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Electricity and Magnetism Lecture 07 - Physics 121 Current, Resistance, DC Circuits: Y&F Chapter 25 Sect. 1-5 Kirchhoff’s Laws: Y&F Chapter 26 Sect. 1 • • • • • • • • • • • • • • Currents and Charge Electric Current i Current Density J Drift Speed, Charge Carrier Collisions Resistance, Resistivity, Conductivity Ohm’s Law Power in Electric Circuits Examples Circuit Element Definitions Kirchhoff’s Rules EMF’s - “Pumping” Charges, Ideal and real EMFs Work, Energy, and EMF Simple Single Loop and Multi-Loop Circuits using Kirchhoff Rules Equivalent Series and Parallel Resistance formulas using Kirchhoff Rules. Copyright R. Janow - Fall 2015 1 Definition of Current : Net flow rate of charge through some area dq i dt or dq i dt q(t) t 0 i(t' )dt' i x t (if i is constant) Units: 1 Ampere = 1 Coulomb per second Convention: flow is from + to – as if free charges are + Charge / current is conserved - charge does not pile up or vanish At any Node (Junction) iin iout i1 i3 i2 i1+i2=i3 Current is the same across each cross-section of a wire i + Kirchhoff’s Rules: (summary) - Current density J may vary [J] = current/area Current (Node) Rule: S currents in = S currents out at any node Voltage (Loop) Rule: S DV’s = 0 for any closed path Energy or Voltage sources (EMFs, electro-motive forces) or current sources can supply energy to a circuit Power Resistances dissipate energy as heat in a circuit: Capacitances and Inductances store energy in E or B fields Copyright R. Janow - Fall 2015 Junction Rule Example – Current Conservation 7-1: What is the value of the current marked i? A. B. C. D. E. 1 A. 2 A. 5 A. 7 A. Cannot determine from information given. 1A 3A 2A 2A 3A 1A 5A 8A iin iout i =7A 6A Copyright R. Janow - Fall 2015 Current density J: Current / Unit Area (Vector) i A’ A + J=i/A (large) Same current crosses larger or smaller Surfaces, current density J varies J’ = i / A’ (small) Small current density in this region i High current density in this region For uniform density: i J A or J i/A For non-uniform density: di J n̂dA units: Amperes/m2 i area conductivity resistivity What makes current flow? Microscopic level J dA JE J E 1/ Drift speed: Do electrons in a current keep accelerating? Yes, for isolated charge in vacuum. No, in a conducting solid, liquid, gas Copyright R. Janow - Fall 2015 COLLISIONS with ions, impurities, etc. cause resistance Charges move at constant drift speed vD: E field in solid wire drives current. • • • + + - APPLIED FIELD = 0 Charges in random motion flow left = flow right APPLIED FIELD NOT ZERO Moving charges collide with fixed ions and flow with drift velocity vd 3 6 Thermal motions (random motions) have speed vth 10 m/s ( 2 kBoltzT) Drift speed is tiny compared with thermal motions. Drift speed in copper is 10-8 – 10-4 m/s. For E = 0 in conductor: no current, vD=0, J = 0, i = 0 For E not = 0 (battery voltage not 0): J E - + n density of charge carriers Units : # /volume nvD # of charge carriers crossing unit area per unit time J qnvD vD net charge crossing area A per unit time E qn i charge/unit time E qnvDA Note: Electrons drift rightward but vd and J are still leftward Copyright R. Janow - Fall 2015 -19 |q| = e = 1.6 x 10 C. EXAMPLE: Calculate the current density Jions for ions in a gas Assume: • Doubly charged positive ions • Density n = 2 x 108 ions/cm3 • Ion drift speed vd = 105 m/s Find Jions – the current density for the ions only (forget Jelectrons) J qnvD 2 1.6 x 10-19 2 108 105 coul/ion J 6.4 ions/cm3 m/s 106 cm3/m3 A./m2 Copyright R. Janow - Fall 2015 Increasing the Current 7-2: When you increase the current in a wire, what changes and what is constant? A. B. C. D. E. The density of charge carriers stays the same, and the drift speed increases. The drift speed stays the same, and the number of charge carriers increases. The charge carried by each charge carrier increases. The current density decreases. None of the above JE J E J qnvD Copyright R. Janow - Fall 2015 Definition of Resistance : Amount of current flowing through a conductor per unit of applied potential difference. E i DV L A R i DV i units : 1 Ohm 1 1 volt/amper e G R -1 " conductance" Units are " Siemens" -1 R depends on the material & geometry Note: C= Q/DV – inverse to R Apply voltage to a wire made of a good conductor. - Very large current flows so R is small. Apply voltage to a poorly conducting material like carbon - Tiny current flows so R is very large. V Circuit Diagram R Resistivity “” : Property of a material itself (as is dielectric constant). Does not depend on dimensions • The resistance of a device depends on resistivity and also depends on shape • For a given shape, different materials produce different currents for same DV • Assume cylindrical resistors R resistance L A For insulators: infinity resistivit y RA for a resistor L resistivit y units : Ohm - meters .m Copyright R. Janow - Fall 2015 Example: calculating resistance or resistivity resistivity resistance L R A proportional to length inversely proportional to cross section area EXAMPLE: Find R for a 10 m long iron wire, 1 mm in diameter L 9.7 x10 -8 .m x 10 m R 1. 2 3 2 2 A x (10 / 2) m Find the potential difference across R if i = 10 A. (Amperes) DV iR 12 V EXAMPLE: Find resistivity of a wire with R = 50 m, diameter d = 1 mm, length L = 2 m 2 RA 50x10-3 x 10 m 3 x 10 / 2 1.96x10 - 8 .m L 2m Use a table to identify material. Not Cu or Al, possibly an alloy Copyright R. Janow - Fall 2015 Resistivity depends on temperature: • Resistivity depends on temperature: Higher temperature greater thermal motion more collisions higher resistance. in .m @ 20o C. SOME SAMPLE RESISTIVITY VALUES 1.72 x 10-8 copper 9.7 x 10-8 iron 2.30 x 10+3 pure silicon Reference Temperature Simple model of resistivity: a = temperature coefficient Change the temperature from reference T0 to T Coefficient a depends on the material 0 (1 a(T - T0 )) a temperature coefficient Conductivity is the reciprocal of resistivity i DV L R Definition: A E L A i 1 J E units : " mho" (.m) -1 Across resistor : DV iR JAR JL EL E/J -1 Copyright R. Janow - Fall 2015 Resistivity Tables Copyright R. Janow - Fall 2015 Current Through a Resistor 7-3: What is the current through the resistor in the following circuit, if V = 20 V and R = 100 ? A. B. C. D. E. 20 mA. 5 mA. 0.2 A. 200 A. 5 A. V Circuit Diagram R DV i R Copyright R. Janow - Fall 2015 Current Through a Resistor 7-4: If the current is doubled, which of the following might also have changed? A. B. C. D. E. The The The The The voltage across the resistor might have doubled. resistance of the resistor might have doubled. resistance and voltage might have both doubled. voltage across the resistor may have dropped by a factor of 2. resistance of the resistor may have dropped by a factor of 2. DV i R Note: there might be more than 1 right answer V Circuit Diagram R Copyright R. Janow - Fall 2015 Ohm’s Law and Ohmic materials (a special case) R V /i (R could depend on applied V) Definitions of resistance: 1/ J/E ( could depend on E) Definition: For OHMIC conductors and devices: • Ratio of voltage drop to current is constant – NO dependance on applied voltage. • i.e., current through circuit element equals CONSTANT resistance times applied V Resistivity does not depend on magnitude or direction of applied voltage Ohmic Materials e.g., metals, carbon,… Non-Ohmic Materials e.g., semiconductor diodes band gap varying slope = 1/R constant slope = 1/R OHMIC CONDITION 1 di R dV is CONSTANT Copyright R. Janow - Fall 2015 Resistive Loads in Circuits Dissipate Power + 1 L O A D V - • Apply voltage drop V across basic circuit element (load) • Current flows through load which dissipates energy • An EMF (e.g., a battery) does work, holding potential V and current i constant by expending potential energy i 2 As charge dq flows from 1 to 2 it loses P.E. = dU dU V dq Vi dt - potential is PE / unit charge - charge = current x time i dU dU dq Power dissipated P Vi dt dq dt V iR P i2R V 2 /R Active sign convention: Voltage rise: + V - [Watts] for any load resistors only Pabsorbed = dot product of Vrise with i element absorbs power for V opposed to i + 1 - 2 element supplies power for V parallel to i + V V 1 i P = - Vi i P = + Vi 2 Note: Some EE circuits texts use passive convention: model voltage drops Vdrop as positive.Copyright R. Janow - Fall 2015 EXAMPLE: EXAMPLE: EXAMPLE: Copyright R. Janow - Fall 2015 Basic Circuit Elements and Symbols Basic circuit elements have two terminals Generic symbol: 2 1 Ideal basic circuit elements: Passive: resistance, capacitance, R C or inductance C L Active: voltage sources (EMF’s), current sources Ideal independent voltage source: • zero internal resistance • constant vs for any load Ideal independent current source: • constant is regardless of load DC vs + - + DC battery E AC vs is Dependent voltage or current sources: depend on a control voltage or current generated elsewhere within a circuit Copyright R. Janow - Fall 2015 Circuit analysis with resistances and EMFs CIRCUIT ELEMENTS ARE CONNECTED TOGETHER AT NODES CIRCUITS CONSIST OF JUNCTIONS (ESSENTIAL NODES) ... and … ESSENTIAL BRANCHES (elements connected in series,one current/branch) i1 i i2 i ANALYSIS METHOD: Kirchhoff’S LAWS or RULES Current (Junction) Rule: Loop (Voltage) Rule: Charge conservation DV 0 (closed loop) Energy conservation DV L R resistivit y i A dW dU P iV P i2R (resistor) dt dt RESISTANCE: R POWER: OHM’s LAW: iin iout i slope = 1/R R is independent of DV or i Copyright R. Janow - Fall 2015 EMFs “pump” charges to higher energy • EMFs (electromotive force) such as batteries supply energy: – move charges from low to high potential (potential energy). – maintain constant potential at terminals – do work dW = Edq on charges (source of the energy in batteries is chemical) – EMFs are “charge pumps” Variable name: script E. • Unit: volts (V). • • Types of EMFs: batteries, electric generators, solar cells, fuel cells, etc. DC versus AC E i work done dW unit charge dq Power supplied by EMF: + E - R i CONVENTION: Current flows CW through circuit from + to – outside of EMF from – to + inside EMF P power dW dt dW E dq E i dt P dt Pemf E i E i Power is dissipated by resistor: PR - i V - i2 R - V2 / R Copyright R. Janow - Fall 2015 Ideal EMF device • Zero internal battery resistance • Open switch: EMF = E no current, zero power • Closed switch: EMF E is also applied across load circuit • Current & power not zero Real EMF device • Open switch: EMF still = E • r = internal EMF resistance in series, usually small ~ 1 • Closed switch: • V = E – ir across load, Pckt= iV • Power dissipated in EMF Pemf = i(E-V) = i2r Multiple EMFs Assume EB > EA (ideal EMF’s) Which way does current i flow? • Apply kirchhoff Laws to find current • Answer: From EB to EA • EB does work, loses energy • EA is charged up • R converts PE to heat • Load (motor, other) produces motion and/or heat R Copyright R. Janow - Fall 2015 The Kirchhoff Loop Rule Generates Circuit Equations • The sum of voltage changes = zero around all closed loops in a multi-loop circuit) • Traversing one closed loop generates one equation. Multiple loops may be needed. • A branch is a series combination of circuit elements between essential nodes. • Assume either current direction in each branch. Minus signs may appear in the result. • Traverse each branch of a loop with or against assumed current direction. • Across resistances, write voltage drop DV = - iR if following assumed current direction. Otherwise, write voltage change = +iR. • When crossing EMFs from – to +, write DV = +E. Otherwise write DV= -E • Dot product i.E determines whether power is actually supplied or dissipated EXAMPLE: Single loop (circuit with battery (internal resistance r) Follow circuit from a to b to a, same direction as i E E - ir - iR 0 EE i r R Potential around the circuit Power in External Ckt P = iV = i(E – ir) P = iE – i2r circuit battery battery dissipation drain dissipation Copyright R. Janow - Fall 2015 Example: Equivalent resistance for resistors in series Junction Rule: The current through all of the resistances in series (a single essential branch) is identical. No information from Junction/Current/Node Rule i i1 i2 i3 Loop Rule: The sum of the potential differences around a closed loop equals zero. Only one loop path exists: - iR 1 - iR 2 - iR 3 0 i R1 R 2 R 3 The equivalent circuit replaces the series resistors with a single equivalent resistance: same E, same i as above - iR eq 0 i R eq CONCLUSION: The equivalent resistance for a series combination is the sum of the individual resistances and is always greater than any one of them. R eq R 1 R 2 R 3 R eq n Ri i 1 inverse of series capacitance rule Copyright R. Janow - Fall 2015 Example: Equivalent resistance for resistors in parallel Loop Rule: The potential differences across each of the 4 parallel branches are the same. Four unknown currents. Apply loop rule to 3 paths. E - i1R1 0 i1 E - i2R 2 0 E E E , i2 , i3 R1 R2 R3 E - i3R 3 0 i not in these equations Junction Rule: The sum of the currents flowing in equals the sum of the currents flowing out. Combine equations for all the upper junctions at “a” (same at “b”). 1 1 1 i i1 i2 i3 E R R R 2 3 1 The equivalent circuit replaces the series resistors with a single equivalent resistance: same E, same i as above. i - iR eq 0 R eq CONCLUSION: The reciprocal of the equivalent resistance for a parallel combination is the sum of the individual reciprocal resistances and is always smaller than any one of them. 1 1 1 1 R eq R1 R 2 R 3 1 R eq n 1 R i 1 i inverse of parallel capacitance rule R eq R 1R 2 R1 R 2 Copyright R. Janow - Fall 2015 Resistors in series and parallel 7-7: Four identical resistors are connected as shown in the figure. Find the equivalent resistance between points a and c. A. 4 R. B. 3 R. C. 2.5 R. D. 0.4 R. E.Cannot determine from information given. c R R R a 1 R eq n 1 R i 1 i R eq R n Ri i 1 Copyright R. Janow - Fall 2015 Capacitors in series and parallel 7-8: Four identical capacitors are connected as shown in figure. Find the equivalent capacitance between points a and c. A. 4 C. B. 3 C. C. 2.5 C. D. 0.4 C. E. Cannot determine from information given. c C C C C a n 1 1 Ceq i1 Ci Ceq n Ci i 1 Copyright R. Janow - Fall 2015 EXAMPLE: Find i, V1, V2, V3, P1, P2, P3 R1= 10 + - i E =7V R2= 7 R3= 8 EXAMPLE: Find currents and voltage drops i i1 + - E =9V i2 R1 R2 i R eq R 1R 2 R1 Copyright R. Janow - Fall 2015 R2 EXAMPLE: MULTIPLE BATTERIES SINGLE LOOP i + - R1= 10 E1 = 8 V + - i E2 = 3 V R2= 15 A battery (EMF) absorbs power (charges up) when I is opposite to E Pemf Ei E i Copyright R. Janow - Fall 2015 EXAMPLE: Find the average current density J in a copper wire whose diameter is 1 mm carrying current of i = 1 ma. i 10-3 amps J 1273 3 2 A x (.5 x 10 m) 2 amps/m Suppose diameter is 2 mm instead. Find J’: i J J' 318 amps/m2 A' 4 Current i is unchanged Calculate the drift velocity for the 1 mm wire as above? J enCu vd where nCu 3 # conduction electrons/m 8.49 x 1028 J 1273 -8 9 . 37 x 10 m / s About 3 m/year !! 19 28 enCu 1.6x10 x 8.49x10 So why do electrical signals on wires seem to travel at the speed of light (300,000 km/s)? vd Calculating n for copper: One conduction electron per atom 6.023 x 1023 atoms/mole 3 6 3 3 nCu 1 electron / atom x x8.96 gm/cm x 10 cm /m 63.5 gm/mole 8.49 x 1028 electrons/m 3 Copyright R. Janow - Fall 2015