Download Solving Electric Circuit Problems using KCL and Matrix Methods - Prof. Don L. Millard and more Study notes Microelectronic Circuits in PDF only on Docsity! 1 ELECTRIC CIRCUITS ECSE-2010 Class 4 Activity 4-1 ACTIVITY 4-1 60 V 40 V4 Ω8 A 12 A 10 Ω R 5 Ω 20 Ω 6 Nodes 2 Voltage Sources 6 2 1(Ref) 3− − = 3 Unknown Node Voltages 1v 2v 3v 40 V 1(v 60) V− 0 V ACTIVITY 4-1 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 outi 0=∑ ACTIVITY 4-1 60 V 40 V4 Ω8 A 12 A 10 Ω R 5 Ω 20 Ω 1v 2v 3v 40 V 1(v 60) V− outi 0=∑ 1 3(v 60) v 10 − − 1(v 40) 5 − 1 2(v v ) R − 0 V ACTIVITY 4-1 1 3 1 1 2 1 v 60 v v 40 v vAt Node v : 0 10 5 R − − − − + + = You Do the Other Two Equations 2 ACTIVITY 4-1 60 V 40 V4 Ω8 A 12 A R 5 Ω 20 Ω 1v 2v 3v 40 V 1(v 60) V− outi 0=∑ 2 1(v v ) R − 0 V 10 Ω 12 A 2(v 0) 4 − 2(v 40) 20 − ACTIVITY 4-1 1 3 1 1 2 1 v 60 v v 40 v vAt Node v : 0 10 5 R − − − − + + = 2 1 2 2 2 v v v 40 vAt Node v : 12 0 R 20 4 − − + + + = ACTIVITY 4-1 60 V 40 V4 Ω8 A 12 A R 5 Ω 20 Ω 1v 2v 3v 40 V 1(v 60) V− outi 0=∑ 0 V 10 Ω 12 A− 8 A 3 1(v (v 60)) 10 − − ACTIVITY 4-1 1 3 1 1 2 1 v 60 v v 40 v vAt Node v : 0 10 5 R − − − − + + = 2 1 2 2 2 v v v 40 vAt Node v : 12 0 R 20 4 − − + + + = 3 1 3 v (v 60)At Node v : ( 12) 8 0 10 − − + − + = ACTIVITY 4-1 1 2 3 1 1 1 1 1v ( ) v ( ) v ( ) 6 8 14 10 5 R R 10 + + + − + − = + = 1 2 33 Equations; Solve for v , v , v Maple, MATLAB, Calculator, Cramer's Rule, etc 1 2 3 1 1v ( ) v (0) v ( ) 12 6 8 2 10 10 − + + = − − = − 1 2 3 1 1 1 1v ( ) v ( ) v (0) 2 12 10 R R 20 4 − + + + + = − = − ACTIVITY 4-1 1 2 3 In Matrix Form: 1 1 1 1 1 10 5 R R 10 v 14 1 1 1 1 + 0 v 10 R R 20 4 2v1 1 0 10 10 ⎡ ⎤+ + − −⎢ ⎥ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥− + = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎣ ⎦⎣ ⎦ ⎢ ⎥− ⎢ ⎥⎣ ⎦ [ ] G siemons [ ] V volts [ ]s I amps = 5 CIRCUIT SOLVER An Interactive Learning Module (ILM) developed by Academy for Electronic Media 1 of many ILM’s developed at Rensselaer Link to Modules on Course WebCT Homepage Or via: http://www.academy.rpi.edu/projects/ccli Click on Circuit Solver “2 Mesh” is same as our Example We’ll do Activity 4-2 using circuit solver (number on ILM is 4-3) Activity 4-2 7 Ω 9 Ω 2 Ω 6 Ω 8 Ω 3 A 4 A10 V1i 2i 3i 5 Meshes 2 Current Sources 1 2 33 Unknown Mesh Currents, i , i , i 5 Ω Ckt Solver 7 Ω 9 Ω 2 Ω 6 Ω 8 Ω 3 A 4 A10 V1i 2i 3i 4 A 3 A 5 Ω 1 1 3 1 Mesh i , Backwards around Arrow : 10 9i 9i 7i 0− + − + = 2 2 2 3 Mesh i , Backwards around Arrow : 5i 5(4) 2i 2i 10 0+ + − + = Ckt Solver 3 3 3 1 3 2 3 Mesh i , Backwards around Arrow : 8i 8(3) 9i 9i 2i 2i 6i 6(4) 0− + − + − + + = 1 2 3 16i 0i 9i 10+ − = 1 2 30i 7i 2i 30+ − = − Ckt Solver 1 2 39i 2i 25i 0− − + = Ckt Solver [ ] [ ] [ ] 1 2 3 s In Matrix Form: i16 0 9 10 0 7 2 i 30 9 2 25 0i R I V ohms amps volts − − ⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − = −⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦⎣ ⎦ = 1 2 3 1 2 3 Solve for i , i , i (e.g. using MAPLE and Gaussjord) i .54 A; i 4.3 A; i .15 A= = − = −
| Ckt Solver
> with (Linalg)
> Rismatrix(3,4, {16,0,-9,10
> Bragausssord(R)
> Hi2 = 130/241 = .5394 Aspe
> #52 = -9130/723 = -4.929 amps
> #43 = 110/723 = -.1521 ampe
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Activity 4-3
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