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# 戴维宁定理 - PowerPoint PPT Presentation

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### 戴维宁定理

（可含受控电源）

R

?

（含独立电源）

4Ω

1

a

1

+

+

+

25V

3A

20Ω

32V

-

-

-

1′

1′

o

-

(b)

I

I

U

U

（a）

, 由

1

+

+

U

-

-

1′

N

No

N

I

+

U

-

(c)

(d)

I = 0

+

Uoc

——开路电压

Req

——等效电阻

——短路电流

Uoc

Req

（可含受控电源）

（含独立电源）

+

U

-

I

1

+

+

U

-

-

1′

N

1. U oc的求法

1）测量：直接测量1-1′得出Uoc

2）计算法：去掉外电路，求一端口的开路电压Uoc

2. Req的求法

1)

2) 独立源置零，求Req

+

+

20

5

I

UOC

a

_

U

_

6

+

Req

I

_

6

90V

140V

_

+

b

b

140 +90

5 –90 = –44V

UOC =Uab=

–44

20+5

= –4.4A

I =

4+6

20

5

_

+

I

6

90V

140V

_

+

[解]

UOC为除6支路外有源二端网络的开路电压，见图b

Req =20  5=4 

Req为除6支路外有源二端网络所有电源都不作用

①分离

②等效

③组合

④求解

– 6V +

a

b

R

– 6V +

I

I

4

2

2

4A

4

+

8V

+

12V

+

8V

3 

+10V -

a

b

2

4

2

4

a

b

[解]

1、求开路电压

2、求等效电阻

3、将待求支路接

– 6V +

a

b

I

3 

+10V -

4、求解

R = 1 

R = 3 

R = 5 

I1

5k

a

b

a

b

+

Uo

+

2.5 k

40V

IC

I2

20k

+

35V

I1

5

a

b

+

40V

IC

I2=0

20k

[解]

1. 求Uoc

I2 = I1+ IC =1.75 I1

I1 = 10 mA

Isc是a,b短路电流

2. 求 Req

I1=40 /（5 103）= 8 mA

Isc=I1+IC=1.75I1

=14 mA

Isc

①分离

②等效

③组合

④求解