rotation of electrons in the core as
magnetic field changing.
Latihan / Exercise
1 Bilangan lilitan dalam gegelung primer dan gegelung sekunder pada transfomer masing-masing ialah 50 dan 250. Apakah beza
keupayaan output apabila transformer menggunakan bekalan kuasa 12 V?
The number of turns in the primary and secondary coil of a transformer are 50 and 250 respectively. What is the output voltage when
the transformer is using a voltage of 12 V?
Penyelesaian / Solution
Diberi / Given N = 250 lilitan / turns, N = 50 lilitan / turns, V = 12 V
s p p
N s V s
=
N p V p
V = N s V p
s
N p
250 lilitan / turns × 12 V
= = 60 V
50 lilitan / turns
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MODUL • Fizik TINGKATAN 5
2 Rajah menunjukkan 12 V, 36 W mentol menyala dengan 4 Rajah menunjukkan 12 V, 48 W mentol menyala dengan
kecerahan normal apabila disambungkan kepada bekalan kecerahan normal apabila disambungkan kepada bekalan
kuasa 240 V melalui transformer. Bilangan lilitan gegelung kuasa 240 V melalui transformer.
primer ialah 500. The diagram shows a 12 V, 48 W bulb light up with
The diagram shows a 12 V, 36 W bulb light up with normal normal brightness when it is connected to a 240 V mains
brightness when it is connected to a 240 V mains supply supply through a transformer.
through a transformer. The number of turns of the primary 0.3 A
coil is 500.
240 V 12 V, 36 W 240 V 12 V, 48 W
Berapakah bilangan lilitan gegelung sekunder? N = 200
s
What is the number of turns of the secondary coil? Hitungkan / Calculate
Penyelesaian / Solution (a) bilangan lilitan gegelung primer.
the number of turns of the primary coil.
Diberi / Given V = 12 V, V = 240 V, N = (b) arus dalam gegelung sekunder.
p
p
s
500 lilitan / turns the current in the secondary coil.
N s = V s (c) kecekapan transformer.
N p V p the efficiency of the transformer.
12 V
N s = Penyelesaian / Solution
500 lilitan / turns 240 V
12 V × 500 lilitan / turns Diberi / Given V = 240 V, N = 200 lilitan / turns,
p
s
N =
s 240 V V = 12 V, I = 0.3 A
s
p
= 25 lilitan / turns UNIT 3
(a) N s = V s
N p V p
UNIT 3
3 Transformer injak-turun disambungkan ke bekalan kuasa 200 lilitan / turns = 12 V
240 V dan memberikan kuasa 90 W kepada komputer N p 240 V
riba, 30 V. 240 V × 200 lilitan / turns
KBAT [Andaikan transformer adalah unggul.] N = 12 V
p
A step-down transformer connected to 240 V mains power N = 4 000 lilitan / turns
supply delivers 90 W of power at 30 V to a notebook computer. p
[Assume that the transformer is ideal.] (b) V = 12 V, P = 48 W
Hitungkan / Calculate s P s s 48 W
(a) arus dalam gegelung sekunder. \ I = = 12 V = 4.0 A
s
the current in the secondary coil. V s
(b) arus dalam gegelung primer. (c) Kuasa output / Output power = 48 W
the current in the primary coil.
Penyelesaian / Solution Kecekapan / Efficiency = P output × 100%
Diberi / Given V = 240 V, P output = 90 W, V = 30 V P input
s
p
48 W
(a) P = I V (b) I V = I V = × 100%
output s s p p s s I p V p
90 W = I × 30 V I × 240 V = 3 A × 30 V 48 W
s p 90 W = 0.3 A × 240 V × 100%
I = 90 W I = 240 V = 66.67%
p
s
30 V
I = 3 A = 0.375 A
s
atau
Kuasa input = Kuasa output
Input power = Output power
Jadi, / Therefore,
I V = 90 W
p
p
\ I = 90 W
p 240 V
I = 0.375 A
p
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MODUL • Fizik TINGKATAN 5
Penjanaan dan Penghantaran Tenaga Elektrik
3.5 The Generation and Transmission of Electricity
Pelbagai kaedah penghasilan tenaga elektrik dan kesan kepada persekitaran (kebaikan dan keburukan)
Various ways of generating electricity and their effects on the environment (advantages and disadvantages)
Jenis Kebaikan Keburukan
Type Advantages Disadvantages
1 Stesen janakuasa terma / Thermal power stations (i) Boleh (i) Pencemaran udara :
membangunkan
Gas buangan panas sebuah Pembakaran bahan api menghasilkan
Hot waste gases Talian penghantaran asap, habuk dan asid toksik di udara.
Transmission lines
Wap air panas Stim bertekanan bandar baru Air
o
Hot water vapour tinggi pada 800 C (menggunakan pollution:
High pressure Burning of fuels produces smoke, dust and
Relau steam at 800 C arang batu,
o
bagas Gegelung petroleum) toxic acid in the air.
Furnace Turbin pemegun
Turbines Stator coils Can develop a new
town (using coal, (ii) Hujan asid :
Rotor
Rotor petroleum) Pembakaran bahan api juga
Air penyejuk Pendandang Transformer injak naik menghasilkan gas berasid seperti sulfur
dibuang semula Boiler Pam Step-up transformer (ii) Bahan buangan dioksida dan nitrogen oksida. Gas ini
Pump
ke sungai atau Penjana boleh dikitar larut dalam air dalam atmosfera untuk
laut Pam Stim tekanan Generator
Cooling water Pump rendah dari Air penyejuk daripada sungai semula (biojisim) membentuk asid yang menyebabkan
discharged back turbin atau laut Can recycle the hujan asid.
into river or sea Low pressure Cooling water from river or sea
steam from waste matter Acid rain :
turbines (biomass) Burning of fuels also produce acidic gases
(a) Bahan api seperti arang batu, petroleum, gas asli dan such as sulphur dioxide and nitrogen
uranium digunakan dalam stesen janakuasa terma. (iii) Memerlukan oxides. These gases dissolve in water in UNIT 3
Fuels such as coal, petroleum, natural gas and uranium bahan radioaktif the atmosphere to form acids which results
are used in the thermal power stations. yang kecil in acid rain.
UNIT 3
Small amount
(b) Tenaga haba dihasilkan oleh pembakaran arang batu of radioactive is (iii) Kesan rumah hijau :
atau minyak dalam relau atau dari pembelahan nukleus required Karbon dioksida berlebihan di atmosfera
uranium dalam teras reaktor nuklear. menyebabkan suhu yang lebih tinggi
Heat energy is produced by burning coal or oil in a furnace terhasil di persekitaran.
or from the fission of uranium nuclei in the core of a Greenhouse
nuclear reactor. effect:
The excessive carbon dioxide in the
(c) Air menyerap tenaga haba di dalam relau atau penukar atmosphere raises the temperature in the
haba dan bertukar menjadi stim pada tekanan tinggi. environment.
Water absorbs the heat energy in a boiler or heat-
exchanger and is changed into steam at a high pressure. (iv) Tidak boleh diperbaharui:
Bekalan terhad dan boleh habis.
(d) Stim bertekanan tinggi menukarkan tenaga haba kepada Non-renewable:
tenaga mekanikal apabila ia memutarkan turbin. Their supply is limited and they will
The high-pressure steam converts heat energy into eventually run out.
mechanical energy as it turns the turbine.
(v) Mahal. / Expensive.
(e) Generator menukarkan tenaga mekanikal kepada tenaga
elektrik. (vi) Kesan sinaran radioaktif yang
The generator converts the mechanical energy into merbahaya kepada manusia dan
electrical energy. persekitaran.
Harmful effects of radioactive radiation on
(f) Kecekapan: 30% – 35% tenaga yang disimpan di dalam humans and environment.
bahan api ditukarkan kepada tenaga elektrik.
Efficiency: 30% – 35% of the energy stored in the fuel is (vii) Berbau busuk:
transformed into electrical energy. Satu bau busuk dilepaskan daripada
biomass.
Bad odour (smell):
A bad odour is released from the biomass.
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MODUL • Fizik TINGKATAN 5
Jenis Kebaikan Keburukan
Type Advantages Disadvantages
2 Stesen kuasa hidroelektrik / Hydroelectric power station (i) Bersih dan tidak (i) Mengganggu keseimbangan
membebaskan ekosistem di persekitaran.
Talian penghantaran Air simpanan
Transmission lines Empangan Stored water bahan cemar ke Satu kawasan hutan yang
Dam
persekitaran. besar perlu dimusnahkan
Clean and does not termasuk flora dan fauna.
emit pollutants to Disturbs the equilibrium of the
Penjana the environment. ecosystem in the environment.
Generator
A large area of forest land has
(ii) Boleh to be destroyed, including flora
and fauna.
diperbaharui.
Turbin Renewable
Turbines . (ii) Mengancam / menyesarkan
Air penduduk tempatan
Water
(iii) Boleh Threatens / Displaces local
(a) Stesen hidroelektrik tidak perlu memanaskan stim. membangunkan population.
In hydroelectric stations, there is no need to heat steam at all. kawasan rekreasi.
Can develop a (iii) Perbelanjaan tinggi.
(b) Air dikumpul dalam takungan (empangan) yang tinggi dan recreation area. High cost.
mempunyai tenaga keupayaan graviti yang tinggi.
Water is collected in a high reservoir and possesses high gravitational (iv) Mengawal banjir.
potential energy.
Control flood.
(c) Apabila air itu mengalir melalui paip, tenaga keupayaan graviti ini
bertukar kepada tenaga kinetik . / When the water flows through a
pipe, its gravitational potential energy is changed to kinetic energy.
(d) Tenaga kinetik ditukar kepada tenaga elektrik semasa air yang UNIT 3
mengalir memutarkan bilah turbin. / The kinetic energy of water
UNIT 3
is changed to electrical energy when the water turns the blades of a
turbine.
3 Tenaga solar boleh berubah kepada tenaga elektrik (i) Bersih dan tidak (i) Memerlukan ruang
melalui dua kaedah: / Solar energy can be converted into membebaskan yang besar untuk
electrical energy by two methods: bahan cemar ke mengumpulkan cahaya
(i) Relau suria/Solar furnace persekitaran. matahari.
Clean and does not
emit pollutants to Requires a large
Titik fokus
Focal point the environment. area to collect the Sun’s rays.
Cahaya matahari (ii) Boleh (ii) Kecekapan penukaran
Sunlight
diperbaharui. adalah agak rendah (kira-
Renewable. kira 5%) dan harga untuk
menghasilkan sel solar agak
(iii) Percuma. mahal. / The efficiency of
Pemantul paraboloid cekung
Concave paraboloidal reflector Free. conversion is quite low (about
(a) Pemantul parabola cekung digunakan untuk memfokus 5%) and the solar cells are
dan menumpukan tenaga haba dari matahari. / A concave expensive to produce.
parabola reflector can be used to focus and concentrate the
Ke sistem pemanas air Penutup transkparensi
Transparent cover
To domestic hot water system (iii) Keamatan sinaran matahari
radiant heat energy from the sun.
Sinar matahari
Sunlight yang dikumpulkan tidak
(b) Tenaga haba digunakan untuk menghasilkan stim konsisten, dan ini bergantung
yang mempunyai tenaga kinetik dan digunakan untuk kepada cuaca dan masa.
Penebat
Panel penyerap
Insulation
Absorbing panel
memutarkan turbin bagi penjana elektrik. The intensity of Sun’s rays
Air sejuk masuk
Cold water in The heat energy is used to produce steam. The high kinetic collected is not consistent as
Pam
Pump
energy of this steam is used to drive the turbine of an electrical this depends on the weather
generator. conditions and the time of day.
Sinar matahari
Penutup kaca Sunlight
lass cover
Sesentuh
grid logam Output
Metal grid Output
contact
© Nilam Publication Sdn. Bhd. Sesentuh logam 116
Metal contact
Silikon jenis-N Silikon jenis-P
N type silicon P type silicon
03-Physic F5 2018.indd 116 26/08/2019 11:23 AM
Turbin angin
Wind turbine
MODUL • Fizik TINGKATAN 5
Jenis Titik fokus Kebaikan Keburukan
Type Focal point Advantages Disadvantages
Cahaya matahari
Ke sistem pemanas air Penutup transparensi Sunlight
To domestic hot water system Transparent cover
Sinar matahari
Sunlight
Pemantul paraboloid cekung
Penebat Concave paraboloidal reflector
Permukaan penyerap
Insulation haba hitam
Air sejuk masuk Black heat-absorbing
Cold water in Pam surface
Ke sistem pemanas air Pump Penutup transkparensi
To domestic hot water system Transparent cover
Sinar matahari
(c) Panel solar menggunakan tenaga matahari untuk menjana
Sunlight
Titik fokus
tenaga terma untuk memanaskan air di rumah.
Focal point
Panel penyerap
Penebat
Solar panels use sun energy to generate thermal energy to heat
Insulation
water at home. Absorbing panel
Cahaya matahari
Air sejuk masuk Sunlight
Cold water in Pam
(ii) Sel solar / Solar cell Pump
Sinar matahari
Penutup kaca Sunlight Pemantul paraboloid cekung
lass cover Concave paraboloidal reflector
Sesentuh
grid logam Output
Metal grid Output
Ke sistem pemanas air Penutup transkparensi
contact
To domestic hot water system Transparent cover
Sinar matahari
Sesentuh logam
Sunlight
Metal contact
Silikon jenis-N Silikon jenis-P UNIT 3
N type silicon P type silicon Panel penyerap
Penebat
Insulation Absorbing panel
UNIT 3
(a) Sel solar diperbuat daripada bahan semikonduktor yang mana
Air sejuk masuk
Pam
Cold water in
mengaruhkan d.g.e. apabila terdedah kepada cahaya matahari.
Pump
Solar cells are made of semiconductor materials which develop
an e.m.f. when exposed to sunlight.
Sinar matahari
Penutup kaca Sunlight
lass cover
(b) Sel solar digunakan dalam kalkulator, lampu, jam tangan,
Sesentuh
pemanas air dan satelit. / Solar cells are used in calculators,
Output
grid logam
lamps, wrist watches, water heaters and in satellites.
Output
Metal grid
Turbin angin
contact Wind turbine
Sesentuh logam
4 Tenaga angin / Wind energy Metal contact (i) Bersih dan tidak (i) Memerlukan kawasan
Silikon jenis-N Silikon jenis-P membebaskan
N type silicon P type silicon luas
bahan cemar ke
persekitaran. untuk membina turbin
Clean and does not angin.
emit pollutants to the Requires a large
environment. area to construct a wind
turbine.
(ii) Boleh diperbaharui.
Renewable. (ii) Kelajuan putaran
pemutar turbin tidak
Turbin angin
Wind turbine konsisten, bergantung
Tenaga kinetik angin memutarkan bilah yang dihubungkan kepada (iii) Percuma. kepada keadaan.
pemutar penjana elektrik untuk menghasilkan tenaga elektrik. Free. The speed of rotation of
The kinetic energy of wind rotates blades connected to the rotor of an the rotor is not consistent,
electrical generator to produce the electrical energy. depending on weather
conditions.
(iii) Membebaskan bunyi
bising. / Produces noise.
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MODUL • Fizik TINGKATAN 5
Jenis Kebaikan Keburukan
Type Advantages Disadvantages
5 Tenaga ombak / Wave energy (i) Bersih dan tidak (i) Memerlukan kos
membebaskan tinggi
Permukaan ledakan Keseimbangan apungan bahan cemar ke untuk
Boom membina dan mengekalkan
Balancing float persekitaran.
Ombak generator gelombang.
Waves Clean and does not high
emit pollutants to the Needs cost
environment. to build and maintain the wave
Gerakan berombak generators.
Rocking motion
(ii) Boleh diperbaharui.
Renewable. (ii) Sangat mudah rosak dan susah
dipertahan.
(iii) Percuma. Very vulnerable and difficult to
Free. protect from damage.
‘Salter duck’ ialah bentuk khas yang terapung dan bergolek ke
depan dan ke belakang apabila ombak menghentam ke atasnya.
Gerakan berombak boleh digunakan untuk memacu turbin dan
menjana arus elektrik. Kesemua penjana tenaga mempunyai
barisan 'salter duck' yang lurus menghadap ombak yang datang.
The ‘salter duck’, is a specially shaped float which rocks back and
forth as the waves strikes it. This rocking motion can be used to
drive a turbine and generates electricity. The whole wave energy
UNIT 3
generator has a row of ducks lined up facing the on coming waves.
UNIT 3
Penghantaran Tenaga Elektrik
Transmission of Electricity
(i) Tenaga elektrik dihantar pada beza keupayaan yang tinggi dan menggunakan arus ulang-alik .
Electrical energy is transmitted at a high voltage and uses an alternating current.
(ii) Transformer injak naik yang menaikkan beza keupayaan pada voltan lampau tinggi iaitu 110 000 V atau 132 000 V. Beza
keupayaan yang ditingkatkan akan menyebabkan penghantaran kuasa dengan arus kecil .
A step-up transformer which increases the voltage to as high a voltage as 110 000 V or 132 000 V. The high voltage will cause the
power to be transmitted with a small current .
(iii) Transformer injak turun digunakan untuk menurunkan beza keupayaan sebelum dihantar kepada pengguna.
Step-down transformers are used to lower the voltage before delivering to the consumers.
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MODUL • Fizik TINGKATAN 5
Tenaga yang hilang dalam kabel
penghantaran tenaga elektrik dan
kelebihan penghantaran voltan tinggi Sistem Rangkaian Grid Nasional Isu Penghantaran
The energy loss in electricity transmission The National Grid network Transmission Issues
cables and the advantage of high
voltage transmission
(i) Kesan pemanasan yang disebabkan Supergrid (275 kV • Penghantaran melalui bawah
oleh rintangan kabel menyebabkan /400 kV) tanah atau di atas tanah
kehilangan tenaga elektrik sebagai Grid Overhead or underground
(132 kV)
tenaga haba .
The heating effect due to the resistance (i) Kabel voltan lampau tinggi
adalah cara yang lebih
of the cable causes loss of electrical Stesen kuasa Transformer Transformer murah untuk menghantar
(275 kV/400 kV
Power station (25 kV/275 kV
energy as heat energy . 25 kV /400 kV) /132 kV) kuasa pada jarak yang jauh.
Industri berat Industri ringan Bandar, kampung dan Extra high voltage cables is
(ii) Apabila arus mengalir dalam Heavy industry Light industry ladang the cheapest way of sending
(11 kV)
(33 kV)
kabel, kehilangan kuasa, P, melalui Towns, villages and farms power over long distances.
(415 V/240 V)
pemanasan adalah:
When a current flows in a cable, the (ii) Untuk mengelakkan
power loss, P, through heating is: percikan elektrik, cara yang
P = I R Transformer Transformer Transformer berkesan ialah menyalut
2
(11 kV/415 V/240 V)
(33 kV/
(132 kV/
33 kV) 11 kV)
penebat pada kabel untuk
(iii) Kuasa yang hilang boleh dikurangkan • Rangkaian Grid Nasional merupakan satu sistem memastikan ruang udara
dengan: rangkaian kabel voltan lampau tinggi dalam yang besar di sekeliling
The power loss can be reduced by: satu rangkaian tertutup menghubungkan semua mereka.
(a) mengurangkan stesen kuasa utama di negara ini dengan semua To prevent sparkling, the
effective way is to insulate
rintangan kabel pengguna utama seperti rumah, pejabat, bandar- the cables to keep huge air UNIT 3
bandar dan kilang-kilang.
reducing the resistance of The National Grid Network is a network system space around them.
the cables of extra high voltage cable transmission lines
UNIT 3
(b) mengurangkan arus mengalir which connects all the major power stations in the (iii) Kabel perlu digantung pada
dalam kabel country with all the major users such as homes, tiang.
reducing the current in the cables offices, cities and factories. The cables have to be
suspended from pylons.
(iv) Oleh itu, untuk mengurangkan • Tenaga elektrik dijanakan di stesen kuasa di
rintangan kabel, wayar kabel tebal mana beza keupayaan ditingkatkan dengan • Kos kabel
digunakan, contoh: wayar kuprum injak naik Cost of cables
atau aluminium. menggunakan transformer (i) Kabel kuprum
In order to reduce resistance of the untuk mengurangkan arus yang mengalir mempunyai rintangan
cables, a thick cable is used e.g copper melalui grid. Ini mengurangkan kehilangan rendah tetapi kos yang
or aluminium. kuasa dalam kabel semasa penghantaran.
Electrical energy is generated in power stations tinggi.
(v) Kuasa yang dihantar melalui kabel where its voltage is increased by using a Copper cables
ialah P = I V. step-up transformer to reduce the power have low resistance but high
The power transmitted by the cables is loss in the cables during transmission by lowering cost.
P= I V. the current flowing through the grid.
(ii) Kabel aluminium sering
(vi) Maka, arus, I, yang mengalir melalui • Di substesen, sebelum penghantaran kepada digunakan kerana ia ringan,
kabel: pelbagai pengguna, voltan lampau tinggi rintangannya rendah dan
So, the current, I, flows in the cables: dikurangkan dengan menggunakan kosnya lebih rendah.
I = P transformer injak turun. Aluminium cables are usually
used as they are light, have
V
At sub-stations, before reaching the various low resistance and its cost
(vii) Ini bermaksud, arus, I yang reduced is low.
consumers, the high voltage is
mengalir melalui kabel adalah by using step-down transformers.
berkadar songsang dengan
beza keupayaan.
This means that the current in the cables
is inversely proportional to the
voltage.
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MODUL • Fizik TINGKATAN 5
Kebaikan Sistem Rangkaian Grid Nasional • Kebocoran cas antara kabel
Advantages of the National Grid Network System dengan bumi
Charge leakage between the cables
1 Mudah dikendalikan / Easy to manage and the earth
Penjanaan tenaga elektrik boleh dikawal dan
dilaraskan mengikut keperluan pada masa-masa (i) Untuk mengelak kebocoran
tertentu. cas, kabel aluminium elektrik
Regional control and switching centres enable power to disokong oleh menara kawat
be sent where and when it is needed. yang tinggi .
To prevent charge leakage, the
2 Pembekalan tenaga elektrik berterusan tanpa aluminium cables are supported
gangguan / Continuous supply of electricity high
by metal
Apabila kawalan serantau dan pusat pensuisan pylons.
membenarkan beberapa stesen dalam rangkaian
tertutup memutuskan rangkaiannya untuk kerja-kerja • Merbahaya jika disambar petir
penyelenggaraan, pengguna masih dapat bekalan Danger of being struck by lightning
tenaga daripada bekalan penjana lain.
While regional control and switching centres allow (i) Kabel ditebat dengan porselin
some stations and lines to be shut down for maintenance dan kaca supaya kabel
work, consumers still get energy supply from other tersebut tidak bersentuhan
generator supplies. dengan menara kawat
elektrik.
3 Mengurangkan kos penjanaan The cables are properly fixed
Reducing cost of generation with porcelain and glass
support so that the cables do
Penghantaran voltan lampau tinggi boleh not touch the pylons.
mengurangkan kuasa yang hilang dalam kabel
penghantaran. Kos penghasilan elektrik dapat • Merbahaya jika dilanggar oleh UNIT 3
dikurangkan. pesawat ringan
UNIT 3
High voltage transmission can reduce the power loss in Danger of being struck by light
transmission cables. The cost of production is reduced. aircraft
4 Mudah untuk dikawal dan diselenggara (i) Lampu dan penanda khas
Easy to control and regulate
mesti dipasang pada menara
Semasa penggunaan tenaga elektrik yang tinggi, kawat elektrik itu.
lebih banyak penjana boleh dihidupkan. Tetapi Lights and special markers
semasa permintaan pengguna kurang, sebahagian must be attached to the pylons.
penjana boleh dipadamkan.
At peak periods of electricity usage, more generators
can be switched on. During periods of low demand,
some generators can be turned off.
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MODUL • Fizik TINGKATAN 5
Latihan / Exercise
1 Stesen kuasa menjana 40 kW kuasa elektrik menggunakan rangkaian penghantaran tenaga berintangan
8 Ω. Berapakah kehilangan kuasa semasa penghantaran akibat rintangan kabel jika tenaga elektrik dihantar pada
A power station generates 40 kW of electric power using a power transmission line with resistance of 8 Ω. What is the power
KBAT dissipated due to the resistance of the transmission cables when the current is transmitted at
(a) 5 000 V (b) 20 kV
Penyelesaian / Solution
(a) Hitungkan nilai arus dalam kabel, I (b) P = I V
Calculate value of current in the cable, I 40 × 10 W = I × (20 × 10 ) V
3
3
3
Kuasa dihantar melalui kabel, P = IV I = 40 × 10 W
3
The power transmitted by the cable, P = IV 20 × 10 V
40 × 10 W = I × 5 000 V = 2 A
3
3
I = 40 × 10 W Maka, kuasa yang hilang disebabkan rintangan,
5 000 V So, the power loss due to the resistance,
= 8 A P = I R
2
Maka, kuasa yang hilang disebabkan rintangan, = (2 A) × 8 Ω
2
So, the power loss due to the resistance, = 32 W
P = I R
2
= (8 A) × 8 Ω
2
= 512 W
UNIT 3
2 Stesen kuasa menjana 80 MW kuasa elektrik pada beza keupayaan 80 kV a.u. melalui rintangan kabel 5 Ω.
A power station generates 80 MW of electric power at a voltage 80 kV through a cable of resistance 5 Ω. Determine:
UNIT 3
(a) kuasa yang hilang semasa penghantaran tenaga melalui kabel.
KBAT
the power loss in the transmission cable.
(b) peratus kuasa yang hilang. / the percentage of the power loss.
(c) kecekapan penghantaran kuasa. / the efficiency of the power transmission.
(d) pengurangan beza keupayaan di dalam kabel. / the voltage drop in the cable.
Penyelesaian / Solution
(a) Hitungkan nilai arus dalam kabel, I (c) Tenaga dihantar = Kuasa dibekal – Kuasa hilang
Calculate value of current in the cable, I Power transmitted = Power supply – Power loss
6
6
Kuasa yang dihantar oleh kabel, P = IV = 80 × 10 W – 5 × 10 W
6
The power transmitted by the cable, P = IV = 75 × 10 W
6
80 × 10 W = I × (80 × 10 V) Kecekapan / So, efficiency = 75 × 10 W × 100%
3
6
6
6
I = 80 × 10 W 80 × 10 W
80 × 10 V = 93.75%
3
= 1 000 A
(d) Pengurangan voltan / Voltage drop = IR
Oleh itu, kuasa hilang akibat rintangan, = 1 000 A × 5 Ω
So, the power loss due to the resistance = 5 000 V
P = I R
2
= (1 000 A) × 5 Ω
2
= 5 × 10 W
6
(b) Peratus kuasa hilang
Percentage of power loss
6
= 5 × 10 W × 100%
80 × 10 W
6
= 6.25%
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MODUL • Fizik TINGKATAN 5
L atihan Pengukuhan / Enrichment Exercise
1 Rajah 1 menunjukkan susunan radas untuk mengkaji 3 Rajah 3 menunjukkan corak medan magnet apabila arus
corak medan magnet yang terbentuk apabila arus elektrik mengalir dalam satu dawai lurus.
mengalir dalam konduktor. Diagram 3 shows a magnetic field pattern when currents
Diagram 1 shows an arrangement of apparatus to study the flows in a straight wire.
pattern of magnetic fields which are formed when the current
flows through conductors. Dawai
6 V 6 V Wire
P
Kadbod
Cardboard Q
Kompas
Compass
Rajah 1 / Diagram 1 Rajah 3 / Diagram 3
Corak medan magnet yang manakah terbentuk pada Antara berikut, yang manakah menunjukkan arah medan
kadbod apabila kedua-dua suis dihidupkan? magnet yang betul bagi kompas P dan Q?
Which magnetic fields pattern is formed on the cardboard Which of the following shows the correct direction of magnetic
when both switches are closed? field for compass P and Q?
A
P Q
B A
C B UNIT 3
UNIT 3
C
D
D
2 Rajah 2 menunjukkan sebatang magnet bersebelahan 4 Rajah 4 menunjukkan sebuah transformer ringkas.
dengan suatu gegelung. Petunjuk galvanometer terpesong Diagram 4 shows a simple transformer.
semasa magnet itu ditolak ke dalam gegelung. / Diagram 2 Teras besi lembut
shows a coil of wire placed next to a magnet. The galvanometer Soft iron core
pointer deflects when the magnet is pushed into the coil.
Input Output
240 V 6 V
Rajah 2 / Diagram 2 Gegelung primer Gegelung sekunder
Secondary coil
Primary coil
Tindakan yang manakah akan menambahkan pesongan Rajah 4 / Diagram 4
petunjuk galvanometer itu?
Which action will increase the deflection of the galvanometer Arus dalam gegelung primer ialah 0.1 A dan kecekapan
pointer? transformer itu ialah 60%. Kuasa output transformer itu
A Kutub magnet itu disongsangkan ialah
The magnet pole is reversed The current in the primary coil is 0.1 A and the efficiency
B Bilangan lilitan gegelung itu ditambah of the transformer is 60%. The output power of the
The number of coil is increased transformer is
C Gegelung dibuat daripada wayar yang bertebat A 6.0 W
The coil is made from insulated wire B 14.4 W
D Magnet itu ditolak perlahan-lahan ke dalam gegelung C 24.0 W
The magnet is pushed slowly into the coil D 40.0 W
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MODUL • Fizik TINGKATAN 5
5 Rajah 5 menunjukkan dua magnet, J dan K, tergantung 7 Rajah 7 menunjukkan struktur binaan motor elektrik.
bebas di kedua-dua hujung solenoid. Diagram 7 shows the structure of an electric motor.
Diagram 5 shows two magnets J and K, hang freely at both
ends of the solenoid.
N S
N S S N Rajah 7 / Diagram 7
Magnet J Magnet K Corak medan magnet yang manakah betul apabila suis
Magnet J Magnet K
lampu dihidupkan?
Which of the magnetic field patterns is correct when the motor
is switched on?
Rajah 5 / Diagram 5 A
Pemerhatian manakah yang betul apabila suis dihidupkan?
Which observation is correct when the switch is turned on?
Magnet J Magnet K
Bergerak menjauhi Bergerak mendekati B
A solenoid / Moves away solenoid / Moves
from the solenoid towards the solenoid
Bergerak menjauhi Bergerak menjauhi
B solenoid / Moves away solenoid / Moves away
from the solenoid from the solenoid C
Bergerak mendekati Bergerak mendekati UNIT 3
C solenoid / Moves solenoid / Moves
UNIT 3
towards the solenoid towards the solenoid
Bergerak mendekati Bergerak menjauhi
D solenoid / Moves solenoid / Moves away D
towards the solenoid from the solenoid
6 Rajah 6 menunjukkan sebiji mentol 24 V, 18 W menyala
pada kecerahan normal. / Diagram 6 shows a bulb 24 V, 18
W lights up with normal brightness. 8 Penghantaran elektrik disalurkan pada voltan yang sangat
Input tinggi supaya
Input 24 V, 18 W Electricity is transmitted at a very high voltage in order to
A meningkatan arus dalam kabel
240 V increase the current in the cables
Output B mengurangkan rintangan kabel
Output
reduce the resistance of the cables
Bilangan lilitan = 100 Bilangan lilitan = 10 C mengurangkan kehilangan kuasa
Number of turns = 100 Number of turns = 10 reduce power loss
D menambahkan rintangan kabel
Rajah 6 / Diagram 6 increase the resistance of the cables
Apakah yang akan berlaku kepada mentol apabila bilangan
lilitan pada gegelung sekunder ditambah kepada 50 lilitan? 9 Berapakah kehilangan kuasa dalam kabel penghantaran
What will happen to the bulb when the number of turns in apabila 50 kW dihantar melalui kabel yang berintangan
secondary coil increases to 50 turns? 4.0 Ω pada voltan 10 kV?
A Malap / Dimmer What is the power loss in a transmission cable when 50 kW
B Lebih terang / Brighter is transmitted through a cable with a resistance of 4.0 Ω at a
C Terbakar / Blown up voltage of 10 kV?
D Kecerahan tidak berubah A 5 W C 100 W
The brightness unchanged B 25 W D 200 W
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MODUL • Fizik TINGKATAN 5
S oalan Struktur / Structure Questions
1 Rajah 1.1 menunjukkan sebuah transformer unggul yang disambung kepada motor elektrik 12 V, 40 W a.u. Motor ini berfungsi
secara normal.
Diagram 1.1 shows an ideal transformer which is connected to a 12 V, 40 W a.c. electric motor. This motor functions normally.
Gegelung Gegelung
240 V a.u. primer sekunder Motor elektrik
240 V a.c. Primary Secondary M Electric motor
coil coil
N = 1 200 lilitan / turns
p
Rajah 1.1 / Diagram 1.1
(a) Nyatakan jenis transformer yang digunakan.
State the type of transformer used.
Transformer injak turun. / Step-down transformer.
(b) (i) Hitung bilangan lilitan gegelung sekunder itu. / Calculate the number of turns of the secondary coil.
N V
—– = —–
S
S
N V
P P
V
S
N = —– × N
S V P P
12 V
= ——–– × lilitan / turns
240 V
UNIT 3
= 60 lilitan / turns
UNIT 3
(ii) Hitung arus yang mengalir dalam gegelung primer itu.
Calculate the current flow in the primary coil.
Kuasa output / Output power = 40 W
\ Kuasa input / Input power = 40 W (Untuk transformer unggul / for idea transformer)
\ P = 40 W, V = 240 V
P
P
P = V I
P P P
P
\ I = —–
P
P V
P
40 W
= ——––
240 V
= 0.1667 A
= 0.17 A
(c) Rajah 1.2 menunjukkan transformer yang sama disambungkan kepada sebuah komponen elektrik dalam kotak Y untuk
menyalakan sebuah mentol a.t.
Diagram 1.2 shows the same transformer being connected to an electrical component in box Y to light up a d.c. bulb.
Y
Gegelung Gegelung
240 a.u. primer sekunder Motor elektrik
240 a.c. Primary Secondary M Electric motor
coil coil
Rajah 1.2 / Diagram 1.2
Namakan komponen elektrik dalam kotak Y. / Name the electrical component in box Y.
Diod. / Diode.
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MODUL • Fizik TINGKATAN 5
2 Rajah (a) dan Rajah (b) menunjukkan sistem penghantaran elektrik yang menghubungkan stesen penjana kuasa dan pengguna
melalui suatu jarak tertentu./ Diagram (a) and Diagram (b) show the electricity transmission system connecting the electric power
generating stations and the user through a certain distance.
Kabel aloi aluminium / Aluminium alloy cable
135 kV
Pilon Transformer Pengguna
Transformer injak-turun
Stesen penjana kuasa Pilion User
Power generator station injak-naik Step-down 18 MW
Step-up
20 MW transformer
transformer
Rajah (a) / Diagram (a)
Kabel aloi aluminium / Aluminium alloy cable
120 kV
Pilon Transformer Pengguna
Stesen penjana kuasa Transformer Pilion injak-turun User
Power generator station injak-naik Step-down 15 MW
Step-up
20 MW transformer transformer
Rajah (b) / Diagram (b)
(a) Lengkapkan jadual di bawah dengan membandingkan aspek-aspek yang dinyatakan:
Complete the table below by comparing the aspects mentioned: UNIT 3
Aspek Rajah (a) Rajah (b)
UNIT 3
Aspect Diagram (a) Diagram (b)
Voltan pada kabel penghantaran Lebih besar Lebih kecil
The voltage on the transmission cable Larger Smaller
Arus dalam kabel penghantaran
The current in the transmission cable Lebih kecil / Smaller Lebih besar / Larger
Guna formula P = IV untuk mencari I (148.15 A) (166.67 A)
Use the formula P = IV to find I
Rintangan kabel penghantaran Sama Sama
The resistance of transmission cable Same Same
Kehilangan kuasa semasa penghantaran Lebih kecil / Smaller Lebih besar / Larger
The loss of power during transmission (20 MW – 18 MW = 2 MW) (20 MW – 15 MW = 5 MW)
Kuasa yang sampai kepada pengguna (Kuasa output) Lebih besar / Larger Lebih kecil / Smaller
The power that reaches the consumer (Output power) (18 MW) (15 MW)
Kuasa input Sama Sama
The input power Same Same
Kecekapan sistem penghantaran elektrik Lebih cekap Kurang cekap
The efficiency of electricity transmission system More efficient Less efficient
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MODUL • Fizik TINGKATAN 5
(b) Hubung kaitkan: / Relate:
(i) voltan pada kabel penghantaran dan arus dalam kabel penghantaran
the voltage on the transmission cable and the current in the transmission cable
Semakin bertambah voltan pada kabel penghantaran, semakin berkurang arus dalam kabel penghantaran
As the voltage on the transmission cable increases, the current in the transmission cable decreases
(ii) arus dalam kabel penghantaran dan kehilangan kuasa semasa penghantaran
the current in the transmission cable and the loss of power during the transmission
Semakin berkurang arus dalam kabel penghantaran, semakin berkurang kehilangan kuasa semasa penghantaran
As the current in the transmission cable decreases, the loss of power during the transmission decreases
(iii) arus dalam kabel penghantaran dan kecekapan sistem penghantaran
current in transmission cable and efficiency in transmission system
Semakin berkurang arus dalam kabel penghantaran, semakin bertambah kecekapan sistem penghantaran
As current in transmission cable decreases, effiency of transmission system increase
(c) Pemboleh ubah:
Variables:
(i) Dimanipulasikan / Manipulated : Arus dalam kabel, I / Current in the cable, I
(ii) Bergerak balas / Responding : Kehilangan kuasa, P / Loss of power, P
UNIT 3
(iii) Dimalarkan / Constant : Rintangan kabel, R / Resistance of cable, R
UNIT 3
Memahami teori melalui pendekatan matematik:
Understanding the theory through the mathematical approach:
Kehilangan kuasa, P[$] = (I [$]) R[malar]
2
Power loss, P[$] = (I [$]) R[constant]
2
Kuasa output[#] = Kuasa input[malar] − Kehilangan kuasa[$]
Output power[#] = Input power[constant] – Power loss[$]
Kuasa output[#]
Kecekapan[#] = × 100%
Kuasa input[malar]
Output power[#]
Efficiency[#] = Input power[constant] × 100%
Apabila rintangan kabel, R, tetap, kehilangan kuasa, P, bergantung pada arus dalam kabel, I
When the resistance of cable, R is constant, the power loss, P depends on the current in the cable. I
Semakin berkurang arus dalam kabel, I, semakin berkurang kehilangan kuasa, P (P α I )
2
As the current in the cable, I decreases, the power loss, P decreases (P α I )
2
Semakin berkurang kehilangan kuasa, semakin bertambah kuasa output
As the power loss decreases, the output power increases
Untuk kuasa input yang tetap, kecekapan bergantung pada kuasa output
When the input power is constant, efficiency depends on the input power
Semakin bertambah kuasa output, semakin bertambah kecekapan (Kecekapan α Kuasa output)
As the input power increases, the efficiecny increasses (Efficiency α Output power) Maklumat tambahan:
Additional information:
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MODUL • Fizik TINGKATAN 5
Elektronik
4
Electronics
Objektif pembelajaran / Learning objective
• Memahami kegunaan osiloskop sinar katod. / Understanding the uses of the cathode ray osciloscope (CRO).
• Memahami diod semikonduktor. / Understanding semiconductor diode.
• Memahami transistor. / Understanding transistor.
• Menganalisis get logik. / Analysing logic gates.
Kegunaan Osiloskop Sinar Katod (OSK)
4.1 The Uses of the Cathode Ray Oscilloscope (C.R.O.)
Terminologi / Terminology Penerangan / Explanation
Pancaran termion Elektron yang dibebaskan daripada permukaan logam yang dipanaskan.
Thermionic emission The release of electrons from a heated metal surface.
Sinar katod Suatu alur elektron yang bergerak dengan laju yang tinggi dalam vakum.
Cathode ray A beam of electrons moving at high speed in a vacuum.
Sinar Katod / Cathode Ray
Pancaran termion / Thermionic emission Sinar katod / Cathode ray
Filamen Anod Vakum UNIT 4
Filament Katod Anode Vacuum
Cathode
UNIT 4
Skrin
berpendarfluor
Elektron 6 V a.u. Fluorescent screen
Electron 6 V a.c.
Alur elektron
Katod Electron beam
Cathode
Voltan lampau tinggi, VLT
Extra high tension, EHT
• Logam mempunyai elektron-elektron yang bebas bergerak
dalam struktur kekisi, tetapi tidak mempunyai tenaga yang • Elektron-elektron yang terhasil dipecutkan dengan
mencukupi untuk melepaskan diri dari permukaan logam menggunakan voltan lampau tinggi (VLT).
kerana daya tarikan nukleus yang kuat. The electrons produced are accelerated by using the extra high
The metal has free electrons that move in the lattice structure, tension (EHT).
but does not have sufficient energy to escape from the metal
surface because of the strong nuclear attraction. • Elektron-elektron mempunyai halaju dan tenaga kinetik yang
tinggi . / The electrons have high velocity and
• Apabila logam dipanaskan, elektron akan memperoleh
tenaga yang mencukupi untuk melepaskan diri dari high kinetic energy.
permukaan logam.
When metal is heated, the electrons will acquire • Elektron-elektron menghentam skrin berpendarfluor.
enough energy to escape from the metal surface. The electrons hits the fluorescent screen.
• Tenaga kinetik Tenaga cahaya
Kinetic energy Light energy
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MODUL • Fizik TINGKATAN 5
Dengan prinsip keabadian tenaga, bagi setiap elektron, tenaga keupayaan elektrik = tenaga kinetik
From the principle of conservation of energy, for each electron, electrical potential energy = kinetic energy
1
eV = mv 2
2
Oleh itu, Hence, v = 2eV
m
v = halaju elektron / velocity of the electrons
V = beza keupayaan antara anod dan katod / potential difference between anode and cathode
e = cas setiap elektron / charge of one electron, 1.6 × 10 C
–19
m = jisim elektron / mass of electron, 9 × 10 kg
–31
Faktor Kesan ke atas kadar pancaran termion
Factor Effect on the rate of thermionic emission
Suhu katod Apabila suhu meningkat, maka kadar pancaran termion juga meningkat.
Temperature of the cathode The rate of thermionic emission increases as the temperature increases.
Luas permukaan katod Jika luas permukaan bertambah, kadar pancaran termion juga akan meningkat.
Surface area of the cathode The rate of thermionic emission increases as the surface area increases .
Beza keupayaan antara anod Jika beza keupayaan meningkat, kadar pancaran termion tidak berubah tetapi halaju elektron yang
dan katod dipancarkan ke arah anod meningkat.
Potential difference between the If the potential difference increases, the rate of thermionic emission is unchanged but the velocity of the
anode and cathode emitted electrons towards the anode increases.
UNIT 4
UNIT 4
Jenis pergerakan sinar katod dalam tiub sinar katod
Types of motion of cathode rays in a cathode ray tube
P Q
Kawasan / Region Jenis gerakan / Types of motion
PQ: Katod ke anod
R Pecutan seragam / Uniform acceleration
Cathode to anode
QR: Anod ke skrin
Anode to screen Halaju seragam / Uniform velocity
Perubahan tenaga pada elektron dalam sinar katod
Energy conversion of electrons in cathode rays
Kawasan / Region Jenis tenaga / Types of energy
Tenaga keupayaan elektrik
P: Katod / Cathode
Electric potential energy
PQR: Katod ke anod dan skrin Tenaga keupayaan elektrik Tenaga kinetik
Cathode to anode and screen Electric potential energy Kinetic energy
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MODUL • Fizik TINGKATAN 5
Mengkaji ciri-ciri sinar katod
To investigate the characteristics of cathode rays
Ciri-ciri sinar katod dikaji dengan menggunakan tiub palang Maltese dan tiub pesongan.
The characteristics of cathode rays are investigated by using the Maltese cross tube and deflection tube.
Tiub palang Maltese Palang Maltese aluminium
Maltese cross tube Bekalan Aluminium Maltese cross
pemanas, 6V
6 V Katod Anod
heater supply Cathode Anode Vakum
Vacuum
Bayang
Shadow
Skrin berpendarfluor
3 kV Fluorescent screen
Prosedur Pemerhatian Penerangan Kesimpulan
Procedure Observation Explanation Conclusion
Pemanas 6 V Bayang palang kelihatan. Bayang dihasilkan oleh Lintasan sinar bergerak dalam
dihidupkan. A shadow of the cross is cahaya dari filamen yang garis lurus
A 6 V heater is seen. dipanaskan. .
switched on. The shadow is formed by the Light rays travel in a straight line .
light rays from the heated
filament.
Sekarang bekalan Bayang palang berwarna Bayang palang dihasilkan Sinar katod bergerak dalam garis lurus.
kuasa 3 kV hijau kelihatan. Ia sama oleh sinar katod. Sinar katod menyebabkan
disambungkan saiz dan sama kedudukan The shadow is formed by the kesan berpendarfluor
antara katod dan dengan bayang yang cathode rays. .
anod. dihasilkan oleh cahaya Sinar katod membawa tenaga kinetik yang UNIT 4
Now, a 3 kV power tadi. ditukarkan kepada tenaga cahaya apabila
UNIT 4
supply is connected A green shadow of the ia menghentam skrin.
between the cathode cross is seen. It is the Cathode rays travel in a straight line.
and anode. same size and at the same
position as the shadow Cathode rays cause fluorescence .
formed by the light earlier. Cathode rays carry kinetic energy which is
converted to light energy when they hit
the screen.
Kutub utara Bayang sinar katod Daya saling tindakan / Daya Sinar katod boleh dipesongkan oleh
magnet bar dibawa bergerak dan terpesong lastik dihasilkan kerana sinar medan magnet
berdekatan dengan ke bawah. katod membawa cas negatif. . Peraturan tangan kiri
sinar katod The cathode ray shadow A catapult force is produced Fleming digunakan untuk menentukan arah
The North pole of a is moved and deflected because the cathode rays gerakan sinar.
bar magnet is brought downward. carry a negative charge. Cathode rays can be deflected by a
close to the cathode magnetic field . The Fleming’s left-hand
rays. rule is used to determine the direction of motion.
Perhatian / Note:
N Arah sinar katod adalah bertentangan dengan
S arah arus. / The direction of the cathode rays is
opposite to that of the electric current.
129 © Nilam Publication Sdn. Bhd.
MODUL • Fizik TINGKATAN 5
Tiub pesongan
Deflection tube
Bekalan Katod Anod
pemanas, 6 V Cathode Anode
6 V Plat logam
heater supply Metal plates
0 – 1 000 V
3 kV
Skrin berpendarfluor
Fluorescent screen
Prosedur Pemerhatian Penerangan Kesimpulan
Procedure Observation Explanation Conclusion
Bekalan pemanas 6 V Tiada medan elektrik di antara Sinar katod bergerak dalam
dan bekalan kuasa 3 kV dua plat logam. garis lurus
disambungkan tetapi plat No electric fields between the .
logam tidak disambung ke two metal plates. Cathode ray travels in a
bekalan kuasa.
6 V heater supply and 3 kV straight line .
power supply are connected but
metal plate is not connected.
Bekalan pemanas 6 V dan + Terdapat medan elektrik di Sinar katod bercas
bekalan kuasa 3 kV antara kedua-dua plat. negatif
disambungkan dan bekalan Electric field exists between two . UNIT 4
kuasa 1 000 V disambungkan plates. Cathode ray is negatively
UNIT 4
kepada plat logam. Sinar katod dipesongkan charged.
6 V heater supply and 3 kV ke atas.
power supply are connected and The cathode ray is is deflected
also 1 000 V power supply is – upward.
connected to the metal plates.
Bekalan pemanas 6 V Terdapat medan elektrik di Sinar katod bercas
dan 3 kV bekalan kuasa antara kedua-dua plat. negatif
disambungkan dan juga – Electric field exists between the .
1 000 V bekalan kuasa two plates. Cathode ray is negatively
disambungkan secara Sinar katod terpesong ke charged.
songsang kepada plat logam. bawah.
6 V heater supply and 3 kV The cathode ray is is deflected
power supply are connected and downward.
also the 1 000 V power supply
is connected to the metal plates +
but in reverse order.
© Nilam Publication Sdn. Bhd. 130
MODUL • Fizik TINGKATAN 5
SINAR KATOD (Menghuraikan) / CATHODE RAY (Describing)
Bergerak secara
garis lurus dalam
vakum
Moves linearly
in the vacuum
Mempunyai halaju,
tenaga kinetik dan Bercas
momentum yang tinggi negatif
Have high velocity, kinetic Ciri-ciri Negatively
energy and momentum sinar katod charged
The characteristics of
the cathode ray
Dipesongkan oleh
medan elektrik dan
medan magnet Menyebabkan skrin
berpendarflour
Deflected by the bercahaya
electric field and Cause fluorescent
magnetic field screen luminous
Contoh / Example
1 Beza keupayaan antara anod dan katod dalam senapang 2 Dalam tiub gambar set televisyen, beza keupayaan 20 kV
elektron ialah 5 kV. Hitungkan tenaga kinetik elektron. merentasi anod dan katod memecutkan sinar elektron. UNIT 4
[e = 1.6 × 10 C] Cas satu elektron ialah 1.6 × 10 C. Berapakah tenaga
–19
–19
UNIT 4
The potential difference between the anode and cathode in kinetik setiap elektron yang menghentam skrin?
an electron gun is 5 kV. Calculate the kinetic energy of the In a picture tube of a television set, a potential difference
–19
electrons. [e = 1.6 × 10 C] of 20 kV is applied across the anode and the cathode to
Penyelesaian / Solution accelerate the electron beam. The charge of each electron
–19
Tenaga keupayaan elektrik = Tenaga kinetik is 1.6 × 10 C. What is the kinetic energy of each electron
striking on the screen?
Electrical potential energy = Kinetic energy
eV = E Penyelesaian / Solution
3
(1.6 × 10 –19 C) × (5 × 10 V) = E E = eV –19 C) × (20 × 10 V)
= (1.6 × 10
3
E = 8.0 × 10 J = 32 × 10 –16
–16
= 3.2 × 10 J
–15
3 Dalam tiub vakum penerima televisyen, sinar katod Penyelesaian / Solution
dihasilkan dan memecut melalui beza keupayaan 7 kV. Tenaga keupayaan elektrik =Tenaga kinetik
Tentukan halaju sinar katod itu. Electrical potential energy = Kinetic energy
[e = 1.6 × 10 C dan m = 9 × 10 kg] 1 2
–19
–31
e
In the vacuum tube of a television receiver, a cathode ray eV = 2 mv
is produced and accelerated through a potential difference 1
3
7 kV. Determine the velocity of the cathode ray. 1.6 × 10 –19 C × 7 × 10 V = 2 × (9 × 10 –31 kg) × v 2
[e = 1.6 × 10 C and m = 9 × 10 kg] (1.6 × 10 –19 C)× (7 × 10 V)
–31
–19
3
e
v = (4.5 × 10 –31 kg)
2
v = 4.99 × 10 m s –1
7
131 © Nilam Publication Sdn. Bhd.
MODUL • Fizik TINGKATAN 5
Struktur dan fungsi bahagian-bahagian yang penting dalam Osiloskop Sinaran Katod (O.S.K.)
Structure and functions of the main parts of the Cathode Ray Oscilloscope ( C.R.O.)
Grid kawalan Anod memfokus Plat-Y
• untuk mengawal • untuk memfokuskan • untuk memesongkan
kecerahan tompok di elektron yang bergerak sinar katod secara Sinar
atas skrin kepada sinar yang halus menegak katod
Control grid Focusing anode Y-plates Cathode Salut grafit / Graphite coating
• to control brightness of • to focus moving • to deflect cathode rays
the spot on the screen electrons into a fine ray rays vertically • Untuk membumikan elektron
yang menghentam skrin
To channel the electrons striking
the screen to the Earth
Filamen Katod
• membekalkan • memancarkan elektron Dasar Vakum
haba kepada katod apabila dipanaskan masa Vacuum
Filament Cathode Terminal y-input Time
• supply heat to the • emit electrons when base
cathode heated Y-input terminal Skrin berpendafluor
Plat-X • untuk menghasilkan
Anod memecut • untuk memesongkan sinar tompok yang cerah apabila
• memecutkan elektron katod secara mendatar dihentam oleh sinar katod
Accelerating anode X-plates Flourescent screen
• to accelerate electrons • to deflect cathode rays • to produce bright spot
horizontally when hit by cathode rays
Senapang elektron Sistem pemesongan Skrin pendarfluor
An electron gun Deflection system Fluorescent screen
UNIT 4
Penggunaan O.S.K. 1 Paparan bentuk gelombang
Uses of C.R.O. Displaying waveforms
UNIT 4
Litar dasar masa
Time-base circuit
Suis kuasa
Skrin Power switch Input-Y Suis Suis
Screen Fokus / Focus Y-input
dipadam
Anjakan-X Switched off dihidupkan
Switched on
X-shift
Anjakan-Y Tiada bekalan kuasa
Gandaan-Y Y-shift disambungkan
Y-gain Kawalan dasar No power supply
Input-Y / Y-input masa connected
Time-base controls
Bumi / Earth
Input-X / X-input Kecerahaan Bekalan kuasa a.t.
Brilliance d.c. power supply
(Brightness)
Suis a.u / a.t / a.c. / d.c. switch
Bekalan kuasa a.u.
a.c. power supply
© Nilam Publication Sdn. Bhd. 132
MODUL • Fizik TINGKATAN 5
2 Mengukur beza keupayaan bekalan kuasa Contoh / Example:
Measuring the potential difference of the power supply
Voltan a.t. / d.c. voltage
V = gandaan-Y × h 1 cm
V = Y-gain × h
h = 2 cm
h = sesaran tegak tompok cahaya
h = vertical displacement of the light spot
–1
Voltan a.u. / a.c voltage Gandaan-Y ditetapkan pada 2.0 V cm .
Y-gain is set at 2.0 V cm .
–1
1
Voltan puncak, V = gandaan-Y× 2 (ketinggian garis Voltan puncak, V = 2.0 V cm × 2 cm
–1
p
menegak) p
1 = 4 V
Peak voltage, V = Y-gain× (height of the vertical line)
p 2 V
V = p = 4 V
1 1 r.m.s.
V = × V V = × V 2 2
p.m.k.d. 2 p r.m.s. 2 p
[ p.m.k.d. punca min kuasa dua ]
V = V = 2.83 V
V r.m.s. = V root mean square
3 Mengukur sela masa yang singkat dan frekuensi Contoh / Example:
Measuring short time interval and frequency
Jarak antara dua denyutan pada skrin = d cm.
The distance between two pulses on the screen = d cm. 1 cm
Kawalan dasar masa = x s cm .
-1
The time-base control setting = x s cm .
-1
d = 4 cm
Tempoh bagi dua denyutan, T = xd
Period for two pulses, T = xd
UNIT 4
[Dalam kes ini, unit T ialah 'saat' Gandaan dasar masa ditetapkan pada x 0.01 s cm –1
=
In this case, the unit of T is 'second' ] The time-base gain is set at x
UNIT 4
Jarak antara dua denyutan pada skrin, d 4 cm
The distance between two pulses on the screen, d =
Oleh itu, tempoh, T 0.01 s cm × 4 cm = 0.04 s
–1
So, period, T =
1
Frekuensi / Frequency, f =
T
1
f = = 25 Hz
0.04 s
Video
OSK
CRO
https://goo.gl/uZTZaW
133 © Nilam Publication Sdn. Bhd.
MODUL • Fizik TINGKATAN 5
Latihan / Exercise
1 Rajah menunjukkan bentuk gelombang yang 2 Rajah menunjukkan bentuk gelombang dihasilkan pada
terhasil daripada bekalan kuasa arus ulang-alik skrin O.S.K di radar stesen lapangan terbang. Titik X
yang disambungkan kepada input-Y pada O.S.K. dan Y menunjukkan masa penghantaran isyarat ke kapal
[Gandaan-Y disetkan = 20 V cm dan kawalan dasar masa KBAT terbang dan masa penerimaan pantulan isyarat melalui
–1
= 5 ms cm ] radar tersebut.
-1
The diagram shows a waveform obtained from an a.c. power [Kawalan dasar masa pada O.S.K. = 50 ms cm ]
–1
supply connected to Y-input of a C.R.O. The diagram shows a waveform obtained on the screen of
[Y-gain setting = 20 V cm and time-base control setting = a C.R.O. at an airport radar station. The points X and Y
–1
5 ms cm ] indicate the time of transmission to an aeroplane and the time
–1
of receiving the reflected signals by the radar station.
[Time-base control setting of the C.R.O. = 50 ms cm ]
–1
Y
X
Tentukan, / Determine, 8 cm
(a) tempoh isyarat yang ditunjukkan Hitung
the period of the signal Calculate
(b) frekuensi isyarat (a) masa isyarat radar dari X ke Y.
the frequency of the signal the time travels of the radar from X to Y.
(c) beza keupayaan puncak ke puncak (b) jarak antara stesen radar dan kapal terbang.
the peak to peak voltage [Halaju cahaya = 3 × 10 m s ]
8
–1
(d) beza keupayaan voltan puncak the distance between the radar station and the
the peak voltage aeroplane.
8
–1
Penyelesaian / Solution [Speed of light = 3 × 10 m s ]
Penyelesaian / Solution
(a) Ketetapan kawalan dasar-masa O.S.K. UNIT 4
Time-base control setting of the C.R.O.
UNIT 4
8 cm = 50 ms cm –1
d = 14 cm Masa penghantaran isyarat radar,
Time travels of the radar,
T = 50 ms cm × 8 cm = 400 ms
–1
(a) Ketetapan kawalan dasar masa = 0.4 s
Time-base control setting = 5 ms cm –1
Tempoh / Period, T = 5 ms cm × 14 cm (b) Halaju cahaya = 3 × 10 m s –1
–1
8
= 70 ms Speed of light = 3 × 10 m s –1
8
= 7.0 × 10 s Menggunakan formula, 2D = vt
–2
1 Using formula, 2D = vt
(b) Frekuensi / Frequency, f = T Jarak antara stesen radar dan kapal terbang
1 Distance between the radar station and the aeroplane,
= 7.0 × 10 s vy
–2
= 14.29 Hz D = 2
(c) Ketetapan gandaan-Y / Y-gain setting = 20 V cm = (3 × 10 m s ) × (0.4 s)
–1
–1
8
Voltan puncak ke puncak 2
7
Peak to peak voltage, V = 20 V cm × 8 cm = 6.0 × 10 m
–1
pp
= 160 V
(d) V = Vpp
2
p
160 V
= 2
= 80 V
© Nilam Publication Sdn. Bhd. 134
MODUL • Fizik TINGKATAN 5
Diod Semikonduktor
4.2 Semiconductor Diodes
Perbandingan antara penebat, semikonduktor dan konduktor:
Comparison between insulator, semiconductor and conductor:
Penebat Semikonduktor Konduktor
Insulator Semiconductor Conductor
Contoh bahan Kaca, seramik, politena Silikon, germanium, selenium Kuprum, aluminium, besi
Example of material Glass, ceramic, polythene Silicon, germanium, selenium Copper, aluminium, iron
Pembawa cas Tiada elektron bebas Elektron bebas dan lohong Elektron bebas
Charge-carriers Free electrons and holes Free electrons
No free electrons
Tinggi Berada antara rintangan penebat dan Rendah
Rintangan High konduktor / Between the resistance of an
Resistance
insulator and conductor Low
Kekonduksian lemah kerana Kekonduksian meningkat apabila Kekonduksian yang baik kerana
Kekonduksian tiada elektron bebas yang suhu terdapat banyak elektron bebas
Conductivity mengalirkan arus elektrik. meningkat. yang mudah bergerak antara
Poor conductivity because there Kekonduksian juga meningkat apabila atom.
are no free electrons to move cahaya matahari menyinarinya atau dengan Good conductivity because
about to conduct electricity. kehadiran bendasing. conductors have many free
Conductivity increases when the electrons that can move easily
temperature increases. Conductivity between atoms.
also increases when light shines on it or with
the presence of impurities.
Terminologi / Terminology Penerangan / Explanation
Bahan yang sifat kekonduksian elektrik terletak di antara penebat elektrik dan UNIT 4
Semikonduktor
UNIT 4
Semiconductor konduktor elektrik. / The material which its properties of electrical conductivity is located
between the electrical insulator and electrical conductors.
Semikonduktor intrinsik Semikonduktor tulen tanpa kewujudan atom bendasing dalam struktur kekisi atom.
Intrinsic semiconductor Pure semiconductor without the existence of impurity atoms in the atomic lattice structure.
Semikonduktor tidak tulen dengan kewujudan atom bendasing dalam struktur kekisi
Semikonduktor ekstrinsik
Extrinsic semiconductor atom. / Non-pure semiconductor with the existence of impurity atoms in the atomic lattice
structure.
Proses penambahan atom bendasing dengan kuantiti yang sangat kecil ke dalam
Pendopan struktur kekisi atom semikonduktor untuk meningkatkan kekonduksian elektrik.
Dopping The process of adding impurity atoms with very small quantities into the atomic lattice
structure of a semiconductor to increase the electrical conductivity.
• Atom yang mempunyai tiga elektron di petala paling luar.
Atom trivalen Atom that has three electrons in the most outer shell.
Trivalent atom • Contoh : boron, indium, gallium / Example : boron, indium, gallium
• Bertindak sebagai atom penerima. / Acting as a recipient atom.
Atom pendop
Dopant atom
• Atom yang mempunyai lima elektron di petala paling luar.
Atom pentavalen Atom with five electrons in the most outer shell.
Pentavalent atom • Contoh : fosforus, arsenik, antimoni / Example: phosphorus, arsenic, antimony
• Bertindak sebagai atom penderma. / Acting as the donor atom.
135 © Nilam Publication Sdn. Bhd.
MODUL • Fizik TINGKATAN 5
Semikonduktor yang kekonduksian elektrik disumbangkan oleh elektron bebas
Semikonduktor jenis-n sebagai pembawa cas majoriti dan lohong sebagai pembawa cas minoriti.
n-type semiconductor The semiconductor which its electrical conductivity is contributed by free electrons
as the majority charge carriers and holes as the minority charge carriers.
Semikonduktor yang kekonduksian elektrik disumbangkan oleh lohong
sebagai pembawa cas majoriti dan elektron bebas sebagai pembawa cas
Semikonduktor jenis-p minoriti.
p-type semiconductor
The semiconductor which its electrical conductivity is contributed by holes as
the majority charge carriers and free electrons as the minority charge carriers.
Elektron (bercas negatif) lebihan yang tidak terlibat dalam pembentukan ikatan kovalen.
Elektron bebas Elektron ini menyumbang kepada peningkatan kekonduksian elektrik.
Free electron The extra electrons (negatively charged) that are not involved in the formation of a covalent
Pembawa cas bond. These electrons contribute to the increasing of electrical conductivity.
Charge carrier
Lohong Lohong yang bercas positif.
Hole Positively charged holes.
• Semikonduktor tulen bersifat seperti penebat dalam keadaan sejuk, tetapi mengalirkan arus yang kecil pada
suhu bilik (bersifat konduktor).
Pure semiconductors act as insulators when they are cold, but conduct a small current when they are at room
temperature.
• Semikonduktor intrinsik seperti silikon dan germanium menunjukkan ciri-ciri semikonduktor dari atom
asalnya.
Ciri-ciri An intrinsic semiconductor such as silicon and germanium derives its semiconductor properties from its own
semikonduktor atoms.
Properties of UNIT 4
semiconductors • Apabila beza keupayaan dikenakan merentasi semikonduktor intrinsik, ia akan mengalirkan arus yang kecil.
When a potential difference is applied across an intrinsic semiconductor, it conducts a very small current.
UNIT 4
• Kekonduksian dalam semikonduktor adalah melalui pergerakan elektron-elektron bebas dan lohong-lohong
dalam arah bertentangan.
Conduction in a semiconductor is by means of movement of free electrons and holes in opposite directions.
• Ia boleh dijadikan konduktor yang baik dengan pendopan semikonduktor dengan sebilangan kecil atom asing.
They can be made to conduct better by doping the semiconductor with a small amount of foreign atoms.
• Tujuan pendopan : Menghasilkan semikonduktor ekstrinsik (jenis-n atau jenis-p) yang kekonduksian
elektriknya lebih baik.
Purpose of doping : To produce extrinsic semiconductor (n-type or p-type) which has better electric conductivity.
• Jenis atom-atom pendop yang digunakan adalah atom trivalen dan atom pentavalen.
The types of dopant atoms used are trivalent atom and pentavalent atom.
Proses pendopan
Doping process • Atom-atom kumpulan III dan kumpulan V adalah sesuai sebagai atom pendop kerana mempunyai saiz atom
yang hampir sama dengan saiz atom semikonduktor di mana atom pendop boleh dimasukkan ke dalam
struktur kekisi atom semikonduktor.
Atoms of group III and group V are suitable as dopant atoms because their atoms have the size nearly the same
with the size of semiconductor atoms where the dopant atoms can be inserted fitly into atomic lattice structure of
the semiconductor.
© Nilam Publication Sdn. Bhd. 136
MODUL • Fizik TINGKATAN 5
Semikonduktor jenis-n / n-type semiconductors
• Atom silikon (mempunyai empat elektron valens) didopkan dengan atom-atom pentavalen (mempunyai lima
elektron valens) seperti antimoni, fosforus atau arsenik untuk meningkatkan bilangan elektron-elektron bebas.
Silicon atoms (with four valence electrons) doped with pentavalent atoms (which have five valence electrons) such
as antimony, phosphorus or arsenic to increases the number of free electrons.
• Atom pentavalen akan menggantikan beberapa atom silikon, kemudian tiap-tiap satu daripada atom
pentavalen akan mempunyai empat ikatan kovalen dan satu elektron bebas lebihan.
The pentavalent atoms will replace some of the silicon atoms, then each of the pentavalent atoms will have four
covalent bonds and one extra free electron.
• Oleh kerana atom pentavalen menderma lebihan elektron, ia dinamakan atom penderma.
Since a pentavalent atom donates an extra electron, it is called donor atom.
• Dalam keadaan ini, terdapat lebih banyak elektron bebas berbanding lohong, oleh itu elektron adalah
pembawa cas majoriti dan lohong adalah pembawa cas minoriti, maka ia dinamakan semikonduktor jenis-n.
In this case, there are more free electrons than holes. Therefore, the electrons are the majority charge-carriers and
Semikonduktor the holes are the minority charge-carriers and is thus known as an n-type semiconductor.
jenis-n • Elektron bebas yang berlebihan menjadi pembawa cas negatif dalam semikonduktor jenis-n.
n-type The excess free electrons become the negative charge-carriers in an n-type semiconductor.
semiconductors
Si Si Si Elektron bebas lebihan
Extra free electron
Si As Si
Ikatan kovalen
Covalent bond
Si Si Si
Atom arsenik Atom silikon Elektron
As Si
Arsenic atom Silicon atom Electron
Semikonduktor jenis-n / n-type semiconductor
Semikonduktor jenis-p / p-type semiconductors
• Silikon yang didopkan dengan atom trivalen (mempunyai tiga elektron valens) seperti boron, indium atau
galium akan meningkatkan bilangan lohong. / Silicon doped with trivalent atoms (which have three valence UNIT 4
electrons) such as boron, indium or gallium increase the number holes.
UNIT 4
• Atom trivalen akan menggantikan beberapa atom silikon, menyebabkan silikon kehilangan satu elektron
daripada empat ikatan kovalen. / The trivalent atoms will replace some of the silicon atoms, resulting in one
electron missing from one of the four covalent bonds.
• Atom trivalen akan menerima satu elektron, maka ia dinamakan atom penerima.
Since a trivalent atom accepts an electron, it is called the acceptor atom.
• Atom trivalen ini menghasilkan semikonduktor jenis-p dengan membentuk kekurangan elektron valens yang
dinamakan lohong. / These trivalent atoms produce p-type semiconductors by creating a shortage of valence
electrons called holes.
• Dalam keadaan ini, terdapat lebih banyak lohong terhasil, oleh itu lohong adalah pembawa cas majoriti
Semikonduktor dan elektron adalah pembawa cas minoriti, semikonduktor ini dinamakan semikonduktor jenis-p. / In this
jenis-p case there are more holes. Therefore, the holes are the majority charge-carriers and the electrons are the minority
p-type charge-carriers and is thus known as p-type semiconductors.
semiconductors • Lohong berlebihan ini menjadi pembawa cas positif dalam semikonduktor jenis-p.
The excess holes become the positive charge carriers in the p-type semiconductors.
Si Si Si Kehilangan elektron membentuk lohong
Missing electron created a hole
Si In Si
Si Si Si
Atom silikon Atom Indium Elektron Lohong
Si Silicon atom In Indium atom Electron Hole
Semikonduktor jenis-p / p-type semiconductor
137 © Nilam Publication Sdn. Bhd.
MODUL • Fizik TINGKATAN 5
Perbandingan antara semikonduktor jenis-n dan jenis-p
The comparison between the n-type semiconductor and p-type semiconductor
Aspek Semikonduktor jenis-n Semikonduktor jenis-p
Aspect n-type semiconductor p-type semiconductor
Ketulenan semikonduktor Tidak tulen Tidak tulen
The purity of semiconductor Not pure Not pure
Jenis atom pendop yang digunakan Atom pentavalen Atom trivalen
The type of dopant atom used Pentavalent atom Trivalent atom
(i) fosforus / phosphorus (i) boron
Contoh-contoh atom pendop yang digunakan (ii) arsenik / arsenic (ii) indium
The examples of dopant atom used
(iii) antimoni / antimony (iii) gallium
Kesan ke atas kekonduksian elektrik Meningkat / menjadi lebih baik Meningkat / menjadi lebih baik
The effect on electrical conductivity Increases / becomes better Increases / becomes better
Bilangan elektron valens atom pendop 5 elektron 3 elektron
The number of valence electron of dopant atom 5 electrons 3 electrons
Pembawa cas majoriti Elektron bebas (cas negatif) Lohong (cas positif)
The majority charge carrier Free electrons (negatively charged) Holes (positively charged)
Pembawa cas minoriti Lohong (cas positif) Elektron bebas (cas negatif)
The minority charge carrier Holes (positively charged) Free electrons (negatively charged)
Diod semikonduktor / Semiconductor diodes
UNIT 4
• Apabila bahan semikonduktor jenis-p bercantum dengan bahan semikonduktor jenis-n, ini akan membentuk
UNIT 4
satu lapisan di tengah-tengahnya yang dinamakan lapisan susutan . Pada lapisan ini, elektron dari
bahan jenis-n hanyut merentasi simpang untuk mengisi lohong dalam bahan jenis-p. Lohong dari bahan jenis-p
menghanyut dalam arah bertentangan ke bahan jenis-n. Hasilnya, lapisan susutan yang nipis terbentuk. Ia
adalah kawasan sempit yang kehilangan semua elektron bebas dan lohong dan bersifat seperti silikon tulen
yang mempunyai kerintangan tinggi. Pengaliran cas seterusnya yang merentasi sempadan lapisan susutan akan
ditolak oleh cas di dalam lapisan tersebut.
When a p-type semiconductor material is in contact with an n-type semiconductor material, a layer called the
Bagaimana depletion layer is formed in the middle. At this junction, electrons from n-type semiconductor drifts
simpang p-n across the junction to fill in the holes in the p-type semiconductor. The holes from the p-type semiconductor drift
berfungsi? in the opposite direction to the n-type semiconductor. As a result, a depletion layer is produced. It is a very narrow
How the region which has lost all its available free electrons and holes and thus behaves almost like a pure silicon, with high
p-n junction resistivity. Any further movement of charges across the boundry in the depletion layer will be repelled by the charges
works? in the layer.
Voltan simpang / Junction voltage
Semikonduktor jenis-p Lapisan susutan Semikonduktor jenis-n
p-type semiconductor Depletion layer n-type semiconductor
© Nilam Publication Sdn. Bhd. 138
MODUL • Fizik TINGKATAN 5
• Lapisan cas negatif dalam bahan jenis-p akan menghalang pembawa cas majoriti dari bahan jenis-n (elektron)
daripada merentasi sempadan. Begitu juga lapisan cas positif dalam bahan jenis-n akan menghalang
pembawa cas majoriti dari bahan jenis-p (lohong) daripada merentasi sempadan dalam arah bertentangan.
Ini menyebabkan suatu beza keupayaan yang bertindak dari bahan jenis-n ke bahan jenis-p. Beza keupayaan
merentasi simpang ini dinamakan voltan simpang . Tiada arus mengalir melalui simpang p-n semasa
cas berada dalam keseimbangan.
The layer of the negative charge in the p-type region will prevent the majority charge carriers from the n-type region
(the electrons) from crossing the boundary. Similarly, the positive charge layer in the n-type region will prevent the
majority charge carriers from the p-type region (the holes) from crossing the boundry in the opposite direction. This
will result in a potential difference acting from the n-type material to the p-type material across the junction. This
potential difference is known as the junction voltage . In its normal state, a p-n junction delivers no current
since the charges are in equilibrium.
Semikonduktor jenis-p Semikonduktor jenis-n
p-type semiconductor n-type semiconductor
Elektron / Electron Lohong / Hole
• Kesan voltan simpang ialah untuk menghalang pembawa cas daripada menghanyut merentasi simpang.
Anggaran voltan simpang bagi germanium dan silikon ialah 0.1 V dan 0.6 V masing-masing.
This junction voltage prevents charge carriers from drifting across the junction. The junction voltages for
germanium and silicon are approximately 0.1 V and 0.6 V respectively.
Simbol diod
semikonduktor
Symbol of a Anod (+) Katod (–)
semiconductor Anode (+) Cathode (–)
diode
Pincang ke depan / Forward-biased UNIT 4
• Untuk membenarkan arus elektrik mengalir melalui diod, beza keupayaan merentasi diod mesti melebihi
UNIT 4
voltan simpang. / In order for an electric current to flow through the diode, the voltage applied across the diode
must exceed the junction voltage.
• Dalam susunan pincang ke depan, beza keupayaan sel lebih besar daripada voltan simpang. Lapisan susutan
menjadi nipis, rintangan diod berkurang. Oleh itu arus yang besar mengalir melalui diod.
In a forward-biased arrangement, the cell voltage is greater than the junction voltage. The depletion layer becomes
narrow, the resistance of diode decreases. Hence a large current flows through the diode.
Voltan simpang
Junction voltage
Rintangan rendah / Low resistance
Fungsi diod Semikonduktor jenis-p Semikonduktor jenis-n
Function of p-type semiconductor n-type semiconductor
diodes
Arus
Lapisan susutan nipis Current
Narrow depletion layer
+ –
Beza keupayaan sel
Cell voltage
• Diod mengalirkan arus kerana lohong dari semikonduktor jenis-p dan elektron dari semikonduktor jenis-n
berupaya merentasi simpang. Mentol akan menyala. / The diode conducts current because the holes from the p-type
semiconductor and electrons from the n-type semiconductor are able to cross over the junction. The bulb will light up.
• Fungsi diod membenarkan arus mengalir dalam satu arah sahaja.
The function of a diode is to allow current to flow in one direction only .
139 © Nilam Publication Sdn. Bhd.
MODUL • Fizik TINGKATAN 5
Pincang songsang / Reverse-biased
• Dalam susunan pincang songsang, beza keupayaan sel lebih rendah daripada voltan simpang. Lapisan susutan
menjadi besar dan rintangan diod meningkat. Oleh itu, tiada arus mengalir melalui diod.
In a reverse-biased arrangement, the cell voltage is lower than the junction voltage. The depletion layer becomes
wide and the resistance of diode increases. Hence, no current flows through the diode.
Voltan simpang
Junction voltage
Rintangan tinggi / High resistance
Bahan Bahan
jenis-p jenis-n
p-type n-type
material material
Lapisan susutan lebar
Wide depletion layer
– +
Beza keupayaan sel
Cell voltage
• Elektron dan lohong ditarik menjauhi simpang. Mentol tidak menyala.
Both electrons and holes are pulled away from the junction. The bulb does not light up.
Graf arus lawan beza keupayaan bagi simpang p-n dalam diod silikon dan diod germanium
Graph of current against voltage for the p-n junction in a silicon and germanium diode
Diod
germanium
Germanium
I / mA diode Diod silikon
Graf Silicon diode
Graph
UNIT 4
V / Volt
0 0.1 0.6
UNIT 4
Pincang songsang Pincang ke depan
Reverse-biased Forward-biased
Diod sebagai rektifier / Diodes as rectifiers
Terminologi Penerangan
Terminology Explanation
Lapisan susutan Ialah kawasan neutral yang tiada pembawa cas
Depletion layer Is the neutral region which has no charge-carriers
Rektifier Ialah peranti elektrik yang menukarkan arus ulang-alik kepada arus terus
Rectifier Is an electrical device that converts alternating current to direct current
Rektifikasi Ialah proses untuk menukarkan arus ulang-alik kepada arus terus dengan menggunakan diod
Rectification Is the process to convert an alternating current into a direct current by using diode
© Nilam Publication Sdn. Bhd. 140
MODUL • Fizik TINGKATAN 5
Jenis
rektifikasi Rajah litar Lakaran graf Penerangan
Types of Diagram of the circuit Sketch of graph Explanation
rectification
Diod Ke O.S.K. V = Beza keupayaan input (a.u.) Proses di mana hanya separuh kitar
in
Diode To C.R.O. V = Input voltage (a.c. current) pertama arus ulang-alik, dibenarkan
in
V in mengalir dalam satu arah.
A process where only half of every
a.c R V cycle of alternating current is made to
flow in one direction.
t / s
0
Dalam separuh kitar pertama, diod
Rektifikasi dipincang ke depan dan arus dapat
separuh mengalir.
gelombang In the first half cycle, the diode is
Half-wave V R forward-biased and current can flow.
Rectification
Arus dihalang mengalir menerusi
t / s diod dalam separuh kitar kedua
0
apabila diod dipincang songsang.
The current cannot flow through the
diode in the second half-cycle when the
V = Beza keupayaan output
R
melalui O.S.K. diode is reverse-biased.
Rectified voltage which shows
in the C.R.O.
(i) Separuh kitaran yang pertama arus a.u. Beza keupayaan output merentasi R Proses di mana kedua-dua separuh
First half of cycle of a.c. current Output voltage across R daripada setiap kitaran arus ulang-
V
R alik, dibenarkan mengalir dalam satu
(i) (ii) (i) (ii) (i) arah / arah yang sama.
D D A process where both halves of every
A 4 1 Ke O.S.K. cycle of an alternating current is made
To C.R.O. 0 t /s to flow in the same direction.
B UNIT 4
D Ketika separuh kitaran pertama, arus
D 2 a.u.,(i), arus mengalir dari A ke D ke
UNIT 4
3 1
R V
R R, ke D ke B.
3
In the first half of the cycle of the a.c.
current, (i), the current flows from A to
Rektifikasi D 1 to R, to D 3 to B.
gelombang
penuh Ketika separuh kitaran kedua, (ii),
Full-wave (ii) Separuh kitaran yang kedua arus a.u. arus mengalir dari B ke D ke R, ke
rectification Second half of cycle of a.c. current 2
D ke A.
4
In the second half, (ii), the current
flows from B to D 2 to R, to D 4 to A.
D D
A 4 1 Ke O.S.K. Arah arus ulang-alik yang mengalir
To C.R.O. melalui perintang R untuk setiap
B separuh kitaran adalah sama.
D 3 D 2 The direction of alternating current
passing through the resistor R for each
R V R half of the cycle is the same.
141 © Nilam Publication Sdn. Bhd.
MODUL • Fizik TINGKATAN 5
Arus mengalir Arus selepas merentasi kapasitor Meratakan gelombang output
Current flows V After the current flows across menggunakan kapasitor
the capacitor Smoothing output wave by using a
capacitor
• Apabila arus melalui resistor dan
D D t / s kapasitor, kapasitor akan dicaskan
A 4 1 Ke OSK 0 dan tenaga disimpan.
To CRO Arus sebelum merentasi kapasitor When a current flows through the
B Before the current flows across resistor and capacitor, the capacitor
the capacitor
D is charged and energy is stored.
D 2 • Apabila tiada arus melalui
3
C R V R perintang dan kapasitor, kapasitor
Arus Kapasitor / Capacitor dinyahcaskan dan tenaganya
Meratakan arus mengalir Kapasitor disambung secara selari digunakan untuk menghasilkan
menggunakan Current dengan perintang. Ia digunakan beza keupayaan merentasi
kapasitor flows untuk: perintang.
Smoothening the A capacitor is connected parallel When there is no current passing
current by using with the resistor. This device can: through the resistor and capacitor,
a capacitor 1 Menyimpan cas elektrik the capacitor is discharged and the
Store electric charge energy from it used to produce a
2 Meratakan bentuk gelombang voltage across the resistor.
bagi arus output yang telah • Hasilnya arus terus output yang
direktifikasi rata terhasil.
Smoothen a waveform in the As a result it produces a smooth d.c.
rectified output output.
3 Memisahkan arus ulang-alik
dan arus terus (sebagai penapis)
Separate the a.c. and d.c. (as a
filter)
UNIT 4
UNIT 4
Transistor
4.3 Transistors
Komponen elektronik yang berfungsi sebagai / The electronic component that function as
Transistor
Transistor 1 suis automatik / automatic switch
2 amplifier / amplifier
n-p-n p-n-p
C C
Jenis transistor dan simbol-
simbol B n C = pengumpul / collector B n C = pengumpul / collector
The types of transistor and p B = tapak / base p B = tapak / base
symbols E = pengeluar / emitter E = pengeluar / emitter
E n E n
transistor n-p-n transistor n-p-n
© Nilam Publication Sdn. Bhd. 142
MODUL • Fizik TINGKATAN 5
n-p-n n-p-n
• Pengumpul (C) disambung kepada terminal • Pengumpul (C) disambung kepada terminal
positif bateri negatif bateri
The collector (C) is connected to the positive The collector (C) is connected to the negative
Cara sambungan transistor terminal of battery terminal of battery
di dalam litar • Pengeluar (E) disambung kepada terminal
The way of connection of • Pengeluar (E) disambung kepada terminal
transistor in circuit negatif bateri positif bateri
The emitter (E) is connected to the negative The emitter (E) is connected to the
terminal of battery positive
terminal of battery
• Sebab : Transistor dipincang ke depan • Sebab : Transistor dipincang ke depan
Reason : Transistor is forward biased Reason : Transistor is forward biased
• Arus tapak (I B), arus pengumpul (I C) dan arus pengeluar (I E)
Base current (I B), collector current (I C) and emitter current (I E)
• I E = I B + I C
• Arus tapak adalah terlalu kecil berbanding dengan arus pengumpul
The base current is too small compare to the collector current
• I B <<<< I C
• Arus tapak dalam lingkungan unit μA, manakala arus pengumpul dalam lingkungan unit mA
The base current is in the unit range of μA, meanwhile the collector current is in the unit range of mA
• Pertambahan yang kecil dalam arus tapak menyebabkan pertambahan yang besar dalam arus
Arus-arus yang terlibat pengumpul / The small increase in base current causes the large increase in collector current
dalam operasi transistor • Nisbah pertambahan arus pengumpul terhadap pertambahan arus tapak dikenali sebagai amplifikasi
The currents involved in the arus atau faktor amplifikasi (tiada unit).
operation of transistor The ratio of increases in collector current to the increase in base current is known as current
amplification or factor of amplification (without unit).
• I C berkadar langsung dengan I B / I C is directly proportional to I B
• Keadaan pensuisan transistor bergantung pada I B dan I C:
The switching condition of transistor depends on I B and I C: UNIT 4
(i) Apabila I B bertambah, I C akan bertambah, transistor akan dihidupkan
UNIT 4
When I B increases, I C will increases, transistor is switched on
(ii) Apabila I B berkurang, I C akan berkurang atau menjadi sifar, transistor akan dimatikan (switched
off).
When I B decreases, I C wil decreases, transistor is switched off.
• Fungsi dua resistor, R dan R yang disambung
x
y
Voltan secara sesiri sepanjang bateri utama adalah
bateri sebagai pembahagi voltan.
Battery R 2 The function of the two resistors, R and R ,
voltage R I X Y
x C c which are connected in series across the main
Pembahagi voltan R B V battery, is as a potential divider.
Voltage divider 1 • Voltan tapak dikira dengan:
Voltan E The base voltage can be calculated as:
tapak R
Base y R y
voltage I b V = R y + R x × V
B
Video
Transistor
Transistor
https://goo.gl/EGmwdr
143 © Nilam Publication Sdn. Bhd.
MODUL • Fizik TINGKATAN 5
(a) Litar memadamkan mentol pada waktu malam dan menyala pada siang hari secara automatik.
Transistor sebagai suis kawalan suhu / The transistor as a temperature controlled switch
Circuit switches off the bulb at night and switches on the bulb at day time automatically.
Transistor sebagai suis kawalan cahaya / The transistor as a light controlled switch
Penerangan / Explanation Apabila seseorang bercakap melalui mikrofon, gelombang bunyi ditukarkan kepada arus ulang-alik. When a person speaks through a microphone, the sound waves are converted into an alternating current. Perubahan yang kecil kepada arus tapak, menyebabkan perubahan yang besar kepada arus pengumpul. Alat pembesar suara menerima arus ulang-alik yang besar daripada litar pengumpul dan menukarkannya kepada gelombang bunyi yang kuat. / The loudspeaker thus receives a large alternating curre
A small change in base current, will cause a big change in the collector current.
Transistor sebagai penguat isyarat / The transistor as an amplifier
UNIT 4
UNIT 4
mengalir. Mentol menyala.
Aplikasi Transistor / Application of Transistor
1 2 3 4 1 2 3 4 5 6 1 2 3
Pembesar suara Loudspeaker Mentol Bulb Mentol Bulb C
Rajah / Diagram C B E C B E R B E
Mikrofon Microphone Temistor Thermistor R Cahaya / Light PPC LDR R 1
© Nilam Publication Sdn. Bhd. 144
MODUL • Fizik TINGKATAN 5
Hence, the potential difference across R is high. The base voltage is high ; and the base current I B , which flows is high enough to produce a
cukup tinggi untuk menghasilkan arus pengumpul, I c yang besar. I c adalah cukup besar untuk menghidupkan transistor dan mengalir
menyala. / The base voltage is high, thus the base current, I B , flows high and causes a large collector current, I c , to flow. The bulb lights up.
Di siang hari, PPC mempunyai rintangan yang rendah seperti R 1 . Voltan tapak adalah rendah menyebabkan arus tapak tidak mengalir
large collector current I c . I c is large enough to switch the transistor on and flows through the relay coil. The relay switch is closed (relay is
apabila arus pengumpul I c menurun ke sifar. / Diode in the heat-controlled circuit is to protect the transistor from being damaged by the
Voltan tapak adalah tinggi, maka arus tapak, I B , mengalir tinggi dan menyebabkan arus pengumpul, I c , yang besar mengalir. Mentol
Diod dalam litar kawalan haba adalah untuk melindungi transistor dari rosak oleh d.g.e yang teraruh besar dalam gegelung geganti
Apabila suhu meningkat, rintangan termistor berkurang. Rintangan termistor menjadi sangat kecil berbanding dengan rintangan R.
transistor tetapi beza keupayaan merentasi perintang R 1 adalah rendah. / In darkness, the LDR has a very high resistance, the potential
Dalam keadaan gelap, PPC mempunyai rintangan yang tinggi, beza keupayaan merentasi PPC cukup tinggi untuk menghidupkan
Maka, beza keupayaan merentasi R adalah tinggi. Beza keupayaan tapak adalah tinggi dan arus tapak yang mengalir, I B , adalah
In daylight, the LDR has a very low resistance as compared to R 1 . The base voltage is too low. The base current does not flow and the
When the temperature rises, the resistance of the thermistor decreases significantly. The resistance of the thermistor is very small as
Oleh itu, beza keupayaan merentasi tapak transistor adalah rendah dan tidak mencukupi untuk menghidupkan suis transistor.
Rintangan termistor berkurangan sebaik sahaja suhu meningkat. / The resistanceof the thermistor falls as the temperature rises.
difference across the LDR is high enough to switch the transistor on and hence the potential difference across resistor R 1 is low.
melalui gegelung geganti. Suis geganti ditutup (suis geganti dihidupkan) dan menyebabkan penggera berbunyi.
Therefore, the potential difference across the base of the transistor is too low and not enough to switch the transistor on.
Termistor dan perintang R dalam rajah menghasilkan pembahagi keupayaan merentasi bekalan kuasa, 6 V.
collector current also does not flow, I B = 0, I c = 0. The transisitor is switched off. The bulb does not light up.
dan arus pengumpul juga tidak mengalir, I B = 0, I c = 0. Transistor terpadam. Mentol tidak menyala.
Suis kawalan haba atau penggera suhu tinggi / Heat-controlled switch or high temperature alarm
(b) Litar menyalakan mentol pada waktu malam dan dipadamkan pada siang hari secara automatik
The thermistor and resistor R in the diagram form a potential divider across the 6 V supply.
Pada suhu bilik, termistor mempunyai rintangan yang tinggi dibandingkan dengan R.
Circuit switches on the bulb at night and switches off the bulb at day time automatically
A thermistor is a resistor in which its resistance changes as the temperature changes.
Termistor ialah perintang yang mana rintangan berubah apabila suhu berubah.
large induced e.m.f. in the relay coil when the collector current I c drops to zero
At room temperature, the thermistor has a high resistance as compared to R.
switched on) and this causes the alarm to be activated.
UNIT 4
compared to the resistance of R.
UNIT 4
1 2 3 1 2 3 4 5 6 7 8
9 V 6 V
Mentol Bulb Suis geganti Relay switch Penggera Alarm
C
E
C
B E
B Maklumat tambahan:
R Additional information:
1 kΩ R B
R 1 A R
Cahaya Light PPC/LDR Termistor Thermistor 10 kΩ
145 © Nilam Publication Sdn. Bhd.
MODUL • Fizik TINGKATAN 5
Latihan / Exercise
1 Rajah menunjukkan litar transistor. Perintang P 2 Rajah menunujukkan simbol transistor n-p-n.
mempunyai rintangan 10 kΩ. Untuk menyalakan mentol, The diagram shows the symbol of an n-p-n transistor.
beza keupayaan merentasi perintang P mestilah sekurang-
kurangnya 2 V. 120 mA
The diagram shows a transistor circuit. Resistor P has a C
resistance of 10 kΩ. In order to light up the bulb, the potential 5 mA
difference across resistor P must be at least 2 V. B
E
x
S I e
Y T 12 V Apakah nilai I ?
e
What is the value of I ?
P Penyelesaian e
Solution
Z
Apakah nilai tertinggi perintang S apabila mentol menyala? I = I + I c
B
E
What is the maximum value of resistor S when the bulb = 5 mA + 120 mA
lights up? = 125 mA
Penyelesaian
Solution
V p = R p
V XZ R p + R s
2 V 10 k Ω
=
12 V 10 k Ω + R s
10 k Ω + R = 60 k Ω
s
R = 50 k Ω
s
UNIT 4
3 M
UNIT 4
X
KBAT
S
Q
R
6 V Y
P Suis
geganti
V Relay
p
switch 240 V a.u.
Z 240 V a.c.
(a) (i) Namakan komponen P.
Name component P.
(ii) Apakah fungsi bagi komponen P?
What is the function of component P?
(b) Apakah fungsi perintang R?
What is the function of resistor R?
(c) (i) Apakah yang akan berlaku kepada beza keupayaan V pada waktu malam?
p
What happens to potential difference V at night?
p
(ii) Terangkan kenapa mentol menyala pada waktu malam.
Explain why the bulb is lighted up at night.
(d) Mentol M berlabel 240 V, 60 W mula menyala apabila V mempunyai nilai minimum 2.0 V dan rintangan bagi perintang
p
P adalah 10 kΩ. Tentukan rintangan perintang S.
Bulb M, labelled 240 V, 60 W starts to light up when V has a minimum value of 2.0 V and the resistance of resistor P is 10 kΩ.
p
Determine the maximum resistance of resistor S.
© Nilam Publication Sdn. Bhd. 146
MODUL • Fizik TINGKATAN 5
Penyelesaian / Solution
(a) (i) Perintang peka cahaya (PPC) / Light-Dependent Resistor (LDR)
(ii) Mengawal voltan tapak secara automatik / Automatically controls the base voltage.
(b) Mengawal dan mengehadkan arus tapak supaya transistor tidak rosak.
Control and limit the base current so that the transistor is not damaged.
(c) (i) Beza keupayaan merentasi V p meningkat
Potential difference across V increases.
p
(ii) Pada waktu malam PPC mempunyai rintangan yang sangat tinggi. Beza keupayaan merentasi PPC sangat tinggi. Maka,
voltan tapak adalah tinggi. Arus tapak mengalir dan ini menyebabkan arus pengumpul yang tinggi mengalir. Transistor
dihidupkan dan mentol menyala.
At night the LDR has a very high resistance. The potential difference across LDR is very high. Hence, the base voltage is high.
The base current flows and produces a large collector current flow. The transistor is switched on and the bulb lights up.
V p R p
(d) =
V XZ R p + R s
10 k Ω
.
. . 2.0 V = 10 k Ω + R s
6.0 V
.
. . 10 k Ω + s = 30 k Ω
.
. . s = 20 k Ω
4 Litar di bawah digunakan sebagai suis kendalian haba.
The circuit below is used as a heat-operated switch.
KBAT
Penyelesaian / Solution
R 5 kΩ
1 (a) (i) Termistor / Thermistor
B (ii)
R 2 A 10 V
C UNIT 4
X
UNIT 4
(b) A: Tapak / Base
B: Pengumpul / Collector
(a) (i) Apakah komponen X? C: Pengeluar / Emitter
What is the component X? 4 V
(ii) Lukis rajah komponen X. (c) (i) R X =
Draw the component X. 5 k Ω + R X 10 V
.
(b) Namakan terminal A, B dan C. . . 10R = 20 k Ω + 4R X
X
.
Name the terminals A, B and C. . . 10R – 4R = 20 k Ω
X
X
(c) Untuk menghidupkan mentol, beza keupayaan . . 6R = 20 k Ω
.
merintangi X mesti paling kurang 4V. . X
In order to light the bulb, the potential difference across . . R = 3.33 k Ω
X
X must be at least 4 V.
(i) Apakah rintangan X apabila mentol menyala? 1 k Ω
What is the resistance of X when the bulb lights (ii) V = (1 + 5) k Ω × 10 V
X
up?
(ii) Apakah yang berlaku kepada mentol X apabila = 1.67 V
rintangan X ialah 1 kΩ?
What happens to the bulb if X has a resistance of V < V
1 kΩ? X minimum
Oleh itu, mentol tidak menyala.
So, the bulb does not light up.
147 © Nilam Publication Sdn. Bhd.
MODUL • Fizik TINGKATAN 5
Get Logik
4.4 Logic Gates
Terminologi Penerangan
Terminology Explanation
Get logik Satu litar elektronik yang mempunyai satu input atau lebih dan satu output.
Logic gate An electronic circuit with one or more inputs and a single output.
Jadual kebenaran Jadual yang menyenaraikan semua kemungkinan input dan output get logik.
Truth table A table that summarises all the possible inputs and outputs of a logic gate.
Algebra Boolean Ungkapan Boolean ialah suatu bentuk ungkapan yang boleh meringkaskan operasi logik suatu get
Boolean algebra logik.
The Boolean algebra is one form of expression which can be used to summarise the logic operation of a
logic gate.
1 Setiap input dan output boleh sama ada tinggi (logik 1) atau rendah (logik 0).
Each input and output can be either high (logic 1) or low (logic 0)
2 Sistem binari "0" mewakili 0 V dan binari "1" mewakili voltan bukan sifar.
A binary "0" represents 0 V, and a binary "1" represents a non-zero voltage.
3 Analogi bagi get-get logik
Analogy for logic gate:
(a) Get TAK / NOT gate Jadual kebenaran:
Truth table:
A X UNIT 4
UNIT 4
A X 0 1
1 0
X = A menyonsangkan
Simbol get TAK Get TAK sentiasa keadaan binari input
The symbol of NOT gate NOT gate always inverse the binary state of the input
Litar Analogi / Analogical circuit Keadaan suis Keadaan mentol Sebab
Condition of switch Condition of bulb Reason
Litar lengkap, arus mengalir
Buka (0) Nyala (1) melalui mentol
Open (0) Lights up (1) The circuit is complete, current
6 V flows through the bulb
Litar pintas, arus tidak mengalir
60 Ω Tutup (1) Padam (0) melalui mentol
Close (1) Off (0) Short circuit, current does not flow
through the bulb
© Nilam Publication Sdn. Bhd. 148
MODUL • Fizik TINGKATAN 5
(b) Get DAN / AND gate Jadual kebenaran: / Truth table:
A B X
A 0 0 0
B X
0 1 0
X = A. B 1 0 0
Simbol get DAN 1 1 1
The symbol of AND gate
Litar Analogi / Analogical circuit Keadaan suis / Condition of switch Keadaan mentol
A B Condition of the bulb
Buka (0) Buka (0) Tidak menyala (0)
Open (0) Open (0) Not lit (0)
A
Buka (0) Tutup (1) Tidak menyala (0)
Open (0) Close (1) Not lit (0)
B Tidak menyala (0)
Tutup (1) Buka (0)
Close (1) Open (0) Not lit (0)
Menyala (1)
Tutup (1) Tutup (1)
Close (1) Close (1) Lights up (1)
(c) Get ATAU / OR gate Jadual kebenaran: / Truth table:
A A B X
X 0 0 0
B 0 1 1 UNIT 4
X = A + B 1 0 1
UNIT 4
Simbol get ATAU 1 1 1
The symbol of OR gate
Litar Analogi / Analogical circuit Keadaan suis / Condition of switch Keadaan mentol
A B Condition of the bulb
Tidak menyala (0)
Buka (0) Buka (0)
Open (0) Open (0) Not lit (0)
A
Menyala (1)
Buka (0) Tutup (1)
Open (0) Close (1) Lights up (1)
B Menyala (1)
Tutup (1) Buka (0)
Close (1) Open (0) Lights up (1)
Menyala (1)
Tutup (1) Tutup (1)
Close (1) Close (1) Lights up (1)
149 © Nilam Publication Sdn. Bhd.
MODUL • Fizik TINGKATAN 5
Latihan / Exercise
1 Rajah menunjukkan get logik dengan input P dan Q.
The diagram shows a logic gate with inputs P and Q.
P
X
Q
Jika input P = 0011010 dan input Q = 1100011, apakah output X?
If the input P = 0011010 and the input Q = 1100011, what is the output X?
Penyelesaian
Solution
Input Output
P Q X
0 1 1
0 1 1
1 0 1
1 0 1
0 0 1
1 1 0
0 1 1
X = 1111101
2 Rajah menunjukkan dua isyarat yang menghubungkan input A dan B dalam suatu get OR.
The diagram shows two signals connected to the inputs A and B of an OR gate. UNIT 4
A
UNIT 4
Input A X
Input B
B
Lukiskan isyarat output X.
Draw the output signal X.
Penyelesaian
Solution
A
Input
Input
B
Output
Output
X
© Nilam Publication Sdn. Bhd. 150
MODUL • Fizik TINGKATAN 5
3 Rajah menunjukkan gabungan get logik dengan input A dan input B. Jika input A = 0011 dan input B = 0101, apakah output X?
The diagram shows the combination of logic gates with inputs A and B. If the input A = 0011 and the input B = 0101, what is the output X?
KBAT A
B
X
Penyelesaian
Solution
0011 0001
A 1110
B
0101 X 1110
0101
0011 1110
X = 1110
Aplikasi Get Logik dalam Sistem Kawalan
Applications of Logic Gates in Control Systems
1 Kipas automatik untuk sistem pendinginan
Automatic fan for cooling system
Pengesan cahaya Input / Input Output / Output
Light detector Pengesan
Kipas Pengesan suhu Kipas
Fan cahaya UNIT 4
Pengesan suhu Light detector Temperature detector Fan
Temperature
UNIT 4
detector 0 0 0
0 1 0
1 0 0
1 1 1
2 Sistem penggera kebakaran
Fire alarm system
Pengesan Input / Input Output / Output
asap
Smoke detector Pengesan asap Pengesan haba Bunyi isyarat
Bunyi isyarat Smoke detector Heat detector Siren
Siren
Pengesan
haba 0 0 Tidak aktif
Heat detector Not activated
Aktif
0 1
Activated
Aktif
1 0
Activated
Aktif
1 1
Activated
151 © Nilam Publication Sdn. Bhd.
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