2. Norah Ali Al-moneef 1

3. Conductors & Semiconductors In conductors, the valence band is only partially-full, so electrons can easily move from being near one atom to being near anotherconductors In semiconductors and insulators, the valence band is completely full, so electrons must gain extra energy to movesemiconductors In semiconductors, the band gap between the full valence band and the empty conduction band is small, so electrons move easily with only thermal energy In insulators, the band gap is larger, so electrons will not easily move into the conduction band Norah Ali Al-moneef 2

4. Conductors & Insulators Electric current moves easily through some materials and less easily through other materials Materials that have very “tightly bound” electrons have few free electrons when an electric force is applied. These materials are insulators (e.g. rubber, glass, dry wood) Materials that allow the movement of a large number of free electrons are called conductors (e.g., silver, copper, aluminum) – Electrical energy is transferred through a conductor by means of the movement of free electrons that move from atom to atom – Displaced electrons continue to “bump” each other – The electrons move relatively slowly but this movement creates electrical energy throughout the conductor that is transferred almost instantaneously throughout the wire (e.g., billiard ball example, wind vs. sound example) Norah Ali Al-moneef 3

5. Electrons in an Electric Field Norah Ali Al-moneef 4 Conduction electrons move randomly in all directions in the absence of a field. If a field is applied, the electric force results in acceleration in a particular direction: F=ma= –eE  a = –eE/m As the charges accelerate, the potential energy stored in the electric field is converted to kinetic energy which can be converted into heat and light as the electrons collide with atoms in the wire This acceleration produces a velocity v = at = –eE t/m

6. ELECTRON MOTION IN A CONDUCTOR WITH AND WITHOUT AN ELECTRIC FIELD Norah Ali Al-moneef 5

7. 27.1 Electric Current Norah Ali Al-moneef 6 Whenever electric charges move, an electric current is said to exist The current is the rate at which the charge flows through a certain cross-section For the current definition, we look at the charges flowing perpendicularly to a surface of area A Charge in motion through an area A. The time rate of the charge flow through A defines the current (=charges per time): Units:1 C/s= 1 A SI unit of the current: Ampere Definition of the current:

8. Electrical current Norah Ali Al-moneef 7 If an electric field points from left to right, positive charge carriers will move toward the right while negative charges will move toward the left The result of both is a net flow of positive charge to the right. Current is the net change in positive charge per time Instantaneous current i = d q / d t Coulomb (C) – represents the total charge of approximately 6.25 x 10 18 electrons

9. Norah Ali Al-moneef 8 The direction of current flow is the direction positive charge would flow This is known as conventional (technical) current flow, i.e., from plus (+) to minus (-) However, in a common conductor, such as copper, the current is due to the motion of the negatively charged electrons It is common to refer to a moving charge as a mobile charge carrier A charge carrier can be positive or negative

10. Charge Carrier Motion in a Conductor The electric field force F imposes a drift on an electron’s random motion (10 6 m/s) in a conducting material. Without field the electron moves from P 1 to P 2. With an applied field the electron ends up at P 2 ’; i.e., a distance v d  t from P 2, where v d is the drift velocity (typically 10 -4 m/s). Norah Ali Al-moneef 9

11. Does the direction of the current depend on the sign of the charge? No! ( a) Positive charges moving in the same direction of the field produce the same positive current as (b) negative charges moving in the direction opposite to the field. Norah Ali Al-moneef 10 E E vdvd vdvd qvdqvd (-q)(-v d ) = qv d

12. Charged particles move through a conductor of cross-sectional area A  n is the number of charge carriers per unit volume V (=“concentration”)  nA  x=nV is the total number of charge carriers in V Norah Ali Al-moneef 11  The total charge is the number of carriers times the charge per carrier, q (elementary charge) ΔQ = (nAΔx)q [unit: (1/m 3 )(m 2 m)As=C] Microscopic model of current

13. Norah Ali Al-moneef 12 The drift speed, v d, is the speed at which the carriers move v d = Δ x/ Δ t Rewritten: Δ Q = (nAv d Δ t)q current, I = Δ Q/ Δ t = nqv d A ΔxΔx If the conductor is isolated, the electrons undergo ( thermal ) random motion When an electric field is set up in the conductor, it creates an electric force on the electrons and hence a current

14. Norah Ali Al-moneef 13  coulombs of charge pass a point in a wire every two seconds. Calculate current. Coulomb (C) – represents the total charge of approximately 6.25 x 10 18 electrons Unit of Current – Ampere (A) = 1coulomb/second  Example:

15. Norah Ali Al-moneef 14 An 18-gauge copper wire (diameter 1.02 mm) carries a constant current of 1.67 A to a 200 W lamp. The density of free electrons is 8.5  10 28 per cubic meter. Find the magnitudes of (a) the current density (b) the drift velocity. (a) A=d 2  /4=(0.00102 m) 2  /4=8.2  10 -7 m 2 J=I /A=1.67 A/(8.2  10 -7 m 2 )=2.0  10 6 A/m 2 (b) From J=I /A=nqv d v d =1.5  10 -4 m/s=0.15 mm/s

16.  Example: 15 Norah Ali Al-moneef

17. If 240 C of charge pass a point in a conductor in 5 min, what is the current through that point in the conductor? Convert 5 min to seconds 5.0min X 60s/1 min = 300s  Example: 16 Norah Ali Al-moneef

18. To the left  Example: 17 Norah Ali Al-moneef

19.  Example: 18 Norah Ali Al-moneef

20. II. Electric current 1. Definition Units: [ I ] = 1A = 1 C/s Conventional current Electron flow 10 20 electrons passed through the electric conductor during 4 seconds. Find the electric current through this conductor.  Example : The electric current of 0.5 A is flowing through the electric conductor. a) What electric charge is passing through the conductor during each second b) What electric charge will pass through the conductor during 1 minute? a) b) Norah Ali Al-moneef 19  Example:

21. 27.2 resistance Norah Ali Al-moneef Norah Ali Almoneef20 I = n q v d A – n = number of free charge carriers/unit volume Current density ( The current per unit cross-section is called the current density J) : Ohm's Law: E =  J J = σ E –  = resistivity –  = 1/  = conductivity – Good conductor: low  and high  Ohm's Law: – R = resistance } Measured in Volt/Ampere = Ohm (  ){ 20

22. In a homogeneous conductor, the current density is uniform over any cross section, and the electric field is constant along the length. Norah Ali Al-moneef 21 a b V=V a -V b =EL The ratio of the potential drop to the current is called resistance of the segment: Unit: 1V/A= 1  ohm  Resistance in a circuit arises due to collisions between the electrons carrying the current with the fixed atoms inside the conductor

23. Ohm’s Law Norah Ali Al-moneef 22 V  I  V=const.  I  V=RI Ohm’s Law is an empirical relationship that is valid only for certain materials Materials that obey Ohm’s Law are said to be ohmic I=V/R R , I  0, open circuit; R  0, I , short circuit The ratio of the potential drop to the current is called resistance of the segment: Unit: V/A=  ohm 

24. Resistivity and Resistance J = E/ρ where ρ is the resistivity I =  V/R Norah Ali Al-moneef 23 Consider a bar or wire of cross- section A and length L, carrying current I and with potential difference  V = V b - V a between the ends. I =  V/R and R = ρL/A is the resistance of the bar. also called Ohm’s Law. We know E =  V/L so I/A = J =  V/Lρ. Thus: ∆V = IR

25. Ohm’s Law, final Plots of V versus I for (a) ohmic and (b) nonohmic materials. The resistance R=V/I is independent of I for ohmic materials, as is indicated by the constant slope of the line in (a). Norah Ali Al-moneef 24 Ohmic Nonohmic

26. Norah Ali Al-moneef 25 Ohmic Resistors Metals obey Ohm’s Law linearly so long as their temperature is held constant Their resistance values do not fluctuate with temperature i.e. the resistance for each resistor is a constant Most ohmic resistors will behave non-linearly outside of a given range of temperature, pressure, etc.

27. On What Does Resistance Depend? If I increase the length of a wire, the current flow decreases because of the longer path If I increase the area of a wire, the current flow increases because of the wider path R =  L/A If I change to a material with better conductivity, the current flow increases because charge carriers move better If I change the temperature, the current flow changes Norah Ali Al-moneef 26

28. More on Resistance I =  V/R Units of Resistivity : ρ is in Ω-m (ohm-meters), so R is in (Ω-m)(m)/(m 2 ) = Ω (ohms) = V/A (volts/ampere) Resistivity ρ depends only upon the material (copper, silver…). Resistance R depends upon the material and also upon the dimensions of the sample (L, A). - R = ρL/A Note: Some devices (e.g.semiconductor diode) do not obey Ohm’s law! Norah Ali Al-moneef 27

29. Resistors are designed to have a specific resistance to reduce the amount of current going to a specific part of a circuit To obey Ohm’s law means a conductor has a constant resistance regardless of the voltage. V (Volts) A (Amps) R (Ohms) 28 Norah Ali Al-moneef

30. 29 Norah Ali Al-moneef

31. V = IR = 2A x 3  = 6v What voltage is required to produce 2a though a circuit with a 3  resistor. V 33 I = 2a  Example: 30 Norah Ali Al-moneef

32. Resistance I V I V Nonohmic device 2. Ohm’s Law Units: [ R ] = 1Ω = 1 V/AOhm’s Law: 31 Norah Ali Al-moneef

33. Resistivity L A I Definition: Temperature dependence of resistivity Example: What is the resistance of 1 m of nichrome wire of 2 mm diameter ? T 32 Norah Ali Al-moneef

34. 33 The drift speed is much smaller than the average speed between collisions When a circuit is completed, the electric field travels with a speed close to the speed of light Although the drift speed is on the order of 10 -4 m/s the effect of the electric field is felt on the order of 10 8 m/s

35. Drift Velocity In a conductor the free electrons are moving very fast in random directions (v ~ 10 6 m/sec) They collide with the atoms of the lattice and are scattered in random directions If an electric field is present, there is a slow net drift of electrons in the direction opposite the electric field v DRIFT ~ mm/sec Norah Ali Al-moneef 34

36. Amperes: the measure of the rate of current flow. – 6.24 × 10 18 electrons passing a point per second is equal to one amp. A current occurs whenever there is a source of electricity, conductors and a complete circuit. Norah Ali Al-moneef 35

37.  Example Norah Ali Al-moneef 36 What is the current flow in a circuit with a voltage of 120 volts and a resistance of 0.23  ?  Example With the increase in the length of the wire, the current increases. A.True B.False

38. V = IR 4v = I x 2  I = 2A A 4v battery is placed in a series circuit with a 2  resistor. What is the total current that will flow through the circuit? 4v 22 I = ?  Example 37 Norah Ali Al-moneef

39. V= IR V = 2A x 3  V = 6v What voltage is required to produce 2A though a circuit with a 3  resistor. V 33 I = 2A  Example 38 Norah Ali Al-moneef

40. V = IR 12 = 4 x R R = 3  What resistance is required to limit the current to 4 A if a 12 V battery is in the circuit? 12v 33 I = 4a  Example 39 Norah Ali Al-moneef

41.  Example Norah Ali Al-moneef 40 A cylindrical copper rod has resistance R. It is reformed to twice its original length with no change of volume. Its new resistance is: 1. R 2. 2R 3. 4R 4. 8R 5. R/2

42. Norah Ali Al-moneef 41 Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1 mm. Conductor B is a hollow tube of inside diameter 1 mm and outside diameter 2 mm. The ratio of their resistances R A /R B is 1. 1/2 2. 1 3. 2 4. 3 5. 4 B A

43. Norah Ali Al-moneef 42 Two cylinders are made of the same material and have the same length but different diameters. They are joined end-to-end and a potential difference is maintained across the combination. Which of the following quantities is the same for the two cylinders? 1. the potential difference 2. the current 3. the current density 4. the electric field 5. none of the above

44. Two cylindrical resistors, R 1 and R 2, are made of identical material. R 2 has twice the length of R 1 but half the radius of R 1. They are connected to a battery V as shown. Compare the currents flowing through R1 and through R2. A. I 1 < I 2 B. I 1 = I 2 C. I 1 > I 2 V I1I1 I2I2 Norah Ali Al-moneef 43  Example

45. Voltage and Current Relationship for Linear Resistors Norah Ali Al-moneef 44 Voltage and current are linear when resistance is held constant.

46. Which one of the following graphs correctly represents Ohm's law, where V is the voltage and I is the current? (a)A (b)B (c)C (d)D (e)A and C 45 Norah Ali Al-moneef

47. If a piece of wire has a certain resistance, which wire made of the same material will have a lower resistance? A )a hotter wire B ) a thicker wire C ) a longer wire D) a thinner wire ANS:B 46 Norah Ali Al-moneef

48. 47 27.4 Resistance and Temperature The resistivity (and hence resistance) varies with temperature. For metals, this dependence on temperature is linear over a broad range of temperatures. An empirical relationship for the temperature dependence of the resistivity of metals is given by Copper Resistance (R) is proportional to resistivity (  ): R =  L / A The resistivity (  ) depends on temperature and the physical properties of the material, so it has a different value for each material

49. Norah Ali Al-moneef 48 Some materials, when very cold, have a resistivity which abruptly drops to zero. Such materials are called superconductors.  is the resistivity at temperature T  0 is the resistivity at some standard temperature T 0  is the “ temperature coefficient ” of electric resistivity for the material under consideration The temperature coefficient of resistivity can be expressed as.

50. Norah Ali Al-moneef 49 In everyday applications we are interested in the temperature dependence of the resistance of various devices. The resistance of a device depends on the length and the cross sectional area. These quantities depend on temperature However, the temperature dependence of linear expansion is much smaller than the temperature dependence of resistivity of a particular conductor. So the temperature dependence of the resistance of a conductor is, to a good approximation, where R 0 and T 0 are the resistance and temperature at a standard temperature, usually room temperature or 20 o C.

51. Reminder: Battery as a “ski lift for charges” : Ski lift raises objects to higher potential energy - flow may vary, but potential energy difference fixed Battery also fixed potential diff., but current may vary 50 Norah Ali Al-moneef

52. 27.6 Electrical Power The chemical energy of the battery is converted to U, electrical potential energy: E chem  U The resulting electric field causes the electrons to accelerate: U  K Collisions in the lattice structure transfer the energy to the lattice as thermal energy: K  E th Thermal energy is a dissipative energy (i.e. can’t be recovered like mechanical energy. 51 Norah Ali Al-moneef

53. 52 Norah Ali Al-moneef

54. the rate at which the system loses electric potential energy as the charge Q passes through the resistor: 53 Norah Ali Al-moneef

55. Electrical Energy = Voltage x Electrical Current x Time Interval energy = V x I (amps) x t (sec) E = V x I x t The system regains this potential energy when the charge passes through the battery, Since a resistor obeys Ohm’s Law: P R = I 2 R = (∆V R ) 2 /R 54 Norah Ali Al-moneef

56. How is Electrical Power calculated? Electrical Power is the product of the current (I) and the voltage (v) The unit for electrical power is watt (W) How much power is used in a circuit which is 110 volts and has a current of 1.36 amps? P = I V Power = (1.36 amps) (110 V) = 150 W  Example 55 Norah Ali Al-moneef

57.  electrical energy: Electrical energy is a measure of the amount of power used and the time of use.Electrical energy is the product of the power and the time.  Example energy = Power X time P = I V P = (2A) (120 V) = 240 W E = (240 W) (4 h) = 960Wh = 0.96 kWh Electrical Energy = Voltage x Electrical Current x Time Interval energy = V x I (amps) x t (sec) E = V x I x t 56 Norah Ali Al-moneef

58. a. 0.44 A b. 2.25 A c. 5 A d. 36 A ANS:B  example A 9-volt battery drives an electric current through a circuit with 4-ohm resistance. What is the electric current running through the circuit? 57 Norah Ali Al-moneef

59. Joule’s Law – States that the rate at which heat produced in a conductor is directly proportional to the square of the current provided its resistance is constant – i.e. P = I 2 R In order to prevent power lines from overheating, electricity is transmitted at a very high voltage From Joule’s law the larger the current the more heat produced hence a transformer is used to increase voltage and lower current i.e. P = V I 58 Norah Ali Al-moneef

60. Power dissipated by a bulb relates to the brightness of the bulb. The higher the power, the brighter the bulb. For example, think of the bulbs you use at home. The 100W bulbs are brighter than the 50W bulbs. 59 Norah Ali Al-moneef

61. 60

62. If an electric fire uses 1.8 MJ of energy in a time of 10 minutes, calculate the power output of the fire. Energy = 1.8 MJ = 1.8x10 6 J t=10 minutes = 600 s Power = Energy / time p = 1.8x10 6 J / 600 =3 10 3 watt  example 61 Norah Ali Al-moneef

63. Calculate the power of a vacuum cleaner if the operating voltage is 120v, and the current flowing through it when it is used is 7.90A. P = V x I P = 120V x 7.9A P = 948 W  example 62 Norah Ali Al-moneef

64. Calculate the voltage of a computer that has 600W of power and 1.9A flowing into the monitor? V = P I V = 600W 1.9A V = 316V  example 63 Norah Ali Al-moneef

65.  If a 500 watt speaker need 10 amps to operate, what is the voltage requirement?  example 64 Norah Ali Al-moneef

66.  Example How much would you be charged for using a 60 Watt light bulb for 10 hours if electricity costs 0.07 $per kWh? E = PT= 0.06kW x 10h = 0.6kWh Cost = 0.6kWh x 0.07 $/kWh= 0.04$ 65 Norah Ali Al-moneef

67. If an electric fire uses 1.8 MJ of energy in a time of 10 minutes, calculate the power output of the fire. E = 1.8 MJ = 1.8x10 6 J t=10 minutes = 600 s 66 Norah Ali Al-moneef

68. Power Power is the rate of doing work. Electrical power is usually expressed in watts or kilowatts In DC and AC circuits, with resistance loads, power can be determined by: Examples of resistance loads are heaters and incandescent lamps.  example Determine the power consumed by a resistor in a 12 volt system when the current is 2.1 amps. Norah Ali Al-moneef 67

69. ELECTRIC POWER When there is current in a circuit as a result of a voltage, the electric power delivered to the circuit is: SI Unit of Power: watt (W) Many electrical devices are essentially resistors: Norah Ali Al-moneef 68

70. Rank in order, from largest to smallest, the powers P a to P d dissipated in resistors a to d. 1. P b > P a = P c = P d 2. P b = P c > P a > P c 3. P b = P d > P a > P c 4. P b > P c > P a > P d 5. P b > P d > P a > P c Norah Ali Al-moneef 69

71.  Example Determine the amount of energy a 100 Watt light bulb will use when operated for 8 hours. What will it cost to operate the light bulb if the electrical energy costs 0.12 $/kWh? Norah Ali Al-moneef 70

72. Energy Use Calculations How much electrical energy will an electric blanket use per month if it is used 8 hours a day? The blanket is on a 120 V circuit and draws 1.5 amp. Norah Ali Al-moneef 71

73. Norah Ali Al-moneef 72 Energy = Power x Time E = (100 W) (300 s) E = 30,000 J E = 30 kJ

74. Norah Ali Al-moneef 73 Rated for 4.2 kW Used 20 h/month Cost of 12 $ per kWh Energy = Power x Time = (4.2 kW) x (20 h) = 84 kWh Cost = Energy x rate per kWh = (84 kWh) x ($0.12) = $10.08

75.  Example Given copper wire 1mm diameter. 100m long has a potential diffidence of 12 V Find a) resistance, b) current in wire, c) current density, d) electric field in wire, e) concentration of electrons (assuming 1electron / atm), f) drift velocity, g) amount of electric charge flowing in 1 minute Resistivity ρ = 1.72x10 -8 Ohm-m Density D = 8.9 E 3 kg/m 3 molecular weight M = 63.546 g/mole Avogadro's # 6.022x10 23 electric charge e = 1.6x10- 19 C r = 5x10 -4 m radius, L=100m, t=60sec 74 Norah Ali Al-moneef

76. Equations:Answers: a) R= ρ L/A, A= π r 2 A=7.85x10 -7, R=2.19 Ω b) V=IRI=V/R I=5.48 A c) J=I/AJ=6.977x10 6 A/m 2 d) E = ρ J0.12 V/m e) n =D Na /M8.434E28 e/m 3 f) I = n q v d A v d = I/nqA = 5.17x10 -4 m/s g) I= dQ/dtQ = It = 329 C 75 Norah Ali Al-moneef

77.  Example 100 W light bulb connected to 110V what is a) current b) resistance c) at 10cents/kwhour how much to illuminate for a year, d) how many can be connected to a 15 ampere circuit breaker, e) how much electric power consumed by all these bulbs, f) if the temperature is 4500K and made from tungsten ( α = 0.0038/K) what is the room temperature resistance at 300K Given P=100 W, V= 100V Imax = 15A price = 0.1 $/kW h T=4500 K To = 300 K α = 0.0038/K 76 Norah Ali Al-moneef

78. EquationsAnswers a) P = IVI=P/V = 0.909 A b) V = IRR=V/I = 121 Ω c) cost = ($0.1)(.1KW)(24 x 365) = $87.60 d) Imax> Nmax INmax = 16 e) Pmax = Nmax PPmax = 1600W f) R=Ro(1 + α (T-To)) = Ro = 7.13 Ω 77 Norah Ali Al-moneef

79. Norah Ali Al moneef 78 OHM’S LAW FORMULAS Current equals voltage divided by resistance V = I x R Voltage equals current multiplied by resistance Resistance equals voltage divided by current Find Current I = VRVR R = V I Find VoltageFind Resistance summary Find current: I=ΔQ/Δt I=nqAv d

80. Norah Ali Al moneef 79 Quantity Unit of Measure Function Name Symbol Voltage V, emf or E Voltage V Pressure which makes current flow Current I Ampere A Rate of flow of electrons Resistance R Ohm  Opposition to current flow

81. Resistance related to physical parameters The dimensions and geometry of the resistor as well as the particular material used to construct a resistor influence its resistance. The resistance is approximately given by Norah Ali Al moneef 80