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ELECTRIC CIRCUITS ECSE-2010 Spring 2003 Class 5 ASSIGNMENTS DUE • Today (Thursday): • • Activities 5-1, 5-2, 5-3 (In Class) Next Monday: • • • • HW #2 Due Experiment #1 Report Due Activities 6-1, 6-2, 6-3, 6-4 (In Class) Next Tuesday/Wednesday: • • Will do Experiment #2 In Class (EP-2) Activities 7-1, 7-2, (In Class) REVIEW • Controlled/Dependent Sources: • VCVS, VCCS, CCVS, CCCS • Used to Model Electronic Devices • Make Circuits Interesting (and Harder) • Equivalent Resistance: • Replace any Load Network with Req • Load Network = R’s, Controlled Sources, No Independent Sources CONTROLLED SOURCES v1 2 10 V Voltage Controlled Voltage Source (VCVS) 3v1 Volts CONTROLLED SOURCES v1 2 10 V Voltage Controlled Current Source (VCCS) 4v1 Amps CONTROLLED SOURCES i1 2 10 V 4 Current Controlled Current Source (CCCS) 5i1 Amps CONTROLLED SOURCES i 2 10 V Current Controlled Voltage Source (CCVS) 6i Volts EQUIVALENT RESISTANCE i i v v [Controlled Sources, R's No Independent Sources] R eq R eq v i Req with Controlled Sources it [with Controlled Source(s)] Connect vt , define i t Use KCL, KVL to express v t in terms of i t vt vt R eq it REVIEW • Req with Controlled Sources: • Connect Test Voltage vt • Define it Using Active Convention • Find it in terms of vt (Use KCL, KVL, Ohm’s Law, etc.) • Req = vt / it WHERE ARE WE? • Circuit Elements: • • • • • Ideal, Independent i and v sources Ideal, Controlled i and v sources Resistors Potentiometers Have Learned How to Define all i’s and v’s using Active/Passive Convention • Have Begun to do Circuit Analysis = Solving for all i’s and v’s WHERE ARE WE? • Circuit Analysis Techniques: • • • • • • Ohm’s Law for R’s KCL and KVL Simplify Circuits using Series/Parallel R’s Replace Load Network with Req Sometimes Hard to Know Where to Start Will Now Develop Systematic Techniques that will Always Work • Node Equations Today • Mesh Equations on Monday NODE EQUATIONS • Systematic Technique for Solving ANY Linear Circuit: • Will Always Work! • Not Always the Easiest Technique • Will Also Learn About Mesh Equations: • Can Use Either Technique; But Cannot Mix • Very Powerful Techniques!! • Will Use For Rest of Course EXAMPLE v1 2 10 V 2 1 v2 3A NODE EQUATIONS • See Example Ckt: • Node Equation Procedure: • Label Unknown Node Voltages (v1, v2, …) EXAMPLE 10 V 10 V 1 vv11 2 2 v2 v2 3A 0V NODE EQUATIONS • Node Equation Procedure: • Label Unknown Node Voltages (v1, v2, …) • # of Unknown Node Voltages = # of Nodes # of Voltage Sources - 1 (Reference) • Example: 4 Nodes - 1 Voltage Source - 1 = 2 Unknown Node Voltages; v1, v2 • Write a KCL at Each Unknown Node Voltage • Sum of Currents Out = 0 • Express i’s in terms of Node Voltages EXAMPLE 10 V i1 10 V 1 vv11 2 i2 2 i3 v2 v2 i4 i5 3A 0V KCL: i1 i 2 i3 0 KCL: i4 i5 0; i5 3 A NODE EQUATIONS • Example: (v1 10) (v1 0) (v1 v 2 ) KCL at v1: 0 2 2 1 2v1 v 2 5 (v 2 v1 ) KCL at v 2 : ( 3) 0 1 ( v1 v 2 ) 3 NODE EQUATIONS • Example: Node v1 2v1 v2 5 Node v2 v1 v2 3 Add: v1 8 V v2 11 V NODE EQUATIONS • Writing a KCL at Each Unknown Node Voltage will Always Provide # of Linear Equations = # Unknowns: • Can Always Solve for v1, v2, …. • Node Equations will ALWAYS work • Can take it to the bank ACTIVITY 5-1 60 V 10 5 R 20 12 A 8A 4 40 V ACTIVITY 5-1 • 6 Nodes - 2 Voltage Sources - 1 (Ref): => 3 Unknown Node Voltages: • v1, v2, v3 ACTIVITY 5-1 v1 60 60 V 10 5 v1 R v 2 20 v3 40 V 12 A 8A 4 0V 40 V ACTIVITY 5-1 • 6 Nodes - 2 Voltage Sources - 1 (Ref): => 3 Unknown Node Voltages: • v1, v2, v3 • Write a KCL at each Unknown Node Voltage, relating Currents to the Node Voltages using Ohm’s Law • Usually best to sum Currents OUT of the Node iout 0 ACTIVITY 5-1 v1 60 60 V 5 v1 i out node 10 R v 2 20 v3 40 V 12 A 8A 4 0V 40 V 0 ACTIVITY 5-1 v1 60 v3 v1 40 v1 v 2 At Node v1: 0 10 5 R v 2 v1 v 2 40 v 2 At Node v2 : 12 0 R 20 4 v3 (v1 60) At Node v3 : (12) 8 0 10 ACTIVITY 5-1 1 1 1 1 1 v1 ( ) v 2 ( ) v3 ( ) 6 8 14 10 5 R R 10 1 1 1 1 v1 ( ) v 2 ( ) v3 (0) 2 12 10 R R 20 4 1 1 v1 ( ) v 2 (0) v3 ( ) 12 6 8 2 10 10 3 Equations; Solve for v1 , v2 , v3 Maple, MATLAB, Calculator, Cramer's Rule, etc ACTIVITY 5-1 In Matrix Form: 1 1 1 1 1 10 5 R R 10 v 14 1 1 1 1 1 + 0 v 2 10 R R 20 4 v3 2 1 1 0 10 10 V I s G volts amps siemons ACTIVITY 5-1 Matrix Equation: G V Is Want to Solve for V Let's Use MAPLE ACTIVITY 5-2 • For Large Matrices, Use MAPLE • Usually for n > 2 • • • • Must Include with (linalg): Two Ways to Solve using MAPLE: See MAPLE Scripts Must Be Careful with Syntax ACTIVITY 5-2 • MAPLE SCRIPT – Classic Method: >with (linalg) >G:=R->matrix(3,3,[1/5+1/10+1/R, -1/R,-1/10,-1/R,1/20+1/4+1/R,0, -1/10,0,1/10]): >G(R); >Is:=matrix(3,1,[14,-10,-2]); ACTIVITY 5-2 >v:=R->evalm(inverse(G(R))&*Is): #&* means matrix multiplication >v(R); >v(1); >ia:=R->(v(R)[2,1]-2): >ia(R); >ia(1); ACTIVITY 5-2 • MAPLE SCRIPT – Gaussjord Method: >with (linalg) >G:=R->matrix(3,4,[1/5+1/10+1/R, -1/R,-1/10,14,-1/R,1/20+1/4+1/R,0, -10,-1/10,0,1/10,-2]): >G(R); >B:=R->guassjord(G(R)): >B(R); #leads to same result ACTIVITY 5-3a 10 k 30 V 8V 6 k v (v 8) V 2 k 30 V 5 mA v 0V 1 Unknown Node Voltage Find v using Node Equations ACTIVITY 5-3a • KCL at Node v: • • • • Sum of Currents Out of Node = 0 (v-30)/6 + (v+8)/2 + (v+8-30)/10 - 5 = 0 v (1/6 + 1/2 + 1/10) = 30/6 – 8/2 + 22/10 + 5 => (23/30) v = 8.2 => v = 10.7 V • Writing a KCL at each Unknown Node always works! ACTIVITY 5-3b 1A 8 2 Unknown Node Voltages 7 ic 10 24 V 24 V v1 3 ib 90 V v2 15 6 0V 2 4 ia Show that v1 24 V; v 2 6 V Find ia , i b , ic ACTIVITY 5-3b • 5 Nodes - 2 Voltage Sources - 1 (Ref) • 2 Unknown Node Voltages, v1, v2 • Write a KCL at each Unknown Node • Sum of Currents Out = 0 ACTIVITY 5-3b KCL at v1 : v1 24 v1 v2 v1 v1 (v 2 90) 0 3 15 6 10 KCL at v2 : v2 (v2 90) v1 v2 v1 v2 24 1 0 24 10 15 7 8 ACTIVITY 5-3b 1 1 1 1 1 1 Node v1: v1 ( ) v 2 ( ) 8 9 17 3 15 6 10 15 10 1 1 1 1 1 1 24 Node v 2 : v1 ( ) v 2 ( ) 9 1 15 10 6 10 15 15 15 ACTIVITY 5-3b Multiply by 30: v1 (20) v2 (5) 510 v1 (5) v2 (12) 192 G: matrix(2,3,[20, 5,510, 5,12, 192]); V: gaussjord(G); v1 24 V v 2 6 V ACTIVITY 5-3b With v 1 24 V; v 2 6 V v2 i a 1 A 24 24 v1 24 v 2 i b 1 3 A 3 87 24 v1 i c i b 3A 3