Sci 1010 chapter 8

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Sci 1010 chapter 8

  1. 1. Chapter 8
  2. 2.  Electric charge is a fundamental quantity – we don’t really know what it is ◦ But we can describe it, use it, control it ◦ Electricity runs motors, lights, heaters, A/C, stereos, TV’s, computers, etc.  Electric Forces – at the microscopic level they hold atoms and molecules together ◦ Electric Forces hold matter together ◦ Gravitational Forces hold the universe together  Magnetism is also closely associated with electricity Intro
  3. 3.  Experimental evidence leads us to conclude that there are two types charges ◦ Positive (+) ◦ Negative (-)  All matter is composed of atoms, which in turn are composed of subatomic particles ◦ Electrons (-) ◦ Protons (+) ◦ Neutron (neutral) Section 8.1
  4. 4.  Note that the Proton and Neutron each have about 1000x more mass than the Electron  If the atom has the same number of protons and electrons it is electrically neutral Section 8.1
  5. 5.  Named after a French scientist – Charles Coulomb (1736-1806)  Note that the charge on a single electron (-) or proton (+) is 1.60 x 10-19 C (very small!)  q usually designates electric charge ◦ Excess of positive charges +q ◦ Excess of negative charges -q Section 8.1
  6. 6.  An electric force exists between any two charged particles – either attractive or repulsive Section 8.1
  7. 7.  Law of Charges – like charges repel, and unlike charges attract ◦ Two positives repel each other ◦ Two negatives repel each other ◦ Positive and negative attract each other  The force between two charged bodies is directly proportional to the product of the two charges & inversely proportional to the square of their distance apart. Section 8.1
  8. 8.  Force = Section 8.1 constant x charge 1 x charge 2 (distance between charges)2 λ In equation form F = kq1q2 / r2 λ k is a constant = 9.0 x 109 Nm2 /C2
  9. 9.  An object with an excess of electrons is said to be negatively charged  An object with a deficiently of electrons is said to be positively charged  Photocopier (xerography) – practical and widespread use of electrostatics Section 8.1
  10. 10. Two negative charges repel Two positive charges repel One negative and one positive attract Repulse Repulse Attract Section 8.1
  11. 11.  Force of attraction/repulsion between two charged bodies is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them  F = (kq1q2) / r2 ◦ F = force of attraction or repulsion ◦ q1 = magnitude of the first charge ◦ q2 = magnitude of the second charge ◦ r = distance between charges ◦ k = constant = 9.0 x 109 N-m2 /C2 Section 8.1
  12. 12.  Equations look similar ◦ F = kq1q2 / r2 & F = Gm1m2/ r2  Both depend on r2  Coulomb’s law can describe either an attractive or repulsive force – gravity is always positive  Electrical charges are much stronger than gravitational forces Section 8.1
  13. 13.  I = charge/time = q/t ◦ I = electric current (amperes) ◦ q = electric charge flowing past a point (coulombs) ◦ t = time for the charge to pass point (seconds)  1 ampere (A) = flow of 1 Coulomb per second  Rearrange equation above: ◦ q = It or 1 coulomb = 1 ampere x 1 second  Therefore, 1 coulomb is the amount of charge that flows past a given point in 1 second when the current is 1 ampere Section 8.2
  14. 14.  Electrical conductor – materials in which an electric charge flows readily (most metals, due to the outer, loosely bound electrons)  Electrical insulator – materials that do not conduct electricity very well due to very tight electron bonding (wood, plastic, glass)  Semiconductor – not good as a conductor or insulator (graphite) Section 8.2
  15. 15.  A wire carries a current of 0.50 A for 2 minutes. a) how much (net) charge goes past a point in the wire in this time? b) how many electrons make up this amount of charge?  GIVEN: I = 0.50 A, t = 2 minutes (120 seconds)  WANTED: q (charge) & n (number of electrons)  (a) q = It = (0.50 A)(120 s) = 60 C (coulombs) Section 8.2
  16. 16.  A wire carries a current of 0.50 A for 2 minutes. a) how much (net) charge goes past a point in the wire in this time? b) how many electrons make up this amount of charge?  To solve for (b), we know that each electron has a charge of 1.6 x 10-19 C and we know the total charge from part (a) = 60 Coulombs  n = 60/(1.6 x 10-19 C / electron) = 3.8 x 1020 electrons Section 8.1
  17. 17.  When work is done to separate positive and negative charges, we have electric potential energy Section 8.2
  18. 18.  Instead of measuring electric potential energy, we measure the potential difference, or voltage  Voltage – the amount of work it would take to move a charge between two points, divided by the value of the charge  Voltage = work / charge = V = W/q  Measured in volts (V) = 1 joule/Coulomb  When we have electric potential energy, this may be used to set up an electrical current Section 8.2
  19. 19.  Whenever there is an electrical current, there is resistance (R) within the conducting material ◦ R is due to atomic/subatomic collisions  Georg Ohm (1787-1854) – formulated a simple relationship between voltage, current, and resistance  Ohm’s Law  V = IR ◦ V = voltage in volts, I = current in amperes, and R = resistance in ohms  1 ohm = 1 volt/1 ampere (R=V/I) Section 8.2
  20. 20.  Electrons flow from negative terminal to positive terminal (provided by the chemical energy of the battery) -- negative to positive  Open switch – not a complete circuit and no flow of current (electrons)  Closed switch – a complete circuit and flow of current (electrons) exists  Closed Circuit Required – to have a sustained electrical current Section 8.2
  21. 21. The light bulb offers resistance. The kinetic energy of the electric energy is converted to heat and radiant energy. Section 8.2
  22. 22. Section 8.2
  23. 23.  Recall: P = W/t (Power = work/time) – Ch. 4  V = W/q (Voltage = work/charge) – Ch. 8  Rearrange V=W/q  W = qV  Substitute W = qV into P = W/t equation  Result  P = qV/t  Recall that I = q/t (Current = charge/time)  ∴ P = IV (Electric power = current x voltage)  also P = I2 R (substitute in V = IR)  P in watts, I in amperes, R in ohms, V in volts Section 8.2
  24. 24.  Find the current and resistance of a 60-W, 120-V light bulb in operation.  Given: P = 60W, V = 120 V  Find: I (current in amperes), R (resistance in ohms)  Since P = IV  I = P/V = 60 W/120 V = 0.50 A  Since V = IR  R = V/I = 120 V/0.50A = 240 Ω  Since P = I2 R  R = P/I2 = 60 W/(0.50 A)2 = 240 Ω (same answer as above) Section 8.2
  25. 25.  A coffeemaker draws 10 A of current operating at 120 V. How much electrical energy does the coffeemaker use each second.  Given: I = 10 A, V = 120 V  Find: P (electrical energy in watts)  P = IV  P = (10 A) x (120 V) = 1200 W or 1200 J/s Section 8.2
  26. 26.  Current: I = q/t ◦ I = current (amperes), ◦ q = electric charge (coulombs), ◦ t = time (seconds)  Coulomb’s Law: F = kq1q2/r2  Voltage: V = W/q [Work is in joules (J)]  Ohm’s Law: V= IR  Electrical Power: P = IV = I2 R Section 8.2
  27. 27.  Direct Current (dc) – the electron flow is always in one direction, from (-) to (+) ◦ Used in batteries and automobiles  Alternating Current (ac) – constantly changing the voltage from positive to negative and back ◦ Used in homes. ◦ 60 Hz (cycles/sec) and Voltage of 110-120 V Section 8.3
  28. 28.  Plugging appliances into a wall outlet places them in a circuit – either in series or in parallel  In a series circuit, the same current (I) passes through all the resistances. ◦ The total resistance is simply the sum of the individual resistances. Section 8.3
  29. 29.  Rs = R1+R2+R3+…  If one bulb were to burn out the circuit would be broken, and all the lights would go out Section 8.3 Same current all the way
  30. 30.  In a parallel circuit, the voltage (V) across each resistance is the same, but the current (I) through each may vary. Section 8.3
  31. 31. Current Divides, Voltage is the same Section 8.3 I = I1+I2+I3+… for R in parallel: 1/Rp = 1/R1+1/R2+… for 2 R in parallel: Rp = (R1R2)/(R1+ R2)
  32. 32.  Three resistors have values of R1=6.0Ω, R2=6.0Ω, and R3=3.0Ω. What is their total resistance when connected in parallel, and how much current will be drawn from a 12-V battery if it is connected to the circuit?  1/Rp = 1/R1+1/R2+1/R3 = 1/6 + 1/6 + 1/3  1/Rp = 1/6 + 1/6 + 2/6 = 4/6 Ω  Rp = (6 Ω)/4 = 1.5 Ω = Total Resistance  I = V/Rp = (12 V)/(1.5 Ω) = 8.0 A = current drawn Section 8.3
  33. 33.  Wired in parallel independent branches any particular circuit element can operate when others in the same circuit do not. Section 8.3
  34. 34.  Wires can become hot as more and more current is used on numerous appliances.  Fuses are placed in the circuit to prevent wires from becoming too hot and catching fire.  The fuse filament is designed to melt (and thereby break the electrical circuit) when the current gets too high.  Two types of fuses: Edison and S-type  Circuit Breakers are generally now used. Section 8.3
  35. 35.  A dangerous shock can occur if an internal ‘hot’ wire comes in contact with the metal casing of a tool.  This danger can be minimized by grounding the case with a dedicated wire through the third wire on the plug. Section 8.3
  36. 36. Section 8.3
  37. 37. Section 8.3
  38. 38.  Closely associated with electricity is magnetism.  In fact electromagnetic waves consists of both vibrating electric and magnetic fields. These phenomena are basically inseparable.  A bar magnet has two regions of magnetic strength, called the poles.  One pole is designated “north,” one “south.” Section 8.4
  39. 39.  The N pole of a magnet is “north-seeking” – it points north.  The S pole of a magnet is “south-seeking” – it points south.  Magnets also have repulsive forces, specific to their poles, called …  Law of Poles – Like poles repel and unlike poles attract ◦ N-S attract ◦ S-S & N-N repel Section 8.4
  40. 40. All magnets have two poles – they are dipoles Section 8.4
  41. 41.  Magnetic field - a set of imaginary lines that indicates the direction in which a small compass needle would point if it were placed near a magnet  These lines are indications of the magnetic force field.  Magnetic fields are vector quantities. Section 8.4
  42. 42.  The arrows indicate the direction in which the north pole of a compass would point. Section 8.4
  43. 43.  The source of magnetism is moving and spinning electrons.  Hans Oersted, a Danish physicist, first discovered that a compass needle was deflected by a current- carrying wire. ◦ Current open  deflection of compass needle ◦ Current closed  no deflection of compass needle  A current-carrying wire produces a magnetic field: stronger current  stronger field  Electromagnet – can be switched on & off Section 8.4
  44. 44.  Most materials have many electrons going in many directions, therefore their magnetic effect cancels each other out  non-magnetic  A few materials are ferromagnetic (iron, nickel, cobalt) – in which many atoms combine to create magnetic domains (local regions of magnetic alignment within a single piece of iron)  A piece of iron with randomly oriented magnetic domains is not magnetic. Section 8.4
  45. 45.  The magnetic domains are generally random, but when the iron is placed in a magnetic field the domains line up (usually temporarily). Section 8.4
  46. 46.  A simple electromagnet consists of an insulated coil of wire wrapped around a piece of iron.  Stronger current  stronger magnet  Electromagnets are made of a type of iron that is quickly magnetized and unmagnetized – termed “soft.” Section 8.4
  47. 47.  Materials cease to be ferromagnetic at very high temperatures – the “Curie temperature” (770o C for iron)  Permanent magnets are made by permanently aligning the many magnetic domains within a piece of material.  One way to create a permanent magnet is to heat the ferromagnetic material above the Curie temperature and then cool the material under the influence of a strong magnetic field. Section 8.4
  48. 48.  This planet’s magnetic field exists within the earth and extends many hundreds of miles into space.  The aurora borealis (northern lights) and aurora australis (southern lights) are associated with the earth’s magnetic field.  Although this field is weak compared to magnets used in the laboratory, it is thought that certain animals use it for navigation. Section 8.4
  49. 49.  Similar to the pattern from a giant bar magnet being present within the earth (but one is not present!) Section 8.4
  50. 50.  The origin of the earth’s magnetic field is not completely understood. ◦ Probably related to internal currents of electrically charge particles in the liquid outer core of the earth, in association with the earth’s rotation  The temperatures within the earth are much hotter than the Curie temperature, so materials cannot be ferromagnetic.  The positions of the magnetic poles are constantly changing, suggesting “currents.” Section 8.4
  51. 51.  Magnetic declination – the angle between geographic (true) north and magnetic north  The magnetic declination varies depending upon one’s location on Earth. ◦ In the northern hemisphere the magnetic North Pole is about 1500 km from the geographic North Pole.  Therefore, a compass does not point to true north, but rather magnetic north. ◦ An adjustment (magnetic declination) must be made to determine true north from a compass. Section 8.4
  52. 52. Section 8.4
  53. 53.  Resistance in Series: Rs = R1+R2+R3+…  Resistance in Parallel: 1/Rp = 1/R1+1/R2+…  For 2 R in parallel: Rp = (R1R2)/(R1+R2) Section 8.4
  54. 54.  Electromagnetism – the interaction of electrical and magnetic effects  Two basic principles: 1) Moving electric charges (current) give rise to magnetic fields (basis for an electromagnet). 2) A magnetic field will deflect a moving electric charge (basis for electric motors and generators). Section 8.5
  55. 55.  Telephone transmission involves the conversion of sound waves into varying electric current  Sound waves at the transmitting end vibrate a diaphragm that puts pressure on carbon grains  The varying pressure on the carbon, results in varying amounts of resistance in the circuit  By Ohm’s law (V=IR), varying resistance results in varying current Section 8.5
  56. 56.  The varying current gives an electromagnet at the receiving end varying magnetic strengths  The electromagnet is drawn toward/away a permanent magnet as a function of the electric current in the line  The electromagnet is attached to a diaphragm that vibrates, creating sound waves that closely resemble the original sound waves Section 8.5
  57. 57. Section 8.5
  58. 58.  When a moving charge (a current) enters a magnetic field, the moving charge will be deflected by the magnetic field  The magnetic force (Fmag) is perpendicular to the plane formed by the velocity vector (v) of the moving charge and the magnetic field (B) Section 8.5
  59. 59.  Electrons entering a magnetic field experience a force Fmag that deflects them “out of the page” Section 8.5
  60. 60.  A beam of electrons is deflected downward due to the introduction of the magnetic field of a bar magnet Section 8.5
  61. 61.  Since a magnetic field deflects a moving charge, a current-carrying wire loop will experience a force Section 8.5
  62. 62. Section 8.5
  63. 63. Section 8.5
  64. 64.  Electric motor – a device that converts electrical energy into mechanical work  When a loop of coil is carrying a current within a magnetic field, the coil experiences a torque and rotates Section 8.5
  65. 65.  The inertia of the spinning loop carries it through the positions where unstable conditions exist  The split-ring commutator reverses the loop current every half-cycle, enabling the loop to rotate continuously Section 8.5
  66. 66.  When a wire is moved perpendicular to a magnetic field, the magnetic force causes the electrons in the wire the move – therefore a current is set up in the wire Section 8.5
  67. 67.  Generator – a device that converts mechanical energy into electrical energy  Generators operate on the principle of electromagnetic induction  Electromagnetic induction was discovered by the English scientist, Michael Faraday in 1831 ◦ When a magnet is moved toward a loop or coil of wire, a current is induced in the wire ◦ The same effect is obtained if the magnetic field is stationary and the loop is rotated within it Section 8.5
  68. 68.  A current is induced in the circuit as the magnet is moved toward and into the coil Section 8.5
  69. 69.  When the loop is rotated within the magnetic field a current is induced in the loop  The current varies in direction every half-cycle and is termed alternating current (ac) Section 8.5
  70. 70.  Transformers serve to step-up or step-down the voltage of electric transmissions  A transformers consists of two insulated coils of wire wrapped around an iron core ◦ The iron core serves to concentrate the magnetic field when there is a current in the input coil  The magnetic field resulting from the primary coil goes through the secondary coil and induces a voltage and current Section 8.5
  71. 71.  Since the secondary coil has more windings, the induced ac voltage is greater than the input voltage.  The factor of voltage step-up depends on the ratio of the windings on the two coils. Section 8.5
  72. 72.  The main reason is to reduce the amount of current going through the lines ◦ As the current is reduced, the amount of resistance is also reduced ◦ If the amount of resistance is reduced, the amount of heat (called ‘joule heat’ or ‘I2 R’ losses)  For example, if the voltage is stepped-up by a factor of 2, the resulting I2 R losses would be reduced four-fold Section 8.5
  73. 73.  The voltage change for a transformer is given by the following:  V2 = (N2/N1)V1  V1 = primary voltage  V2 = secondary voltage  N1 = Number of windings in primary coil  N2 = Number of windings in secondary coil Section 8.5
  74. 74.  A transformer has 500 windings in its primary coil and 25 in its secondary coil. If the primary voltage is 4400 V, find the secondary voltage.  Use equation V2 = (N2/N1)V1  Given: N1 = 500, N2 = 25, V1 = 4400 V  V2 = (25/500)(4400 V) = 220 V  This is typical of a step-down transformer on a utility pole near homes Section 8.5
  75. 75. Voltage is dramatically stepped-up at the generating plant to minimize joule heat loss during long-distance transmission. The voltage must then be stepped-down for household use. Section 8.5
  76. 76.  F = kq1q2/r2 Coulomb’s Law  I = q/t Current  V= W/q Voltage  V= IR Ohm’s Law  P= IV = I2 R Electric Power Review
  77. 77.  Rs = R1 + R2 + R3… Resistance in Series  1/Rp = 1/R1 + 1/R2 + 1/R3 …. Resistance in Parallel  Rp = (R1R2)/(R1 + R2) Two Resistances in Parallel  V2 = (N2/N1)V1 Transformer (Voltages and Turns) Review

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