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Physics KSSM F5 Zen Notes by Cikgu Heery

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Heery’s Zen Notes

Physics KSSM F5

Chapter 3: ELECTRICITY


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A. ELECTRICAL FIELD/ MEDAN ELEKTRIK

1. What is electical field/ apa itu medan elektrik?

Area around a charge, where a force is generated with another charge Kawasan sekitar satu cas, dimana satu daya terhasil dengan cas lain

2. Pattern of electrical field/ Corak medan elektrik

- pattern is like magnetic field/ corak macam medan magnet - direction is always from positive to negative charge/ arah sentiasa daripada cas positif ke negatif


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Extra pattern/ corak tambahan:

3. Calculating the electrical field strength, E/ Kira kekuatan medan elektrik, E

BETWEEN 2 CHARGED POINTS ANTARA 2 TITIK CAS

BETWEEN 2 CHARGED PLATES ANTARA 2 PLAT BERCAS

E = force (N)/ total charges (C) Unit of strength is N/C or NC-1

E = Potential difference (V) / distance,d (m) Unit of strength is V/m or Vm-1

E = daya (N)/ jumlah cas (C) Unit kekuatan ialah N/C atau NC-1

E = Beza keupayaan (V)/ jarak,d (m) Unit kekuatan ialah V/m atau Vm-1


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QUESTION EXAMPLE/ CONTOH SOALAN:

Force = 700 N Charge for one proton = 1.6 x 10-19 C Calculate the electrical field strength with 20 protons Daya = 700N Cas untuk satu proton = 1.6 x 10-19 C Kirakan kekuatan medan eletrik dengan 10 proton

Answer/ jawapan: 1 charge is 1.6 x 10-19 C So, 20 charges = 1.6 x 10-19 x 20 = 3.2 x 10-18 C Strength = force / total charges = 700N / 3.2 x 10-18 C = 2.19 x 1020 NC-1

1 cas ialah 1.6 x 10-19 C Maka, 20 cas = 1.6 x 10-19 x 20 = 3.2 x 10-18 C Kekuatan = daya / jumlah cas = 700N / 3.2 x 10-18 C = 2.19 x 1020 NC-1

Using the info above, calculate the electrical field strength Menggunakan maklumat di atas, kira kekuatan medan elektrik

Answer/ jawapan: Strength = potential difference/ distance Kekuatan = beza keupayaan/ jarak = 30V/ 0.5m = 60Vm-1


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4. 2 experiments of electrical field/ 2 eksperimen medan elektrik

a. Ping pong:

a. First, positive plate attracts electrone from the ball Mula-mula, plat positif menarik elektron dalam bola.

b. Negative charges in the ball flow into positive plate Cas negatif dalam bola mengalir ke plat positif.

c. Ball will lose electrones, and become positively charged (diagram) Bola akan kehilangan cas negatif dan menjadi bercas positif (rajah)

a. The positive ball attracted to negative plate Bola positif tertarik ke plat negatif.

b. Negative charges from plate flow into the ball Cas negatif dari plat mengalir ke bola

c. Ball will receive electrones, and become negatively charged (diagram) Maka, bola akan menerima cas negatif dan menjadi bercas negatif (gambar sebelah).


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b. Candle experiment/ eksperimen lilin

a. Candle flame ionizes the air into positive and negative charges b. Heavier positive charges are attracted to negative plate c. Lighter negative charges are attracted to positive plate d. So, larger flame is seen more spread out to the negative plate

a. Api lilin mengionkan udara kepada cas positif dan negatif b. Cas positif yg lebih berat tertarik kpd plat negatif c. Cas negatif yg lebih ringan tertarik kpd plat positif d. Maka, api kelihatan lebih besar dan marak ke plat negatif


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5. Current/ arus:

Rate of charge flow over time/ Kadar aliran cas per masa

Formula: Current = Total charges / time I = Q/t

Q = It

Unit of current is Ampere (A)

Formula: Arus = Jumlah cas / masa I = Q/t

Q = It Unit arus ialah Ampere (A)


1.5 x 105 electrons flow in a conductor for 2 minutes. Calculate the current in the conductor (charge of one electron = 1.6 x 10-19C) 1.5 x 105 elektron mengalir dalam satu konduktor selama 2 minit. Kirakan arus dalam konduktor tersebut (cas satu elektron = 1.6 x 10-19C)

Calculation: 1 charge of electron = 1.6 x 10-19 C So, 1.5 x 105 electrons = 1.6 x 10-19 x 1.5 X 105 = 2.4 x 10-14 C Current = Total charges / time = 2.4 x 10-14 C/ 120s = 2 x 10-16 A

Pengiraan: 1 cas elektron = 1.6 x 10-19 C Maka, 1.5 X 105 elektron = 1.6 x 10-19 x 1.5 X 105 = 2.4 x 10-14 C Arus = Jumlah cas / masa = 2.4 x 10-14 C/ 120s = 2 x 10-16 A


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6. Potential difference/ Beza keupayaan:

Energy/work used to move 1C of total charges between 2 points Tenaga/ kerja digunakan utk menggerakkan 1C cas antara 2 titik

Potential difference = Work or energy Total charges V = E/Q

E = QV Unit of PD is Volt (V)

Beza keupayaan = Kerja atau tenaga Jumlah cas V = E/Q

E = QV

Unit BK ialah Volt (V)

*We can insert Q = It into E = QV to create new formula E = VIt

E = Q x V E = It x V

E = VIt


1. 2.6 x 10-10J of energy is used to move 3.4 x 107 electrons flow in a conductor for 5 minutes. Calculate: a. potential difference b. current (charge of one electron = 1.6 x 10-19C) 1. 2.6 x 10-10J tenaga digunakan utk menggerakkan 3.4 x 107 elektron dalam satu konduktor selama 5 minit. Kirakan:


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a. beza keupayaan b. arus (cas satu elektron = 1.6 x 10-19C)

Calculation: 1 charge of electron = 1.6 x 10-19 C So, 3.4 x 107 electrons = 1.6 x 10-19 x 3.4 X 107 = 5.44 x 10-12 C a. PD = Energy / Total charges = 2.6 x 10-10J / 5.44 x 10-12 C = 47.8V b. Current = Total charges / time = 5.44 x 10-12 C / 300s = 1.81 x 10-14 A

Pengiraan: 1 cas elektron = 1.6 x 10-19 C Maka, 3.4 X 107 elektron = 1.6 x 10-19 x 3.4 X 107 = 5.44 x 10-12 C a. BK = Tenaga / Jumlah cas = 2.6 x 10-10J / 5.44 x 10-12 C = 47.8V b. Arus = Jumlah cas / masa = 5.44 x 10-12 C / 300s = 1.81 x 10-14 A

2. A PS5 labelled 240V of voltage and 2A of current is turned on for 2 hours. Calculate the total energy consumed 2. Satu PS5 berlabel voltan 240V dan arus 2A dihidupkan selama 2 jam. Kirakan jumlah tenaga yang digunakan

Energy/Tenaga, E = VIt = 240V X 2A X 7200s = 3.46 x 106 J


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B. RESISTANCE/ RINTANGAN

1. Every conductor is either Ohmic conductor or non-Ohmic conductor

Setiap konduktor adalah samada konduktor Ohm atau konduktor bukan Ohm

- Experiment/ eksperimen:

Adjust rheostat to change ammeter reading, then observe the voltmeter reading Selaraskan reostat untuk ubah bacaan ammeter, kemudian, perhatikan bacaan voltmeter Build table and plot graph voltan agains current Bina jadual dan plotkan graf voltan melawan arus

Result/ Keputusan:

Voltage is directly proportional to current Voltan berkadar langsung dengan arus

Voltage increases with current Voltan bertambah dengan arus

Conclusion: conductor is Ohmic Kesimpulan: konduktor bersifat Ohm

Conclusion: conductor is non Ohmic Kesimpulan: konduktor bukan bersifat Ohm


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2. 3 factors affecting resistance/ 3 faktor mempengaruhi rintangan:

Length of conductor

Panjang konduktor, ℓ

Area of conductor’s cross section Luas keratan rentas

konduktor, A

Wire resistance

Rintangan dawai, ρ

a. Conductor length experiment/ eksperimen panjang konduktor:

SHORT CONDUCTOR LONG CONDUCTOR

Ammeter = 4A Voltmeter = 12V

Resistance = 12V/4A = 3Ω


Resistance = 6V/4A = 1.5Ω

Hypothesis = The longer the conductor, the higher the resistance

Hipotesis = makin panjang konduktor, makin tinggi rintangan

b. Conductor thickness experiment/ eksperimen ketebalan konduktor:

THIN CONDUCTOR THICK CONDUCTOR


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Hypothesis = The thicker the conductor (larger cross section), the lower the

resistance

Hipotesis = makin panjang konduktor (keratan rentas makin besar), makin rendah rintangan

c. Wire resistance/ rintangan dawai

- all conductor have different wire resistance

- can be calculated using formula:

p = wire resistance (Ωm) R = resistance (Ω)

A = area of cross section area (m2) ℓ = conductor length (m)

*so, wire resistance is directly proportional to resistance – the higher the p, the

higher the resistance!

*maka rintangan wayar adalah berkadar langsung dengan rintangan – makin besar

rintangan wayar, makin tinggi rintangan!

- Application of wire resistance/ aplikasi rintangan wayar:


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heater coil = high wire resistance lingkaran pemanas = rintangan wayar tinggi

cable = low wire resistance kabel = rintangan wayar rendah

- to generate more heat to heat up water - can stand high temperature - utk jana lebih banyak haba utk panaskan air - dapat bertahan pada suhu tinggi

- more current flow into the heater - prevent cable from becoming too hot - utk alirkan lebih arus ke pemanas - elakkan kabel jadi terlalu panas

C. ELECTROMOTIVE FORCE (EMF)/ DAYA

GERAK ELEKTRIK (DGE)

1. What is emf?/ Apa itu dge?

Energy/work used to move 1C of total charges in a complete circuit Tenaga/ kerja digunakan utk menggerakkan 1C cas dalam litar lengkap

Complete circuit = dry

cell + conductor circuit

Litar lengkap = sel

kering + litar konduktor


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2. Let’s compare potential difference and emf/ Jom bezakan beza keupayaan dan

dge:

POTENTIAL DIFFERENCE EMF

Use energy to move 1C between 2 points in conductor

Use energy to move 1C in complete circuit

Symbol = V

Symbol = Ɛ

*so, PD is a part of EMF/ maka, BK ialah sebahagian drpd DGE

3. Eksperiment to find out emf/ eksperiment utk mengira dge:

- adjust rheostat until ammeter reading zero / bulb turns off

- now, voltmeter reading shows the emf

- selaraskan reostat hingga bacaan ammeter jadi sifar/ lampu padam

- sekarang, bacaan voltmeter menunjukkan dge


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4. Graph:

Information from the graph/ maklumat daripada graf:

a. Graph shows linear form with negative gradient Graf menunjukkan corak linear dengan kecerunan negatif

b. y-intercept shows the emf, Ɛ (highest voltage possible) pintasan-y menunjukkan dge, Ɛ (nilai voltan paling tinggi)

c. Gradient = y/x = V/I = r = internal resistance Kecerunan = y/x = V/I = r = rintangan dalam r = resistance in the dry cell circuit r= rintangan dalam litar sel kering

d. We can arrange the variables to find out the formula of emf Kita boleh susun pembolehubah utk mendapatkan formula dge:

Use linear graph formula, y = mx + c

y = y axis

V

m = gradient

V/I = -r

x = x axis

I

c = y-intercept

Ɛ

V = -Ir +Ɛ Ɛ = V + Ir


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5. How to calculate internal resistance / Cara kira rintangan dalam:

Dry cells arranged in a serial circuit Sel kering disusun secara litar siri

Arranged in paralled circuit Disusun secara litar selari

Total, rT

rT = r1 + r2 + r3

Total internal resistance, rT 1/rT = 1/r1 + 1/r2 + 1/r3 Advantage of arrangment/ kelebihan susunan - lower internal resistance/ rintangan dalam lebih rendah

D. ELECTRICAL ENERGY AND POWER/

TENAGA DAN KUASA ELEKTRIK

1. How to get power formula/ Bagaimana formula kuasa diterbitkan

Use power formula, Power = Work/ Energy Time

Electrical power, P = Electrical energy

Time

P = VIt t

P = IV

E = VIt

E = Pt

Unit of

power is

Watt (W)

Unit of

energy is

Joule (J)


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Replace V with IR to get another formula:

P = I x V = I x IR

P = I2R

2. Calculate the cost of electrical energy/ Mengira kos tenaga elektrik:

We use E = Pt and convert it into E = kWh

Energy cost, E = Power (kW) x time (hour) x cost per unit

QUESTION EXAMPLE:

1. A PS5 is labelled 350W and 240V. Calculate the ideal fuse to be installed Satu PS5 berlabel 350W dan 240V. Kirakan nilai fius yg sesuai From P= IV Current, I = P/V = 350W/ 240V = 1.46A So, suitable fuse = 1.5A

2. A light bulb is labelled 100W and 240V. Calculate: a. Resistance b. Energy consumed if turned on for 2 hours Satu mentol berlabel 10W dan 240V. Kirakan: a. Rintangan b. Tenaga digunakan jika dihidupkan selama 2 jam

a. Find current first, I = P/V = 10W/240V = 0.42A Then calculate resistance, R P = I2 R R = P/I2 = 10W/ 0.422 = 24Ω

b. Use E = Pt = 100W x 7200s = 7.2 x 105 J


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3. A fan labelled 30W and 250V is turned on for 3 hours every day. Calculate the cost of fan’s electrical energy for June bill if the cost per unit is 20 cent. Satu kipas berlabel 30W dan 250V dihidupkan selama 3 jam setiap hari. Kirakan kos penggunaan tenaga elektrik kipas untuk bil bulan Jun jika kos per unit ialah 20 sen Use formula E = kWh x cost per unit Energy cost, E = 0.03kW x 3 hours x 30 days x 20 cent = RM 0.54

--DONE_CHAPTER_3--

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