70 3070 30 21

70 3020 2

20 5 42

450 8

21 4

0 5

OV V

由分壓定理：

Q1. (1/2) Find Vo and Io

1 2 1 3 1 2 3

2 1 2 1 2

3 1 3 1 3

mesh I :

50 70( ) 20( ) 0 90 70 20 50-------(1)

70( ) 30 0 70 100 0-------------(2)

20( ) 5 0 20 25

O

i i i i i i i

i i i i i

i i i i i

利用 求解

3 1 2 1

1 1 1

3

1

1 2 1 3 1

2

0---------------(3)

4 7(3) (2) (1) :

5 107 4

90 70 20 5010 5

(90 49 16) 50

50 7 42A =1.4A

0.2

=1.6A25 10 5

O

i i i i

i i i

i

i i

I i

i i

A

i

i

由式 得 ；由式 得 ；代回式 得

； ；

1i2i

3i

Q1. (2/2) Find Vo and Io

-3 3

3

3 -3 3

3 -3 3 6

-3

3

:

400 401

30 - 2 30 10 - 401 2 10

1 1 10

10 30 10 -802 10

30803 10 30 10 400 5 2000 10 10

803

30 10 30

803 10 803

a

a

o

node a

I I I I

m I k II

k

I I

I V I K

I A

考慮

74.60

72 80

03

mV

Q2. Find Vo

Q3. (1/2) Find Vo by using superposition

2 Ω

1 Ω

4 Ω2 A10 V

0.5 Vo

+Vo-

2 Ω

1 Ω

4 Ω10 V

0.5 V02

Vx +Vo1

-

Step1

11

1 11

o1

For 10V Voltage Source

100.5 0..........(1)

2 1

0.5 0..........(2)1 4

(2) (1) V = 8 V

oo

o oo

Vx Vx VV

V Vx VV

For 2A Current Source

22 0.5 2 0.....(3)

2 12

0.5 2 0.....(4)1 4

(4) (3) Vo2=3.2 V

Vy Vy VoVo

Vo Vy VoVo

2 Ω

1 Ω

4 Ω2 A

0.5 Vo1 Vy

+Vo2

-

Step2

Vo = Vo1+Vo2 = 8 + 3.2 = 11.2 V

Step3

Q3. (2/2) Find Vo by using superposition

2 Ω

8 Ω

1 Ω12 V

3 A

4 Ω

5vO

v1 v2

+ vo -

i2

i1 i3

1 1 3

1 2 2

0 3 1 1 2 3

1

3 2

1: 12 2 4( ) 0........(1)

2 : 8 ........(2)

3 : 5 4( ) ( ) 1 0........(3)

2 ........(4)

3........(5)

o

Loop i i i

Loop v v i

Loop v i i v v i

v i

i i

Q4. (1/2) Determine v1 and v2 using mesh

1 3

1 2 3

2 3

1 1

2 2

3 3

1 1

2

(2) & (4) (3)

3 2 6

14 8 5 0

2 3

126

113 0 2 6123

14 8 5 0 , 11

0 1 1 3 156

11

12012 2 10.91

11

then

i i

i i i

i i

i i

i i

i i

v i V V

v

1 21104

8 100.36 11

v i V V

Q4. (2/2) Determine v1 and v2 using mesh

1A 6Ω

14Ω

5Ω3A

14Va

b

6Ω

14Ω

5Ω

a

b

(i) Find Rth

th (i) R = 5 // (14+6) = 4

Q5. (1/2) Find the Thevenin equivalent ckt

6Ω 14Ω

5Ω

14Va

b

i6V15V

(ii) Find Vth

(ii)

(6-14+15) - (6+14+5) i = 0

77-25i=0 , i=

25 Vab=Vth =5i-15= -13.6 V

Q5. (2/2) Find the Thevenin equivalent ckt

18V

6Ω

3Ω

2Ω

0.25vo

a

b

vo

6//3 = 2Ω

2Ω

0.25vo

a

vo

Vx

Ix

(i) Find RN(i)

( 0.25 ) 2............(1)

2 ............(2)

(2) (1)

3N

Vx Vo Ix Vo

Vo Ix

VxR

Ix

Q6. (1/2) Find the Norton equivalent ckt

6Ω

3Ω

2Ω

0.25vo

a

b

vo

Isc=IN

(ii) Find IN

(ii)

180.25

6 2 34

0.25 12

N

Vo Vo VoVo

Vo V

VoI Vo A

Q6. (2/2) Find the Norton equivalent ckt

11

1

1

4 4

2 3 1

30 6020

30 601

( ) ( )1 1

20 ( ) ( )

20 10

2 30 20

3 :

(0) (0) (0) 30

eq

s

eq

eqs s

eq

t t

C F

S Ci t i t

S C S C

Ci t i t

C C

e e mA

V V V V

由初始條件

1 1 11 0

4

4

0

4

1( ) (0)

20 20

30

2

166.667 146.6

0 1 ( )( 1) 20

30 4

67 t

t

tt

t

V i t dt VC

me dt

me

e V

2 1 22 0

4

4

0

4

1( ) (0)

20 30

60

20

83.333 113.33

1 ( )( 1) 30

60 4

3

t

tt

t

t

V i t dt VC

me dt

me

e V

Q7. Find 1 2( ) and ( )V t V t

3

1 2

1

1 1 31

1 1 1

1

0 ( )

01 3 5

1 10

3 515

23

i

i

i

OP OP

V V

V V

node V

V V V VV

K K K

V V V V

V V

負回授虛接地

考慮

1 3 2 3 3

1 1

1

05 2 10

05 2 10

15 1057 7

23 23105

4 9105

1.141

2

92

O

O

O i i

OO

O

i

i

V V V V V V

K K KVV V

V V V V

VI

KmA

V

KI

V

V

Q8. (1/2) Find Io/Vi

1 2 3 4

7 3 4

72 5

5

74 4

6

1 3

31 1

1

33 2

2

2 3 4

10

R 103 =3 R

R 3

R 54 =4 R

R 2

(2 )

R2 =2 R 5

R

R1 =1 R 10

R

OV V V V V

R R R K

V K

V K

Va V V

V K

V K

放大 倍

放大 倍

令

放大 倍

放大 倍

1 2 3 4Design 2 3 4OV V V V V Q9.