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- 1. ELECTRICITY
- 2. LEARNING OBJECTIVE <ul><li>1.1 Analysing electric field and charge flow. </li></ul><ul><li>1.2 Analysing the relationship between electric current and potential difference. </li></ul><ul><li>1.3 Analysing series and parallel circuit. </li></ul><ul><li>1.4 Analysing electromotive force and internal resistance. </li></ul><ul><li>1.5 Analysing electrical energy and power </li></ul>
- 3. <ul><li>Matter is made up of tiny particles called atoms . </li></ul>Negative charge Positive charge No charge / Neutral -1.6 × 10 -19 C +1.6 × 10 -19 C
- 4. Electric charge, Q <ul><li>Unit : coulomb, C </li></ul><ul><li>Charge on one electron is = -1.6 ×10 -19 C </li></ul><ul><li>Charge on one proton is = 1.6 ×10 -19 C </li></ul><ul><li>A body is : </li></ul><ul><li>Neutral : if it has equal number of +ve and -ve charge </li></ul><ul><li>Charge negative : if has more –ve than +ve charge (atoms gain s electron) </li></ul><ul><li>Charge positive : if has more +ve than –ve charge (atoms losses electron) </li></ul>Q = ne e = charged for 1 electron ( -1.6 ×10 -19 C) n = no.of electron
- 5. Electric current, I <ul><li>Electric current is the rate of flow of electric charge . </li></ul><ul><li>Unit : Ampere, A </li></ul>t = time (seconds)
- 6. Relationship between electron flow and electric current <ul><li>A flow of electric charge (electron) through a conductor produces an electric current. </li></ul>
- 8. Exercise 1 <ul><li>The current flows in a light bulb is 0.5 A. </li></ul><ul><li>(a) Calculate the amount of electric charge that flows through the bulb in 2 hours. </li></ul><ul><li>(b) If one electron carries a charge of 1.6 x 10 -19 C, find the number of electrons transferred through the bulb in 2 hours. </li></ul>Exercise 2 Electric charges flow through a light bulb at the rate of 20 C every 50 seconds. What is the electric current shown on the ammeter? Exercise 3 When lightning strikes between two charged clouds, an electric current of 400 A flows for 0.05 s. What is the quantity of charge transferred? 3600 C 2.25×10 22 0.4 A 20 C
- 9. Electric Field <ul><li>Electric field is a region which an electric charge experience an electric force (attraction or repulsion). </li></ul><ul><li>An electric field is represented by a series of arrowed lines called electric field lines . </li></ul><ul><li>The lines indicates both magnitude and direction of the field. </li></ul><ul><li>Electric field lines NEVER CROSS each other. </li></ul><ul><li>Electric field lines are most dense around objects with the greatest amount of charge . </li></ul>
- 11. + - + + - - - + + + + + + - a) d) c) b) e) f) h)
- 12. 1.2 Analysing the relationship between electric current and potential difference.
- 13. Potential difference, V <ul><li>Definition: Work done when 1 C of charge moves between two points in an electric field . </li></ul><ul><li>Unit : Volt, V </li></ul><ul><li>Potential difference can be measured using voltmeter </li></ul><ul><li>Voltmeter must be connected in parallel between the two point concerned. </li></ul>OR
- 14. What is 1 volt? <ul><li>1 Volt = 1 joule per coulomb. </li></ul><ul><li>The potential difference across two points in a circuit is 1 Volt if 1 Joule of work is done in moving 1 Coulomb of charge from one point to the other. </li></ul>
- 15. Exercises 1 <ul><li>In a closed circuit, a 6 V battery is used to drive 40 C of electric charge through a light bulb. How much work is done to drive the electric charge through the bulb? </li></ul>Exercises 2 If 72 J of work has to be done to carry 6 C of charge across two parallel metal plates, what is the potential difference across the metal plates? 240 J 12 V
- 16. Relationship between current and potential difference <ul><li>The greater the potential difference or voltage, the greater the current flow. </li></ul><ul><li>The potential difference is directly proportional to the current flowing through it. </li></ul>
- 17. Ohm’s Law <ul><li>States that the electric current, I flowing through a conductor is directly proportional to the potential difference, V across it if the temperature and other physical conditions are constant. </li></ul>
- 18. Ohmic conductor and non-ohmic conductor <ul><li>Conductors that obey Ohm’s Law are called ohmic conductors . </li></ul><ul><li>Conductors that do not obey Ohm’s Law are called non-ohmic conductors . </li></ul>Non-Ohmic conductor Ohmic conductor
- 19. Resistance, R <ul><li>Definition: Ratio of the potential difference (V) across the conductor to the current (I) flowing through it. </li></ul><ul><li>Unit : Ohm, Ω // VA -1 </li></ul><ul><li>A measure of how much a conductor resists the flow of electricity. A good conductor has a low resistance and a poor conductor has a high resistance. </li></ul>
- 20. Disadvantage of resistance <ul><li>Resistance causes some of the electrical energy to turn into heat , so some electrical energy is lost along the way if we are trying to transmit electricity from one place to another through conductor. </li></ul>Advantage of resistance <ul><li>Resistance allows us to use electricity for heat and light. </li></ul><ul><li>The heat is generated from electric heaters or the light that we get from light bulbs is due to the resistance of the wire. </li></ul><ul><li>In a light bulb, the current flowing through a resistance filament causes it to become hot and then glow. </li></ul>
- 21. Factors affecting resistance <ul><li>Length of wire </li></ul><ul><li>Cross-sectional area of wire </li></ul><ul><li>Temperature </li></ul><ul><li>Type of material </li></ul>
- 22. a) Length of wire <ul><li>R is directly proportional to its length, l </li></ul><ul><li>R ∝ l </li></ul>b) Cross-sectional area of wire <ul><li>R is inversely proportional to its cross sectional area, A. </li></ul>
- 23. c) Temperature <ul><li>The resistance, R generally increases with temperature. </li></ul>d) Type of wire <ul><li>The resistance of a wire depends on the material it is made from. </li></ul>
- 24. Superconductor <ul><li>A superconductor is a material whose resistance becomes zero when its temperature drops to a certain value called the critical temperature . </li></ul>
- 26. 1.3 Analysing Series and Parallel circuit.
- 27. SERIES CIRCUIT PARALLEL CIRCUIT I 1 = I 2 = I 3 I = I 1 +I 2 + I 3 V = V 1 + V 2 + V 3 V = V 1 = V 2 =V 3 R = R 1 + R 2 + R 3 1/R = 1/R 1 + 1/R 2 + 1/R 3
- 28. Compare the brightness
- 29. SERIES CIRCUIT PARALLEL CIRCUIT The brightness of each bulb is equally the same. Brightness of each bulb is greater If one bulb blow out, the circuit is broken. V = V 1 = V 2 =V 3 R = R 1 + R 2 + R 3 1/R = 1/R 1 + 1/R 2 + 1/R 3
- 30. Electromotive force (e.m.f), E <ul><li>The electromotive force, E (e.m.f.) is defined as the work done by a source in driving one coulomb of charge around a complete circuit. </li></ul><ul><li>Unit : volt, V @ JC -1 </li></ul>What does the label 1.5 V on the battery mean? <ul><li>The voltage label on a battery or cell indicates its e.m.f that is 1.5 V. </li></ul><ul><li>A cell has an e.m.f. of 1.5 V if a flow of 1 C of charge produces 1.5 J of electrical energy to the whole circuit. </li></ul>
- 31. Compare e.m.f. and potential difference
- 32. Electromotive force, E Potential difference, V Indicates the electrical energy given to 1 C of charge flowing through the cell or source . Indicates the electrical energy that is transformed to other forms of energy when 1 C of charge passes between two point in a circuit. Represented by the voltmeter reading in an open circuit (when switch is opened) Represented by the voltmeter reading in a closed circuit (when switch is closed) Measured in JC -1 or Volts,V Measured in JC -1 or Volts,V Formula E = I (R + r) (R+r) = total external resistance and total internal resistance Formula V = IR R = Total external resistance
- 33. Internal resistance, r <ul><li>“ Definition: Internal resistance of a source or cell is the resistance against the moving charge due to the electrolyte” </li></ul>
- 34. How to measure internal resistance? <ul><li>E = V + Ir </li></ul><ul><li>E = IR + Ir @ I(R+r) </li></ul>
- 35. Graph V against I <ul><li>E = V + Ir </li></ul><ul><li>V = -Ir + E </li></ul><ul><li>Compare y = mx + c </li></ul><ul><li>∴ E = y-intercept </li></ul><ul><li>r = gradient </li></ul>V /V I /A E - V E V I
- 36. Exercises <ul><li>1. A bulb M is connected to a battery by means of a switch. A voltmeter is also connected across the battery. When the switch is open, the voltmeter reads 6.0 V. When the switch is closed, the voltmeter reads 4.8 V. </li></ul><ul><li>(a) What is the e.m.f. of the battery? </li></ul><ul><li>(b) If the resistance of the bulb M is 8 Ω, what is the current passing through M when the switch is closed? </li></ul><ul><li>(c) Find the value of the internal resistance, r, of the battery. [2 Ω] </li></ul>
- 37. Exercises <ul><li>When switch S is opened, the voltmeter reading is 1.5 V. When the switch is closed, the voltmeter reading is 1.35 V and the ammeter reading is 0.3 A. </li></ul><ul><li>Calculate: </li></ul><ul><li>(a) e.m.f </li></ul><ul><li>(b) internal resistance </li></ul><ul><li>(c) resistance of R </li></ul>
- 38. Exercises <ul><li>Determine: </li></ul><ul><li>(a) the e.m.f of the cell </li></ul><ul><li>(b) the internal resistance of the cell </li></ul>
- 39. Comparison between total e.m.f and total internal resistance in a series and parallel circuit <ul><li>Each cell has E = 1.5 V and r = 0.5 Ω </li></ul>E 1 E 2 E 1 E 2 E = E 1 + E 2 Total e.m.f = 2 x 1.5 = 3.0 V E = E 1 = E 2 Total e.m.f = 1.5 V r = r 1 + r 2 Total r = 2 x 0.5 = 1.0 Ω 1/r = 1/ r 1 + 1/ r 2 Total 1/r = 1/0.5 + 1/0.5 r = 0.25 Ω
- 40. These appliances are labeled with values such as 240 V 60 W..What do these labels mean?
- 41. Electrical energy and Electrical Power
- 42. E /J V/ V Q /C E /J P/W t/s Electrical energy, E Electrical power, P defined as the ability of the electric current to do work. defined as the rate of energy dissipated or transferred Since V = E/Q Q =It V = IR Since P = E/t E = VIt V = IR Then , E = VIt Then, P = VI and E = I 2 Rt and P = I 2 R P = V 2 R
- 43. Exercise <ul><li>1) An electric kettle is connected across a 240 V power supply. If the resistance of the heating element is 40 Ω, calculate </li></ul><ul><li>(a) the current flowing through the element [6 A] </li></ul><ul><li>(b) the quantity of heat produced in 10 minutes [8.64×10 5 J] </li></ul><ul><li>2) An immersion heater has a power rating of 240 V, 750 W. </li></ul><ul><li>(a) What is the meaning of its power rating? </li></ul><ul><li>(b) What is the resistance of the immersion heater? [76.8 Ω ] </li></ul><ul><li>(c) What is the electrical energy consumed in 15 minutes? [6.75×10 5 J] </li></ul>
- 44. Power rating and Energy Consumption of Electrical Appliances <ul><li>An electrical kettle which is marked 240 V 1500 W means that the electric kettle will consume 1500 J of electrical energy every 1 second if it is connected to the 240 V. </li></ul><ul><li>Energy consumed = Power rating × Time </li></ul><ul><li> E = Pt </li></ul>E :unit Joule P :unit Watt T :unit seconds
- 45. <ul><li>The larger the power rating in the electrical appliance, the more energy is used every second. </li></ul><ul><li>The longer the usage time, the more electrical energy is consumed. </li></ul>
- 46. How to calculate the cost of electrical energy? <ul><li>Cost = number of units × cost per unit </li></ul>Example An appliance with a power of 2 kW is used for 10 minutes, three times a day. If the cost of electricity is 25 cents per unit, what is the cost of operating the appliance in the month of April? [RM 7.50]
- 47. Efficient Use of Electrical Energy <ul><li>Efficiency is a percentage of the output power to the input power. </li></ul><ul><li>Efficiency = Energy output x 100% </li></ul><ul><ul><li> Energy input </li></ul></ul><ul><li> = Power output x 100% </li></ul><ul><li> Power input </li></ul><ul><li>The efficiency of an electrical appliance is always less than 100% as some energy is lost in the form of heat and sound. </li></ul>
- 48. Exercise <ul><li>1) A lamp is marked ‘240 V, 100 W’. What is the efficiency of the lamp if it produces a light output of 12 W? [12 %] </li></ul>

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