Department of Electronic EngineeringNational Taipei University of Technology

••• (Node Voltage Analysis)

• (Mesh Current Analysis)

Department of Electronic Engineering, NTUT2/26

•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

vi

R R R=

+ + +⋯

1 21

N

s N nn

R R R R R=

= + + + =∑⋯

+

+

−+

−+

−

v

i

v1

−

vN

v2

R1

R2

RN

3/26

•N

•

Department of Electronic Engineering, NTUT

1 21 2, , , N

Ns s s

RR Rv 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

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( )230 Vte−

(b)(b)(b)(b)

+−

+

−

+

−

i

a b

′a ′b

2 Ω 12 Ω

16 Ω 4 Ωv1 v2( )230 Vte− (a) (b)

5/26

1

Department of Electronic Engineering, NTUT

(b)(b)(b)(b)

+−

+

−

+

−

i

a b

′a ′b

2 Ω 12 Ω

16 Ω 4 Ωv1 v2( )230 Vte−

(c)(c)(c)(c)

+−

+

−

i

a

′a

( )230 Vte−

2 Ω

8 Ω v1

(b) (c)

• (c) KLC

230 2 8 0te i i−− + + =23 ti e−=

( )2 21

830 24 V

2 8t tv e e− − = ⋅ = +

( )22 1

46 V

12 4tv v e− = ⋅ = +

( )23 2

84 V

8 4tv v e− = ⋅ = +

6/26

• :

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

iv

G G G=

+ + +⋯

11 2

1 1 1 1 1N

np N nR R R R R=

= + + + =∑⋯

1 21 2

1 2

||eq

R RR R R

R R= =

+

+

−vi G1 G2 GN

i1 i2 iN

7/26

•

Department of Electronic Engineering, NTUT

1 21 2

1 2

, , , p p pNN

p p p N

R R RGG Gi i i i i i i i i

G R G R G R= = = = = =⋯

1 2 1 21 2

1 1 1: : : : : : : : :N N

N

i i i G G GR R R

= =⋯ ⋯ ⋯

+

−vi G1 G2 GN

i1 i2 iN

8/26

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

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

212 3 A

2 6i = ⋅ = +

• (c)

( )2 1

4 3 A

4 6 6 4i i = ⋅ = + +

• (a)

10/26

(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

i3i5

i4i1

G1 G4 c

is2

G5

11/26

•

• 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

3

• 3: v1 , v2 , v3 , i1 , i2 , i3 , i4

Department of Electronic Engineering, NTUT

:

Cramer’s rule

A2

3 Av1

i22 Ω

1 Ω

4 Ω

v2 v3

i3i4i1

11 3 G = 22 5 G = 33 5 G =

12 21 1 G G= =

13 31 2 G G= =

23 32 2 G 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

− −∆ = − = −

− −

11 1.3 Vv

∆= = −∆

22 0.34 Vv

∆= =∆

33 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

Ω3

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 Av1

i22 Ω

1 Ω

4 Ω

v2 v3

i3i4i1

14/26

Ω3

•

M (M-1)(M-1) (M-1)

(Supernode)KCL

Department of Electronic Engineering, NTUT15/26

• v1 v2 v3 v4 v5 5v1 = 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

34

5vs1

G1

G4

G6

G5G3

G2 vs2

16/26

•

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

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

V 100

4v1

Ω

1 Ω

4

Ω4Ω

22v2

v3v4

i5i1 i2

v2

i3

18/26

(Mesh Current Analysis)

•KVL

IRKVL

• 2 2Ia 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

•

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

•

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

•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

•

KVL

Department of Electronic Engineering, NTUT23/26

5

• 5: i1 , i2 , i3

Department of Electronic Engineering, NTUT

KVL

KVL

+−

V 2

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 21 11 7

3 4 5 0 A , A , A3 9 9

0 1 1 2

i

i i i i

i

− − − − = ⇒ = = =

−

24/26

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 Ω 3 Ω 4 Ω KVL 2i1 + 3i2 + 4i3 = 0

4 Ω

+ −A1

+

−

+

−

+ −

+ −

v4

2 Ω 3 Ω

v5

v2

v1

v3

1 Ωi3

i1 i22v3 A

( ) ( ) ( )1

2 1 2 3

3

1 0 1 128 4 17

1 7 0 0 A , A , A45 45 45

2 3 4 0

i

i i i i

i

− − − = ⇒ = = =

25/26

•

(KCL)(KVL)

•

•

( ) ( )

Department of Electronic Engineering, NTUT26/26