Chapter1: Electricity


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Chapter1: Electricity

  2. 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. 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. 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. 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. 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>
  7. 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
  8. 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>
  9. 11. + - + + - - - + + + + + + - a) d) c) b) e) f) h)
  10. 12. 1.2 Analysing the relationship between electric current and potential difference.
  11. 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
  12. 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>
  13. 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
  14. 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>
  15. 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>
  16. 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
  17. 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>
  18. 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>
  19. 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>
  20. 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>
  21. 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>
  22. 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>
  23. 26. 1.3 Analysing Series and Parallel circuit.
  24. 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
  25. 28. Compare the brightness
  26. 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
  27. 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>
  28. 31. Compare e.m.f. and potential difference
  29. 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
  30. 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>
  31. 34. How to measure internal resistance? <ul><li>E = V + Ir </li></ul><ul><li>E = IR + Ir @ I(R+r) </li></ul>
  32. 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
  33. 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>
  34. 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>
  35. 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>
  36. 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
  37. 40. These appliances are labeled with values such as 240 V 60 W..What do these labels mean?
  38. 41. Electrical energy and Electrical Power
  39. 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
  40. 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>
  41. 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
  42. 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>
  43. 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]
  44. 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>
  45. 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>