電路學 - [第二章] 電路分析方法
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The document discusses different circuit analysis techniques including node voltage analysis, mesh current analysis, and the use of conductance matrices. It provides examples of applying these techniques to solve for unknown voltages and currents in circuits containing multiple nodes and meshes. Key steps include setting up systems of equations using Kirchhoff's laws and the conductance matrix representation of the circuit to solve for the unknown variables. Solutions are obtained using techniques like Cramer's rule.
電路學 - [第二章] 電路分析方法
- 1. Department of Electronic Engineering National Taipei University of Technology
- 2. • • • (Node Voltage Analysis) • (Mesh Current Analysis) Department of Electronic Engineering, NTUT2/26
- 3. • N KVL N = Department of Electronic Engineering, NTUT 1 2 Nv v v v= + + +⋯ 1 1 2 2, , , N Nv R i v R i v R i= = =⋯ 1 2 Nv R i R i R i= + + +⋯ 1 2 N v i R R R = + + +⋯ 1 2 1 N s N n n R R R R R = = + + + = ∑⋯ + + − + − + − v i v1 − vN v2 R1 R2 RN 3/26
- 4. • N • Department of Electronic Engineering, NTUT 1 2 1 2, , , N N s s s RR R v v v v v v R R R = = =⋯ 1 2 1 2: : : : : :N Nv v v R R R=⋯ ⋯ + + − + − + − v i v1 − vN v2 R1 R2 RN 4/26
- 5. 1 • 1: i v1 v2 v3 ? Department of Electronic Engineering, NTUT (a)(a)(a)(a) + − 16 Ω + − 6 Ω + − 8 Ω + − 2 Ω 12 Ω 4 Ω v1 i ′a b c ′b ′c a v2 v3 ( )2 30 Vt e− (b)(b)(b)(b) + − + − + − i a b ′a ′b 2 Ω 12 Ω 16 Ω 4 Ωv1 v2( )2 30 Vt e− (a) (b) 5/26
- 6. 1 Department of Electronic Engineering, NTUT (b)(b)(b)(b) + − + − + − i a b ′a ′b 2 Ω 12 Ω 16 Ω 4 Ωv1 v2( )2 30 Vt e− (c)(c)(c)(c) + − + − i a ′a ( )2 30 Vt e− 2 Ω 8 Ω v1 (b) (c) • (c) KLC 2 30 2 8 0t e i i− − + + = 2 3 t i e− = ( )2 2 1 8 30 24 V 2 8 t t v e e− − = ⋅ = + ( )2 2 1 4 6 V 12 4 t v v e− = ⋅ = + ( )2 3 2 8 4 V 8 4 t v v e− = ⋅ = + 6/26
- 7. • : N i = i1 + i2 + ... + iN • R1 R2 Department of Electronic Engineering, NTUT 1 1 2 2, , , N Ni G v i G v i G v= = =⋯ 1 2 Ni G v G v G v= + + +⋯ 1 2 N i v G G G = + + +⋯ 11 2 1 1 1 1 1N np N nR R R R R= = + + + = ∑⋯ 1 2 1 2 1 2 ||eq R R R R R R R = = + + − vi G1 G2 GN i1 i2 iN 7/26
- 8. • Department of Electronic Engineering, NTUT 1 2 1 2 1 2 , , ,p p pN N p p p N R R RGG G i i i i i i i i i G R G R G R = = = = = =⋯ 1 2 1 2 1 2 1 1 1 : : : : : : : : :N N N i i i G G G R R R = =⋯ ⋯ ⋯ + − vi G1 G2 GN i1 i2 iN 8/26
- 9. 2 • 2: i1 i2 ? Department of Electronic Engineering, NTUT 6 Ω3 Ω 6 Ω 3 Ω 4 Ω 6 Ω a b a′ b′ (a)(a)(a)(a) (A)12 i1 i2 3 Ω 3 Ω3 Ω6 ΩA12 a b a′ b′ (b)(b)(b)(b) i1 (a) (b) 9/26
- 10. 2 Department of Electronic Engineering, NTUT 3 Ω 3 Ω3 Ω6 ΩA12 a b a′ b′ (b)(b)(b)(b) i1 a A12 a′ (c)(c)(c)(c) 6 Ω2 Ω i1 ( )1 2 12 3 A 2 6 i = ⋅ = + • (c) ( )2 1 4 3 A 4 6 6 4 i i = ⋅ = + + • (a) 10/26
- 11. (Node Voltage Analysis) • KCL • • va , vb , vc va , vb , vc Department of Electronic Engineering, NTUT a : 1 2 1 0si i i+ − = ( )1 2 1 0a b a sG v v G v i− + − = 1 3 4 0i i i− + + = ( ) ( )1 3 4 0a b b b cG v v G v G v v− − + + − =b : 4 5 2 0si i i− + + = ( )4 5 2 0b c c sG v v G v i− − + + =c : ( )1 2 1 10a b sG G v G v i+ − + = ( )1 1 3 4 4 0a b cG v G G G v G v− + + − = ( )4 4 5 20 b c sG v G G v i− + + = − is1 a b d G2 i2 G3 i3 i5 i4i1 G1 G4 c is2 G5 11/26
- 12. • • G Department of Electronic Engineering, NTUT v1 . . . vN N ( ) Gjj j j Gjk = Gkj , j ≠ k ， j k j k ij j G G G G G G G G G v v v i i i N N N N NN N N 11 12 1 21 22 2 1 2 1 2 1 2 −−−− −−−− −−−− −−−− −−−− −−−− ==== ...... ...... ...... ...... ... ... ... ... ... 12/26
- 13. 3 • 3: v1 , v2 , v3 , i1 , i2 , i3 , i4 Department of Electronic Engineering, NTUT Cramer’s rule A2 3 A v1 i22 Ω 1 Ω 4 Ω v2 v3 i3i4 i1 11 3G = ℧ 22 5G = ℧ 33 5G = ℧ 12 21 1G G= = ℧ 13 31 2G G= = ℧ 23 32 2G G= = ℧ 1 2 3 3 1 2 2 1 5 0 3 2 0 5 3 v v v − − − − = − − 3 1 2 1 5 0 50 2 0 5 − − ∆ = − = − 2 2 1 2 3 5 0 65 3 0 5 − − − ∆ = = − − 3 3 2 2 1 3 0 17 2 3 5 − − ∆ = − = − − 4 3 1 2 1 5 3 56 2 0 3 − − ∆ = − = − − − 1 1 1.3 Vv ∆ = = − ∆ 2 2 0.34 Vv ∆ = = ∆ 3 3 1.12 Vv ∆ = = − ∆ ( )1 1 21 1.64 Ai v v= − = − ( )2 1 21 0.36 Ai v v= − = 3 3.36 Ai = − 4 1.36 Ai = 13/26 Ω 3
- 14. 3 • Department of Electronic Engineering, NTUT = 3 × 5 × 5 + (-1) × 0 × (-2) + (-1) × 0 × (-2) −(-2) × 5 × (-2) − 0 × 0 × 3 − (-1) × (-1) × 5 = 75 + 0 + 0 − 20 − 0 − 5 = 50 502 051 213 − − −− =∆ (4) 3 1 2 1 5 0 2 0 5 2 3 3 1 2 3 −−−− −−−− −−−− −−−− ==== −−−− −−−− v v v (1) For node 1： 2+1× (v1 v2)+2×(v1 v3) = 0 ⇒ (1+2) × v1-1×v2 2 × v3 = 2 ⇒ G11v1+G12v2+G13 = vs1 (2) For node 2： 1× (v1 v2) + 4 × (v2 0) 3= 0 ⇒ −1× v1+(1+4)v2+ 0 × v3 = 3 ⇒ G21v1+G22v2+G23v3 = vs2 (3) For node 3： 2× (v3 v1) + 3 × (v3 0) + 3= 0 ⇒ −2× v1+ 0 × v2 +(2+3)v3 = 3 ⇒ G31v1+G32v2+G33v3 = vs3 A2 3 A v1 i22 Ω 1 Ω 4 Ω v2 v3 i3i4 i1 14/26 Ω 3
- 15. • M (M-1) (M-1) (M-1) (Supernode) KCL Department of Electronic Engineering, NTUT15/26
- 16. • v1 v2 v3 v4 v5 5 v1 = vs1 , v5 – v4 = vs2 v2 v3 v4 KCL Department of Electronic Engineering, NTUT 2 3 ( )1 2 4 2 1 1 2 3 4 5 0G G G v G v G v G v+ + − − − = ( )2 3 5 3 2 2 5 4 0G G G v G v G v+ + − − = ( ) ( )4 5 2 5 4 3 6 5 0G v v G v v G v− + − + = + − 1 2 3 4 5vs1 G1 G4 G6 G5 G3 G2 vs2 16/26
- 17. • KVL KCL KCL 1 2 3 KCL ( ) Department of Electronic Engineering, NTUT + − 1 2 3 44 sv v= ( )1 2 3 1 2 2 3 3 1 0sG G G v G v G v G v+ + − − − = ( ) ( )2 1 2 5 2 1 3 0G v G G v v vβ− + + + − = ( ) ( )3 1 3 4 3 1 3 0G v G G v v vβ− + + − − = vs G1 G2 G3 G4 ( )1 3v vβ − G5 17/26
- 18. 4 • 4: v2 , v3 , v4 v1 = 100 V Department of Electronic Engineering, NTUT 3 (a) 2 (b) 3 (c) 4 ( ) ( )1 2 3 2 2 2 41 100 2 4i i i v v v v= + ⇒ − = + − 3 22v v= − ( ) ( ) ( )3 4 5 2 4 3 4 40 4 4 4 0 0i i i v v v v v+ + = ⇒ − + − + − = 2 2 3 3 4 4 7 0 4 100 12 V 2 1 0 0 24 V 1 1 3 0 4 V v v v v v v − = = ⇒ = − − = − + − + − 1 Ω 4 32 V100 4v1 Ω 1 Ω 4 Ω 4 Ω 2 2v2 v3 v4 i5i1 i2 v2 i3 18/26
- 19. (Mesh Current Analysis) • KVL IR KVL • 2 2 Ia Ib KVL Department of Electronic Engineering, NTUT 1 (a-b-e-f-a) ⇒ 2 (b-c-d-e-b) ⇒ + − + − vg1 a b c def I1 vg2 I2 I3 R1 R3 R2 Ia Ib 1 1 1 3 3 0gv I R I R− + + = 3 3 2 2 2 0gI R I R v− + + = 1 1 3 3 1gI R I R v+ = 2 2 3 3 2gI R I R v+ = 19/26
- 20. • Department of Electronic Engineering, NTUT b + − + − vg1 a b c def I1 vg2 I2 I3 R1 R3 R2 Ia Ib 3 1 2I I I= − 1 2,a bI I I I= = 3 a bI I I= − ( ) ( ) 1 3 3 1 3 2 3 2 a b g a b g R R I R I v R I R R I v + − = − + + = − 11 3 3 23 2 3 ga gb vR R R I vR R R I + − = −− + 20/26
- 21. • Department of Electronic Engineering, NTUT M v1 j i1 j Rjj j Rjk=Rk j j k j k ij , ik Rjk( Rk j) ij , ik Rjk( Rk j) 。 + − + − vg1 a b c def I1 vg2 I2 I3 R1 R3 R2 Ia Ib 11 12 1 1 1 21 22 2 2 2 1 2 3 M M M M M MM M M R R R i v R R R i v R R R R i v ± ± ± ± = ± ± ± ⋯ ⋯ ⋮ ⋮ ⋱ ⋮ ⋮ ⋮ 21/26
- 22. • Ia , Ib R3 R3 R3 Ia Ib I3 = Ia + Ib Department of Electronic Engineering, NTUT + − + − vg1 a b c def I1 vg2 I2 I3 R1 R3 R2 Ia Ib 11 3 3 23 2 3 ga gb vR R R I vR R R I + = + 22/26
- 23. • KVL Department of Electronic Engineering, NTUT23/26
- 24. 5 • 5: i1 , i2 , i3 Department of Electronic Engineering, NTUT KVL KVL + − V2 2 Ω 1 Ω 3 Ω 2 Ω + − + − A2 + − + − 3 Ω + −+ − v1 v3 v5 v3v2 v6i1 i2 i3 ( )1 1 22 3 2i i i+ − = ( ) ( )1 2 2 3 3 2 3 3 2 2 A i i i i i i − = + + − = ( ) ( ) ( ) 1 2 1 2 3 3 5 3 0 2 1 11 7 3 4 5 0 A , A , A 3 9 9 0 1 1 2 i i i i i i − − − − = ⇒ = = = − 24/26
- 25. 6 • 6: i1 , i2 , i3 Department of Electronic Engineering, NTUT (a) 1: KVL i1 – i3 = 1 (b) 2: KVL i1 – i2 = 2v3 = 2(3i2) (c) 3: KVL 2 3 4 KVL 2i1 + 3i2 + 4i3 = 0 4 Ω + − A1 + − + − + − + − v4 2 Ω 3 Ω v5 v2 v1 v3 1 Ω i3 i1 i2 2v3 A ( ) ( ) ( ) 1 2 1 2 3 3 1 0 1 1 28 4 17 1 7 0 0 A , A , A 45 45 45 2 3 4 0 i i i i i i − − − = ⇒ = = = 25/26
- 26. • (KCL) (KVL) • • ( ) ( ) Department of Electronic Engineering, NTUT26/26