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1 Norah Ali Al-moneef Conductors & Semiconductors • In conductors, the valence band is only partially-full, so electrons can easily move from being near one atom to being near another • In semiconductors and insulators, the valence band is completely full, so electrons must gain extra energy to move • 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 2 Norah Ali Al-moneef • 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) 3 Norah Ali Al-moneef Electrons in an Electric Field 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 = –eEt/m 4 Norah Ali Al-moneef ELECTRON MOTION IN A CONDUCTOR WITH AND WITHOUT AN ELECTRIC FIELD 5 Norah Ali Al-moneef 27.1 Electric Current 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 Definition of the current: Charge in motion through an area A. The time rate of the charge flow through A defines the current (=charges per time): I av Q t Units:1 C/s= 1 A SI unit of the current: Ampere 6 Norah Ali Al-moneef Electrical current 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 Q t I Instantaneous current av i=dq/dt • Coulomb (C) – represents the total charge of approximately 6.25 x 1018 electrons 7 Norah Ali Al-moneef 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 8 Norah Ali Al-moneef Charge Carrier Motion in a Conductor The electric field force F imposes a drift on an electron’s random motion (106 m/s) in a conducting material. Without field the electron moves from P1 to P2. With an applied field the electron ends up at P2’; i.e., a distance vdt from P2, where vd is the drift velocity (typically 10-4 m/s). Norah Ali Al-moneef 9 Does the direction of the current depend on the sign of the charge? No! qvd E vd (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. E vd Norah Ali Al-moneef (-q)(-vd) = qvd 10 Microscopic model of current Charged particles move through a conductor of cross-sectional area A n is the number of charge carriers per unit volume V (=“concentration”) nAx=nV is the total number of charge carriers in V The total charge is the number of carriers times the charge per carrier, q (elementary charge) ΔQ = (nAΔx)q [unit: (1/m3)(m2 m)As=C] Norah Ali Al-moneef 11 The drift speed, vd, is the speed at which the carriers move vd = Δx/Δt Rewritten: ΔQ = (nAvdΔt)qΔx current, I = ΔQ/Δt = nqvdA 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 12 Norah Ali Al-moneef Example: 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 1018 electrons Unit of Current – Ampere (A) = 1coulomb/second Q 3C I 1.5 C/s 1.5 A t 2s 13 Norah Ali Al-moneef Example: 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 8.51028 per cubic meter. Find the magnitudes of (a) the current density (b) the drift velocity. (a) A=d 2p/4=(0.00102 m)2p/4=8.210-7 m2 J=I /A=1.67 A/(8.210-7 m2)=2.0106 A/m2 (b) From J=I /A=nqvd J 2.0 10 6 A / m 2 vd nq (8.5 10 28 m 3 )(1.60 10 19 C) vd=1.510-4 m/s=0.15 mm/s 14 Norah Ali Al-moneef Example: 15 Norah Ali Al-moneef Example: • 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 Q 240 C I 0.80 A t 300 s 16 Norah Ali Al-moneef 5.0min X 60s/1 min = 300s Example: Q Ne 5.6 1014 1.6 10-19 I 2mA t t 0.04 Q Ne 6.4 1021 1.6 10-19 c) I 8.53 A To the left t t 2.00 60 Q I t 0.835 A 5 s b) Q Ne N e e 1.6 10-19 N 2.611019 electrons 17 Norah Ali Al-moneef Example: a) 18 Q 1.67C I 0.835 A t 2.00 s Q I t 0.835 A 5 s e e 1.6 10-19 N 2.611019 electrons b) Q Ne c) Q Ne 6.4 1021 1.6 10-19 I 8.53 A t t 2.00 60 Norah Ali Al-moneef N Example: II. Electric current 1. Definition Q I t Conventional current Units: [ I ] = 1A = 1 C/s Electron flow 1020 electrons passed through the electric conductor during 4 seconds. Find the electric current through this conductor. q (1.6 1019 C )(1020 ) I t 4s 4A 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? q It (0.5 A)(1 s) 0.5 C a) q It (0.5 A)(60 s) 30 C b) 19 Norah Ali Al-moneef 27.2 resistance • I = n q vd A – n = number of free charge carriers/unit volume • Current density • (The current per unit cross-section is called the current density J) : J I ne(vd A) nevd A • Ohm's Law: E = J A J=σE – = resistivity – = 1/ = conductivity – Good conductor: low and high • Ohm's Law: – R = resistance } Measured in Volt/Ampere = Ohm (){ 20 Ali Almoneef Norah Ali Norah Al-moneef 20 In a homogeneous conductor, the current density is uniform over any cross section, and the electric field is constant along the length. b a The ratio of the potential drop to the current is called resistance of the segment: Unit: 1V/A= 1 (ohm) V=Va-Vb=EL Resistance in a circuit arises due to collisions between the electrons carrying the current with the fixed atoms inside the conductor Norah Ali Al-moneef 21 Ohm’s Law 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, I0, open circuit; R0, I, short circuit • The ratio of the potential drop to the current is called resistance of the segment: V R I Unit: V/A= (ohm) Norah Ali Al-moneef 22 Resistivity and Resistance J = E/ρ where ρ is the resistivity Consider a bar or wire of crosssection A and length L, carrying current I and with potential difference V = Vb - Va between the ends. We know E = V/L so I/A = J = V/Lρ. Thus: I = V/R also called Ohm’s Law. ∆V = IR I = V/R and R = ρL/A is the resistance of the bar. 23 Norah Ali Al-moneef Ohm’s Law, final Ohmic 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). Nonohmic Norah Ali Al-moneef 24 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. 25 Norah Ali Al-moneef 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 26 Norah Ali Al-moneef More on Resistance I = V/R Units of Resistivity : ρ is in Ω-m (ohm-meters), so R is in (Ω-m)(m)/(m2) = 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! 27 Norah Ali Al-moneef 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) 28 Norah Ali Al-moneef R (Ohms) 29 Norah Ali Al-moneef Example: What voltage is required to produce 2a though a circuit with a 3 resistor. 3 V = IR = 2A x 3 = 6v 30 Norah Ali Al-moneef V I = 2a 2. Ohm’s Law Nonohmic device I I V V Resistance V R I V I R 31 Norah Ali Al-moneef Units: [ R ] = 1Ω = 1 V/A V IR Ohm’s Law: R const Resistivity I Definition: A L L R A A R L Example: What is the resistance of 1 m of nichrome wire of 2 mm diameter ? L 1m 6 3 R 10 m 3 18 10 2 3 A p (10 m) Temperature dependence of resistivity T 0 1 (T T0 ) T 0 0 (T T0 ) 0 (T T0 ) 32 0 T 0 (T T0 ) Norah Ali Al-moneef T 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 108 m/s 33 Norah Ali Al-moneef Drift Velocity • In a conductor the free electrons are moving very fast in random directions (v ~ 106 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 • vDRIFT ~ mm/sec 34 Norah Ali Al-moneef • Amperes: the measure of the rate of current flow. – 6.24 × 1018 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. 35 Norah Ali Al-moneef Example What is the current flow in a circuit with a voltage of 120 volts and a resistance of 0.23 ? V = IR V 120 V I= = = 521.7 A R 0.23 Example With the increase in the length of the wire, the current increases. A. True B. False 36 Norah Ali Al-moneef Example A 4v battery is placed in a series circuit with a 2 resistor. What is the total current that will flow through the circuit? V = IR 2 4v = I x 2 4v I = 2A 37 Norah Ali Al-moneef I=? Example What voltage is required to produce 2A though a circuit with a 3 resistor. V= IR 3 V = 2A x 3 V V = 6v 38 Norah Ali Al-moneef I = 2A Example What resistance is required to limit the current to 4 A if a 12 V battery is in the circuit? V = IR 3 12 = 4 x R R = 3 39 Norah Ali Al-moneef 12v I = 4a Example 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 Norah Ali Al-moneef 40 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 RA/RB is 1. 1/2 2. 1 3. 2 4. 3 A 5. 4 B 41 Norah Ali Al-moneef 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 42 Norah Ali Al-moneef Example Two cylindrical resistors, R1 and R2, are made of identical material. R2 has twice the length of R1 but half the radius of R1. They are connected to a battery V as shown. Compare the currents flowing through R1 and through R2. 2 1 L A A2 A1 / 4 L2 2 L1 R r2 r1 / 2 A pr 2 I 2 / I1 ? C. I1 > I2 B. I1 = I2 A. I1 < I2 I1 V L2 2 L1 R2 8 R1 A2 A1 / 4 V V 1 I2 I1 R2 8 R1 8 43 Norah Ali Al-moneef I2 Voltage and Current Relationship for Linear Resistors Current (A) Voltage versus Current for a 10 ohm Resistor 0.6 0.5 0.4 0.3 0.2 0.1 0 0 1 2 3 4 5 6 Voltage (V) Voltage and current are linear when resistance is held constant. 44 Norah Ali Al-moneef Which one of the following graphs correctly represents Ohm's law, where V is the voltage and I is the current? (a) (b) (c) (d) (e) 45 A B C D A and C Norah Ali Al-moneef 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: 46 B Norah Ali Al-moneef 27.4 Resistance and Temperature •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 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 0 [1 (T T0 )] 47 Norah Ali Al-moneef Copper • is the resistivity at temperature T • 0 is the resistivity at some standard temperature T0 • is the “temperature coefficient” of electric resistivity for the material under consideration • Some materials, when very cold, have a resistivity which abruptly drops to zero. Such materials are called superconductors. T 0 0 (T T0 ) T 0 1 (T T0 ) 0 (T T0 ) T 0 • The temperature coefficient of resistivity can be expressed as. 0 T 48 Norah Ali Al-moneef • 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, R R0 1 (T T 0) where R0 and T0 are the resistance and temperature at a standard temperature, usually room temperature or 20o C. 49 Norah Ali Al-moneef 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 27.6 Electrical Power • The chemical energy of the battery is converted to U, electrical potential energy: Echem U • The resulting electric field causes the electrons to accelerate: UK • Collisions in the lattice structure transfer the energy to the lattice as thermal energy: KEth • Thermal energy is a dissipative energy (i.e. can’t be recovered like mechanical energy. 51 Norah Ali Al-moneef 52 Norah Ali Al-moneef the rate at which the system loses electric potential energy as the charge Q passes through the resistor: du d dQ ( Q V ) V I V dt dt dt 53 Norah Ali Al-moneef •The system regains this potential energy when the charge passes through the battery, • Since a resistor obeys Ohm’s Law: PR = I2R = (∆VR)2/R Electrical Energy = Voltage x Electrical Current x Time Interval energy = V x I (amps) x t (sec) E = V x I x t 54 Norah Ali Al-moneef 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) Example How much power is used in a circuit which is 110 volts and has a current of 1.36 amps? P=IV 55 Norah Ali Al-moneef Power = (1.36 amps) (110 V) = 150 W 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. Electrical Energy = Voltage x Electrical Current x Time Interval energy = V x I (amps) x t (sec) E = V x I x t Example energy = Power X time P = (2A) (120 V) = 240 W P = IV E = (240 W) (4 h) = 960Wh = 0.96 kWh 56 Norah Ali Al-moneef 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? 0.44 A b. 2.25 A c. 5 A d. 36 A a. ANS: 57 B Norah Ali Al-moneef • 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 2R 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 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 60 Norah Ali Al-moneef example 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 = 1.8x106 J t=10 minutes = 600 s Power = Energy / time p = 1.8x106 J / 600 =3 10 3 watt 61 Norah Ali Al-moneef example 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=VxI P = 120V x 7.9A P = 948 W 62 Norah Ali Al-moneef example 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 63 Norah Ali Al-moneef example If a 500 watt speaker need 10 amps to operate, what is the voltage requirement? 500 V 10 V 50V 64 Norah Ali Al-moneef 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 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 = 1.8x106 J t=10 minutes = 600 s 66 Norah Ali Al-moneef 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: P = IV • Examples of resistance loads are heaters and incandescent lamps. P = Watts example I = Amps V = Volts • Determine the power consumed by a resistor in a 12 volt system when the current is 2.1 amps. P = IV = 2.1 A x 12 V = 25.2 W 67 Norah Ali Al-moneef ELECTRIC POWER When there is current in a circuit as a result of a voltage, the electric power delivered to the circuit is: P IV SI Unit of Power: watt (W) Many electrical devices are essentially resistors: P I (IR ) I R 2 V2 V P V R R 68 Norah Ali Al-moneef Rank in order, from largest to smallest, the powers Pa to Pd dissipated in resistors a to d. 69 1. Pb > Pa = Pc = Pd 2. Pb = Pc > Pa > Pc 3. Pb = Pd > Pa > Pc 4. Pb > Pc > Pa > Pd 5.Norah PbAli> Al-moneef Pd > Pa > Pc Example • Determine the amount of energy a 100 Watt light bulb will use when operated for 8 hours. Energy = Power x Time = 100 watts x 8 hour = 800 wh • What will it cost to operate the light bulb if the electrical energy costs 0.12 $/kWh? $ 1 kW $ = 0.12 x 800 W x x 8 h = 0.77 $ kWh 1, 000 W 70 Norah Ali Al-moneef 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. 8 h 30 day 1 kW Energy (kWh) = (120 V x 1.5 A) W x x x = 43.2 kWh day month 1,000 W 71 Norah Ali Al-moneef Energy = Power x Time E = (100 W) (300 s) E = 30,000 J E = 30 kJ 72 Norah Ali Al-moneef Energy = Power x Time = (4.2 kW) x (20 h) = 84 kWh Cost = Energy x rate per kWh = (84 kWh) x ($0.12) Rated for 4.2 kW Used 20 h/month Cost of 12 $ per kWh 73 Norah Ali Al-moneef = $10.08 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/m3 molecular weight M = 63.546 g/mole Avogadro's # 6.022x10 23 electric charge e = 1.6x10-19 C r = 5x10-4m radius, L=100m, t=60sec 74 Norah Ali Al-moneef Equations: a) b) c) d) e) f) g) 75 Answers: R= ρL/A, A=πr2 A=7.85x10-7, R=2.19 Ω V=IR I=V/R I=5.48 A J=I/A J=6.977x106 A/m2 E = ρJ 0.12 V/m n =D Na /M 8.434E28 e/m3 I = n q vd A vd = I/nqA = 5.17x10-4 m/s I= dQ/dt Q = It = 329 C Norah Ali Al-moneef 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 Equations Answers a) P = IV I=P/V = 0.909 A b) V = IR R=V/I = 121 Ω c) cost = ($0.1)(.1KW)(24 x 365) = $87.60 d) Imax> Nmax I Nmax = 16 e) Pmax = Nmax P Pmax = 1600W f) R=Ro(1 + α (T-To)) = Ro = 7.13 Ω 77 Norah Ali Al-moneef summary OHM’S LAW FORMULAS Find current: I=ΔQ/Δt I=nqAvd Find Current Find Voltage Find Resistance V I= R V=IxR V R= Current equals voltage divided by resistance Voltage equals current multiplied by resistance Resistance equals voltage divided by current 78 Norah Ali Al moneef I Unit of Measure Quantity Name Symbol Voltage V, emf or E Current I Resistance 79 R Norah Ali Al moneef Name Voltage Ampere Ohm Symbol Function V Pressure which makes current flow A Rate of flow of electrons Opposition to current flow 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 R L A T 0 1 (T T0 ) R R0 1 (T T 0) P IV PI R 2 80 Norah Ali Al moneef V2 P R