General Physics (PHY 240) Lecture 5 Electrodynamics Electric current temperature variation of resistance electrical energy and power Direct current circuits emf resistors in series http://www.physics.wayne.edu/~alan/240website/main.htm Chapter 7-8
Department of Physics and Astronomy announces the Spring-Summer 2007 opening of The Physics Resource Center on Monday, May 2 in Room 72 of Physics Research Building. Hours of operation: Monday and Tuesday, Wednesday and Thursday Friday, Saturday and Sunday 0:00 AM to 5:00 PM 0:00 AM to 4:00 PM Closed Undergraduate students taking PHY020, PHY230, PHY240, PHY270/275 and PHY280/285 will be able to get assistance in this Center with their homework, labwork and other issues related to their physics course. The Center will be open: Monday, May 2 to Thursday, August 2, 2007. 2
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Exam this Wednesday 5/23 2-4 4 questions You must show your work for credit. Closed book. You may bring an 8 ½ x inch page of notes. Bring a calculator and a pen/pencil ( don t t have extras). 4
Suggested Course Sequence FRESHMAN YEAR Fall Physics 230/23 or 270/27 Mathematics 800 English (BC) UGE 000 General Education Req. Winter Physics 240/24 or 280/28 Chemistry 220/230 Mathematics 200 English (C) SOPHOMORE YEAR Fall Physics 3X00: Application of Mathematics in Biomed. Chemistry 240/50 Mathematics 2020 Biology 500 Winter Physics 4X00: ntroduction to Biomedical Physics Biology 50 College Foreign Lang. General Education Req. JUNOR YEAR Fall Physics 5340/4: Optics Chemistry 2220/2230 or 2280/2290 Biology Elective College Foreign Lang. Winter Physics 5620: Electronics Physics 6Z00: Biomedical Seminar # Elective College Foreign Lang. SENOR YEAR Fall College Group Req. Physics 6X00: Biological Physics Elective Elective General Education Req. Winter Physics/Radiology 6Y00: Physics in Medicine Physics 6W00: Biomedical Research # College Group Req. General Education Req. # These will also be offered in the summer For more info, please contact Prof. Peter Hoffmann at hoffmann@wayne.edu New Degree Program in BOMEDCAL PHYSCS 5
Lightning Review Last lecture:. Capacitance and capacitors Capacitors with dielectrics (C if κ ) 2. Current and resistance Current and drift speed Resistance and Ohm s s law is proportional to Resistivity material property ΔQ = Δ t = nqv A ρ = R RA l Review Problem: Consider two resistors wired one after another. f there is an electric current moving through the combination, the current in the second resistor is a. equal to b. half c. smaller, but not necessarily half the current through the first resistor. = d 2 Q U = Q = = C 2 2C 2 A C = κε0, C = κc0 d R R 2 2 a b c 6
7.5 Resistivity - Example (a) Calculate the resistance per unit length of a 22-gauge nichrome wire of radius 0.32 m. A= πr = π 0.32 0 m = 3.24 0 m 2 3 7 2 Cross section: ( ) 2 Resistivity (Table):.5 x 0 6 Ωm. Resistance/unit length: 6 R ρ.5 0 Ωm = = = 7 2 l A 3.24 0 m 4.6 Ω m 7
7.5 Resistivity - Example (b) f a potential difference of 0.0 is maintained across a.0-m length of the nichrome wire, what is the current? Δ 0.0 = = = R 4.6Ω 2.2A 8
7.6 Temperature ariation of Resistance -ntro The resistivity of a metal depends on many (environmental) factors. The most important factor is the temperature. For most metals, the resistivity increases with increasing temperature. The increased resistivity arises because of larger friction caused by the more violent motion of the atoms of the metal. 9
For most metals, resistivity increases approx. linearly with temperature. ρ ρ= ρo + α T T ( ) o T Metallic Conductor ρ is the resistivity at temperature T (measured in Celsius). ρ ο is the reference resistivity at the reference temperature T ο (usually taken to be 20 o C). α is a parameter called temperature coefficient of resistivity. For a conductor with fixed cross section. R = Ro + α T T ( ) o ρ T Superconductor 0
7.6 Temperature ariation of Resistance - Example Platinum Resistance Thermometer A resistance thermometer, which measures temperature by measuring g the change in the resistance of a conductor, is made of platinum and has a resistance of 50.0 Ω at 20 o C. When the device is immersed in a vessel containing melting indium, its resistance increases to 76.8 Ω.. Find the melting point of ndium. Solution: Using α=3.92x0-3 ( o C) - from table 7..
Platinum Resistance Thermometer A resistance thermometer, which measures temperature by measuring g the change in the resistance of a conductor, is made of platinum and has a resistance of 50.0 Ω at 20 o C. When the device is immersed in a vessel containing melting indium, its resistance increases to 76.8 Ω.. Find the melting point of ndium. Solution: Using α=3.92x0-3 ( o C) - from table 7.. R o =50.0 Ω. T o =20 o C. R=76.8 Ω. T T R R 76.8Ω 50.0Ω o o = = αr 3 o T = 37 o = 57 o C C R = Ro + α T T ( ) ( o C) [ ] 3.92 0 50.0Ω o 2
7.7 Superconductivity 9: : H. K. Onnes,, who had figured out how to make liquid helium, used it to cool mercury to 4.2 K and looked at its resistance: At low temperatures the resistance of some metals 0, measured to be less than 0-6 (i.e., ρ<0-24 Ωm)! 6 ρ conductor Current can flow, even if E=0. Current in superconducting rings can flow for years with no decrease! 957: : Bardeen (UUC!( UUC!), Cooper, and Schrieffer ( BCS( BCS ) ) publish theoretical explanation, for which they get the Nobel prize in 972. t was Bardeen s second Nobel prize ( st: 956 transistor) 3
7.8 Electrical energy and power n any circuit, battery is used to induce electrical current chemical energy of the battery is transformed into kinetic energy of mobile charge carriers (electrical energy gain) Any device that possesses resistance (resistor) present in the circuit will transform electrical energy into heat kinetic energy of charge carriers is transformed into heat via collisions with atoms in a conductor (electrical energy loss) = R C D + - B A 4
Electrical energy Consider circuit on the right in detail AB: charge gains electrical energy form the battery Δ E =ΔQ Δ B C (battery looses chemical energy) CD: electrical energy lost (transferred into heat) Back to A: same potential energy (zero) as before Gained electrical energy = lost electrical energy on the resistor A D 5
Power Compute rate of energy loss (power dissipated on the resistor) Use Ohm s s law ΔE ΔQ P= = Δ = Δ Δt Δt P= Δ = R= ( Δ ) 2 2 R Units of power: S: watt delivered energy: kilowatt-hours ( )( ) 3 6 kwh = 0 W 3600 s = 3.60 0 J 6
Example Power Transmission line A high-voltage transmission line with resistance of 0.3 Ω/km carries 000A, starting at 700 k, for a distance of 60 km. What is the power loss due to resistance in the wire? Given: =700000 ρ=0.3 Ω/km L=60 km =000 A Find: P=? Observations:. Given resistance/length, compute total resistance 2. Given resistance and current, compute power loss ( )( ) R = ρl = 0.3 Ω km 60 km = 49.6 Ω Now compute power 2 ( ) ( ) 2 6 P = R= 000 A 49.6 Ω = 49.6 0 W 7
Mini-quiz Why do the old light bulbs usually fail just after you turn them on? When the light bulb is off, its filament is cold, and its resistance is R 0. Once the switch it thrown, current passes through the filament heating it up, thus increasing the resistance, ( ) R= Ro + α T T This leads to decreased amount of power delivered to the light bulb, as P= 2 / R Thus, there is a power spike just after the switch is thrown, which burns the light bulb. o 8
Chapter 8: Direct Current Circuits 9
ntroduction: elements of electrical circuits A branch: A branch is a single electrical element or device (resistor, etc.). b b b b b A circuit with 5 branches. A junction: A junction (or node) is a connection point between two or more branches. b b b A circuit with 3 nodes. f we start at any point in a circuit (node), proceed through connected electric devices back to the point (node) from which we started, without crossing a node more than one time,, we form a closed-path (or loop). 20
8. Sources of EMF Steady current (constant in magnitude and direction) requires a complete circuit path cannot consist only of resistances: cannot have only potential drops in direction of current flow Electromotive Force (EMF) provides increase in potential E converts some external form of energy into electrical energy Single emf and a single resistor: emf can be thought of as a charge pump = R + - = R = E E 2
8. Sources of EMF (continued) mportant note: in our complete circuit, conservation of energy requires that as we travel around the circuit we should encounter a number of voltage drops and increases so that we get back to the same potential we started with! = R + - E = R = E Start with 0 here and move clockwise 22
EMF Each real battery has some internal resistance AB: potential increases by E, the source of EMF, then decreases by r (because of the internal resistance) Thus, terminal voltage on the battery Δ is Δ = E r Note: E is the same as the terminal voltage when the current is zero (open circuit) B r E A C D R 23
EMF (continued) Now add a load resistance R Since it is connected by a conducting wire to the battery terminal voltage is the same as the potential difference across the load resistance Δ = E r = R, or E = r + R Thus, the current in the circuit is = E R + r r E B A Power output: 2 2 E= r+ R C D R Note: we ll assume r negligible unless otherwise is stated 24
Measurements in electrical circuits oltmeters measure Potential Difference (or voltage) across a device by being placed in parallel with the device. Ammeters measure current through a device by being placed in series with the device. A 25
Direct Current Circuits Two Basic Principles: Conservation of Charge Conservation of Energy Resistance Networks a R ab eq = R eq ab b R eq 26
8.2 Resistors in series A + By definition, v 2 R 2 v + _ + i R Thus, R eq would be B v C Δ = R. Because of the charge conservation, all charges going through the resistor R 2 will also go through resistor R. Thus, currents in R and R 2 are the same, eq = 2 = 2. Because of the energy conservation, total potential drop (between A and C) equals to the sum of potential drops between A and B and B and C, Δ R + R = = + Δ = R+ R2 2 Req R R2 R = R + R eq 2 27
Resistors in series: notes Analogous formula is true for any number of resistors, R = R + R + R + (series combination) eq 2 3... t follows that the equivalent resistance of a series combination of resistors is greater than any of the individual resistors. 28
Resistors in series: example n the electrical circuit below, find voltage across the resistor r R in terms of the resistances R, R 2 and potential difference between the battery s terminals. A + v 2 R 2 v + _ + i R B v C Energy conservation implies: = + 2 = R and = R with 2 2 = R + R,so = Then, ( ) Thus, 2 R = R R + 2 R + R 2 This circuit is known as voltage divider. 29
Resistors in series Conservation of Charge = = 2 = 3 Conservation of Energy ab = + 2 + 3 R R eq eq = = R ab + = 2 3 + + + R oltage Divider: 2 + R 3 2 + 3 = + R = 2 2 ab R eq + 3 3 b a R = R R 2 2 = 2 R 2 R 3 3 = 3 R 3 30
Resistors in parallel Conservation of Charge = + 2 + 3 Conservation of Energy ab = = 2 = 3 R eq R eq = ab ab = R = + R + 2 2 ab + 2 ab + + R 3 + 3 ab 3 = + 2 2 R + 3 3 R 2 R 3 = R 2 = 2 R 2 3 = 3 R 3 b a Current Divider: = R R eq 3
Example: Determine the equivalent resistance of the circuit as shown. Determine the voltage across and current through each resistor. Determine the power dissipated in each resistor Determine the power delivered by the battery E=8 + R =4Ω R 2 =3Ω R 3 =6Ω 32