Bài giảng Electric circuit theory - Chapter III: Electrical Circuit Analysis - Nguyễn Công Phương

Electric Circuit Theory Electrical Circuit Analysis Nguyễn Công Phương Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 2 Contents I. Basic Elements Of Electrical Circuits II. Basic Laws III. Electrical Circuit Analysis IV. Circuit Theorems V. Active Circuits VI. Capacitor And Inductor VII. First Order Circuits VIII.Second Order Circuits IX. Sinusoidal Steady State Analysis X. AC Power Analysis XI. Three-phase Circuits XII. Magnetically Coupled Circuits XIII.Frequency Response XIV.The Laplace Transform XV. Two-port Networks Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 3 Electrical Circuit Analysis 1. Branch current method 2. Node voltage method 3. Mesh current method Branch current method (1) R1 R2 R3 R4E1 E2 J + – + – i1 i2 i3 i4+ –v1 + –v3 + v2 – a b c + v4 – Node a: i1 + i2 – i3 = 0 Node b: i3 – i4 + J = 0 A B Loop A: – E1 + R1i1 – R2i2 + E2 = 0 Loop B: – E2 + R2i2 + R3i3 + R4i4 = 0 i1 + i2 – i3 = 0 i3 – i4 = –J R1i1 – R2i2 = E1 – E2 R2i2 + R3i3 + R4i4 = E2 i1 i2 i3 i4 Ex. 1 4Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn Branch current method (2) A B i1 + i2 – i3 = 0 i3 – i4 = –J R1i1 – R2i2 = E1 – E2 R2i2 + R3i3 + R4i4 = E2 nKCL = n – 1 nKVL = b – n + 1 Ex. 1 R1 R2 R3 R4E1 E2 J + – + – i1 i2 i3 i4+ –v1 + –v3 + v2 – a b c + v4 – 5Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn Branch current method (3) A B i1 + i2 – i3 = 0 i3 – i4 = –J R1i1 – R2i2 = E1 – E2 R2i2 + R3i3 + R4i4 = E2 1. Find: nKCL = n – 1, and nKVL = b – n + 1 2. Apply KCL at nKCL nodes 3. Apply KVL at nKVL loops 4. Solve simultaneous equations Ex. 1 R1 R2 R3 R4E1 E2 J + – + – i1 i2 i3 i4+ –v1 + –v3 + v2 – a b c + v4 – 6Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn Branch current method (4) 1. Find: nKCL = n – 1, and nKVL = b – n + 1 2. Apply KCL at nKCL nodes 3. Apply KVL at nKVL nodes 4. Solve simultaneous equations nKCL = 4 – 1 = 3; nKVL = 6 – 4 + 1 = 3 A: R1i1 + R5i5 + R2i2 – E1 = 0 B: R3i3 + R5i5 – R4i4 = 0 C: R2i2 + R6i6 + R4i4 – E6 = 0 a: – i1 + i2 – i6 = 0 b: i1 – i5 + i3 + J = 0 c: – i3 – i4 + i6 – J = 0 A B C Ex. 2 +– + – R1 R3 R2 R4 R5 R6 E1 E6 i1 i2 i3 J i4i5 + + + + + – – – – – a b c d 7Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn i6 + – Branch current method (5) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 8 A B C2 3 6: 0b i i i+ + = 4 3 5: 0c i i i− + = i1 i5 i2 i6 i3i41 4: 0d i i J− − − = 1 1 5 5 4 4 1:A R i R i R i E+ − = 3 3 6 6 5 5 3 6:B R i R i R i E E− + = − 6 6 2 2 6:C R i R i E− = R1 R2 R3R4 R5 R6 E6 E3 E1 J d a bc + – + – Ex. 3 Branch current method (6) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 9 + – +– R1 E1 R2 J R3 E3 i1 i2 i3 Ex. 4 R1 = 10Ω, R2 = 20Ω, R3 = 15Ω, E1 = 30V, E3 = 45V, J = 2A. Find currents? 1 2 3 0i i i J+ − + = 1 1 2 2 1R i R i E− = 2 2 3 3 3R i R i E+ = 1 2 3 1 2 2 3 2 10 20 30 20 15 45 i i i i i i i + − = −  → − =  + = 1 2 3 1 2 3; ;i i i ∆ ∆ ∆ = = = ∆ ∆ ∆ 1 1 1 10 20 0 ; 0 20 15 − ∆ = − 1 2 1 1 30 20 0 ; 45 20 15 − − ∆ = − 2 1 2 1 10 30 0 ; 0 45 15 − − ∆ = 3 1 1 2 10 20 30 0 20 45 − ∆ = − Branch current method (7) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 10 + – +– R1 E1 R2 J R3 E3 i1 i2 i3 R1 = 10Ω, R2 = 20Ω, R3 = 15Ω, E1 = 30V, E3 = 45V, J = 2A. Find currents? 1 2 3 0i i i J+ − + = 1 1 2 2 1R i R i E− = 2 2 3 3 3R i R i E+ = 1 2 3 1 2 2 3 2 10 20 30 20 15 45 i i i i i i i + − = −  → − =  + = 1 1 1 10 20 0 0 20 15 − ∆ = − 20 0 1 1 1 1 1 10 0 20 15 20 15 20 0 − − − = − + − 1( 20 15 20 0) 10[1 15 20( 1)] 0[1 0 ( 20)( 1)]= − × − × − × − − + × − − − 650= − Ex. 4 Branch current method (8) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 11 + – +– R1 E1 R2 J R3 E3 i1 i2 i3 R1 = 10Ω, R2 = 20Ω, R3 = 15Ω, E1 = 30V, E3 = 45V, J = 2A. Find currents? 1 2 3 0i i i J+ − + = 1 1 2 2 1R i R i E− = 2 2 3 3 3R i R i E+ = 1 2 3 1 2 2 3 2 10 20 30 20 15 45 i i i i i i i + − = −  → − =  + = 1 2 3 1 2 3; ;i i i ∆ ∆ ∆ = = = ∆ ∆ ∆ 1 2 3650; 1350; 300; 2350∆ = − ∆ = − ∆ = ∆ = − 1 2 3 1350 2.08 A 650 300 0.46 A 650 2350 3.62 A 650 i i i − = = −   → = = − − − = = − Ex. 4 12 Electrical Circuit Analysis 1. Branch current method 2. Node voltage method 3. Mesh current method Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn Node voltage method (1) 1 1 1 0acE R i v− + + = Ex. 1 Node a: i1 + i2 – i3 = 0 Node b: i3 – i4 + J = 0 R1 R2 R3 R4 E1 E2 J + – + – i1 i2 i3 i4+ –v1 + –v3 + v1 – a b c + v4 – R1 E1 + – i1 + –v1 vac a c 1 1 1 ( ) 0a cE R i v v→ − + + − = 0 c v = 1 1 1 aE vi R − → = 13Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn Node voltage method (2) 2 2 2 0acE R i v− + + = Ex. 1 R2 E2 + – i2 + –v2 vac a c 2 2 2 ( ) 0a cE R i v v→ − + + − = 0 c v = 2 2 2 aE vi R − → = Node a: i1 + i2 – i3 = 0 Node b: i3 – i4 + J = 0 R1 R2 R3 R4 E1 E2 J + – + – i1 i2 i3 i4+ –v1 + –v3 + v1 – a b c + v4 – 14Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn Node voltage method (3) Ex. 1 R3i3 + –v3 a b 3 3 3 vi R = 3 3 a bv vi R − → = Node a: i1 + i2 – i3 = 0 Node b: i3 – i4 + J = 0 R1 R2 R3 R4 E1 E2 J + – + – i1 i2 i3 i4+ –v1 + –v3 + v1 – a b c + v4 – 15Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn Node voltage method (4) Ex. 1 4 4 4 4 b cv vvi R R − = = 4 4 bvi R → = R4 i4 + – v4 c b 0cv = Node a: i1 + i2 – i3 = 0 Node b: i3 – i4 + J = 0 R1 R2 R3 R4 E1 E2 J + – + – i1 i2 i3 i4+ –v1 + –v3 + v1 – a b c + v4 – 16Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn Node voltage method (5) Ex. 1 4 4 bvi R = 3 3 a bv vi R − = 1 1 1 aE vi R − = 2 2 2 aE vi R − = Node a: i1 + i2 – i3 = 0 Node b: i3 – i4 + J = 0 1 2 1 2 3 3 4 0 0 a a a b a b b E v E v v v R R R v v v J R R − − − + − =   − − + =  R1 R2 R3 R4 E1 E2 J + – + – i1 i2 i3 i4+ –v1 + –v3 + v1 – a b c + v4 – 17Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn Node voltage method (6) Ex. 1 1 2 1 2 3 3 1 2 3 3 4 1 1 1 1 1 1 1 a b a b E E v v R R R R R R v v J R R R     + + − = +        →      − + + =        1 2 1 2 3 3 4 0 0 a a a b a b b E v E v v v R R R v v v J R R − − − + − =   − − + =  R1 R2 R3 R4 E1 E2 J + – + – i1 i2 i3 i4+ –v1 + –v3 + v1 – a b c + v4 – a b v v  →   1 2 1 2 3 4 1 2 3 4 ; ; ;a a a b b E v E v v v vi i i i R R R R − − − = = = = 1 2 3 4 i i i i    →    18Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn Node voltage method (7) 1. the reference node 2. the sum of the reciprocals of all resistances connected to each node 3. the negative sum of the reciprocals of the resistances of all branches joining each pair of nodes 4. current source(s) for each node 5. node voltage equations 6. node voltages 7. branch currents Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn Ex. 1 1 2 1 2 3 3 1 2 3 3 4 1 1 1 1 1 1 1 a b a b E E v v R R R R R R v v J R R R     + + − = +             − + + =        R1 R2 R3 R4 E1 E2 J + – + – i1 i2 i3 i4+ –v1 + –v3 + v1 – a b c + v4 – 19 Node voltage method (8) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn Ex. 1 R1 R2 R3 R4 E1 E2 J + – + – a b c 20 3 4 4 1 2 4 1 2 4 1 2 1 1 1 : 1 1 1 1 : b c b c b v v J R R R E E c v v J R R R R R R      + − =              − + + + = − − −        Node voltage method (9) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn Ex. 1 R1 R2 R3 R4 E1 E2 J + – + – a b c 21 1 2 1 2 3 2 1 2 1 2 2 1 2 4 1 2 1 1 1 1 : 1 1 1 1 : a c a c E E a v v R R R R R R E E c v v J R R R R R R      + + − = +              − + + + = − − −        Node voltage method (10) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 22 +– + – R1 R3 R2 R4 R5 R6 E1 E6 J a b c d Ex. 2d is the ground reference 1 6 1 2 6 1 6 1 6 1 1 1 3 5 3 1 6 6 3 3 4 6 6 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 a b c a b c a b c E E v v v R R R R R R R E v v v J R R R R R R E v v v J R R R R R R      + + − − = − −                − + + + − = +                    − − + + + = − +             a b c v v v   →    1 6 1 2 3 4 5 6 1 2 3 4 5 6 ( ) ( ) ; ; ; ; ;b a a c b c b c a E v v v v v v v E v vi i i i i i R R R R R R − − − − − → = = − = = = = i1 i2 i3 i4i5 + + + + + – – – – – + – Node voltage method (11) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 23 +– + – R1 R3 R2 R4 R5 R6 E1 E6 J a b c d Ex. 2c is the ground reference 1 6 1 2 6 1 2 1 6 1 1 1 3 5 5 1 2 5 2 4 5 1 1 1 1 1 : 1 1 1 1 1 : 1 1 1 1 1 : 0 a b d a b d a b d E E a v v v R R R R R R R Eb v v v J R R R R R R d v v v R R R R R        + + − − = − −                  − + + + − = +                   − − + + + =            Node voltage method (12) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 24 R1 R2 R3R4 R5 R6 E6 E3 E1 J d a bc + – + – Ex. 3 ( ) ( ) 3 6 2 3 6 3 3 6 3 3 3 4 5 4 3 1 4 1 4 1 1 1 1 1 : 0 1 1 1 1 1 : 1 1 1 : 0 b c d b c d b c d E Eb v v v R R R R R R E c v v v R R R R R R Ed v v v J R R R R      + + − − = +                − + + + − = −                  − − + + = − −         Node voltage method (13) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 25 R1 R2 R3R4 R5 R6 E6 E3 E1 J d a bc + – + – Ex. 3 ( ) ( ) 61 1 2 5 6 2 6 1 1 6 3 6 2 6 2 3 6 3 6 1 1 1 4 1 1 1 1 1 1 1 1 : 1 1 1 1 1 : 0 1 1 1 : 0 a b d a b d a b d EE a J R R R R R R R R R E Eb R R R R R R R Ed J R R R R ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ        + + + − + − = − +                 − + + + + − = +              − − + + = − −         Node voltage method (14) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 26 +– + – E1 E2 R1 R2 R3 R4 a b d c 2 1b dv v E E− = − 3 4 0b dv v J R R + − = 0 c v = b d v v  →   ( )1 2a d bv v E v E→ = − = − 1 2 3 4 1 2 3 4 ; ; ;a d a b b d v v v v v vi i i i R R R R − − → = = = = Ex. 4 Node voltage method (15) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 27 + – +– R1 E1 R2 J R3 E3 i1 i2 i3 Ex. 5 R1 = 10Ω, R2 = 20Ω, R3 = 15Ω, E1 = 30V, E3 = 45V, J = 2A. Find currents? 1 1 1 30 452 10 20 15 10 15a v  + + = + −    9.23 Vav→ = 1 2 3 30 9.23 2.08 A 10 9.23 0.46 A 20 45 9.23 3.62 A 15 i i i − = =  − → = = −  + = = a Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 28 Electrical Circuit Analysis 1. Branch current method 2. Node voltage method 3. Mesh current method Mesh current method (1) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 29 5A 3A 2A Branch currents + – 5A 3A Mesh currents + – Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 30 Mesh current method (2) 6A –3A 5A2A 8A ?6A ?–3A ?–8A ?5A?–2A ?6 – (–3) = 9A ?–3 – 8 = –11A ?–6 – 2 = –8A ? ? ? Mesh current method (3) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 31 R1 R2 R3 R4 E1 E2 J + – + – i1 i2 i3 i4+ –v1 + –v3 + v1 – a b c + v4 –A: R1i1 – R2i2 = E1 – E2 B: R2i2 + R3i3 + R4i4 = E2 iA iB J i1 = iA i2 = iB – iA i3 = iB i4 = iB + J R1iA – R2(iB – iA) = E1 – E2 R2(iB – iA) + R3iB + R4(iB + J) = E2 (R1+ R2)iA – R2iB = E1 – E2 – R2 iA + (R2 + R3 + R4)iB = E2 – R4J A B i i  →   Ex. 1 Mesh current method (4) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 32 R1 R2 R3 R4 E1 E2 J + – + – a b c iA iB J R1iA + R2(iA – iB) = E1 – E2 R2(iB – iA) + R3iB + R4(iB + J) = E2 Ex. 1 R1(iA ) R2(iB – iA) + R2(iA – iB) = E1 – E2 + R3(iB ) + R4(iB + J) = E2 Mesh current method (5) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 33 R1 R2 R3 R4 E1 E2 J + – + – a b c iA iB J Ex. 1    1( )AR i 2 ( )A BR i i J+ − + 1 2E E= − 2 ( )B AR i i J− − 3( )BR i j+ − 4 ( )BR i+ 2E= i1 i2 i3 i4 A B i i  →   1 ;Ai i→ = 2 ;B Ai i i J= − − 3 ;Bi i J= − 4 Bi i= Mesh current method (6) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 34 R1 R2 R3 R4 E1 E2 J + – + – a b c iA iB J Ex. 1    1( )AR i J− 2 ( )A BR i i+ − 1 2E E= − 2 ( )B AR i i− 3( )BR i J+ − 4 ( )BR i+ 2E= i1 i2 i3 i4 A B i i  →   1 ;Ai i J→ = − 2 ;B Ai i i= − 3 ;Bi i J= − 4 Bi i= Mesh current method (7) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 35 R1 R2 R3 R4 E1 E2 J + – + – a b c iA iB J Ex. 1 1( )AR i J− 3( )A BR i i J+ + − 4 ( )A BR i i+ + 2 ( )BR i 3( )B AR i i J+ + − 4 ( )B AR i i+ + 2E= 1E=   i1 i2 i3 i4 A B i i  →   1 ;Ai i J→ = − 2 ;Bi i= 3 ;A Bi i i J= + − 4 A Bi i i= + Mesh current method (8) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 36 iA iB iC J A: R1(iA ) B: R3(iB + J ) C: R2(iC – iA) + R5(iA – iB) + R2(iA – iC) = E1 + R4(iB – iC) + R5(iB – iA) = 0 + R4(iC – iB) + R6(iC ) = –E6 iA iC iB +– + – R1 R3 R2 R4 R5 R6 E1 E6 J + + + + + – – – – – a b c d i1 i2 i3 i4i5 i6 i1 = iA; i2 = iA – iC; i6 = – iCi3 = – iB – J; i4 = iB – iC; i5 = iA – iB; Ex. 2 Mesh current method (9) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 37 iA iB iC J +– + – R1 R3 R2 R4 R5 R6 E1 E6 J + + + + + – – – – – a b c d Ex. 2      1( )AR i 3( )A BR i i+ + 6 ( )A CR i i+ + 1 6E E= − 3( )B AR i i+ 4 ( )B CR i i J+ − − 5( )BR i J+ − 0= 2 ( )CR i 4 ( )C BR i i J+ − + 6 ( )C AR i i+ + 6E= − Mesh current method (10) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 38 R1 R2 R3R4 R5 R6 E6 E3 E1 J d a bc + – + – Ex. 3 A B CJ 1 5 4 1 3 6 5 3 6 2 6 6 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) A A B A B B C B A C C B R i J R i i R i E R i R i i R i i E E R i R i i E − + + + =  + − + + = −  + − = A B C i i i   →    i1 i5 i2 i6 i3i4 1 ;Ai i J= − 2 ;Ci i= − 3 ;Bi i= 4 ;Ai i= − 5 ;A Bi i i= + 6 C Bi i i= − Mesh current method (11) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 39 R1 R2 R3R4 R5 R6 E6 E3 E1 J d a bc + – + – Ex. 3 A B C J 1 5 4 1 3 6 5 3 6 2 6 6 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) A A B A B B C B A C C B R i R i i J R i J E R i R i i R i i J E E R i R i i E + + − + − =  + − + + − = −  + − = Mesh current method (12) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 40 + – + –E1 E2 R1 R2 J iA iB 1 2 1 2A BR i R i E E+ = + A Bi i J− = − Ex. 4 Mesh current method (13) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 41 + – +– R1 E1 R2 J R3 E3 i1 i2 i3 Ex. 5 R1 = 10Ω, R2 = 20Ω, R3 = 15Ω, E1 = 30V, E3 = 45V, J = 2A. Find currents? iA iB J 10 20( 2) 30 20( 2) 15 45 A A B B A B i i i i i i + − + =  − − + = 30 20 10 20 35 85 A B A B i i i i − = − →  − + = 2.08 A 3.62 A A B i i = →  = 1 2 3 2.08 A 2 0.46 A 3.62 A A A B B i i i i i i i = =  → = − + − = −  = = Mesh current method (14) Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 42 + – +– R1 E1 R2 J R3 E3 i1 i2 i3 Ex. 5 iA iB J 10 20( 2) 30 20( 2) 15 45 A A B B A B i i i i i i + − + =  − − + = 2.08 A 3.62 A A B i i = →  = 1 2 3 2.08 A 2 0.46 A 3.62 A A A B B i i i i i i i = =  → = − + − = −  = = + – +– R1 E1 R2 J R3 E3 i1 i2 i3 iA iB J 10( ) 20( 2) 30 10( ) 15 30 45 A B A B A B i i i i i i + + + =  + + = + 1.54 A 3.62 A A B i i = − →  = 1 2 3 2.08 A 2 0.46 A 3.62 A A A B B i i i i i i i = =  → = − + − = −  = = Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 43 Electrical Circuit Analysis 1. Branch current method 2. Node voltage method 3. Mesh current method