1

Thevenin Equivalent Circuit with a dependent Source Example: Find the Thevenin equivalent for the circuit containing dependent sources shown in figure1.

figure1

Solution : The first step is to recognize that the current ix must be zero.(Note the absence of a return path for ix to enter the left-hand portion of the circuit).

• Thevenin voltage VTh The Thevenin voltage will be the voltage across the 25 W resistor. With ix = 0,

VVv

V

obtainweequationsthesecombineweWhen

vVBecausevvi

isicurrentThe

iivV

ThTh

Th

ThTh

abTh

52000

35500

,22000

352000

35

500)25)(20(

−=⇒−

×−=

=−

=−

=

−=−==

• Thevenin resistance RTh The Thevenin resistance will be the ratio of the Thevenin voltage VTh to the short-circuit current isc (current away terminals a, b when they are connected together).

sc

ThTh i

VR =

25Ω

5V +

- 3v 20i v

a

b

-

+

vab

2kΩ

-

+ i

ix

+ -

2

Therefore, with the short in place, the circuit shown in figure1 becomes the one shown in figure2. with the short circuit shunting the 25Ω resistor, all the current from the dependent current source appears in the short, so isc = -20 i

Figure2 The current controlling the dependent current source is

Ω=×−−

==

−=×−=⇒

==

− 10010505

505.220

5.22000

5

3

sc

ThTh

scTh

sc

iV

R

getweiandVFrommAi

mAi

Figure3 illustrates the Thevenin equivalent for the circuit shown in figure1

Figure3

25Ω +

- 3v

20i v=0

a

b

+ _ 5V

2kΩ

-

+ i

ix

isc

100Ω a

- + 5V

b

3

Superposition with a independent Source Example: Find the branch currents in the circuit shown in figure4 by using the superposition principle.

Figure4

Solution :

• We begin by finding the branch currents resulting from the 120V voltage source.

We denote those currents with prime. Replacing the ideal current source with an

open circuit deactivates it, figure5 shows this.

Figure5

We can easily finding the branch currents in the circuit in figure5 once we know the node

voltage v1 across the 3Ω resistor. We write

AiiAiAi

iiiicurrentsbranchtheforressionthewritecanweNow

VvwhichFrom

vvvv

56

30103

30156

30120exp

30

04236

120

'4

'3

'2

'1

'4

'3

'2

'1

1

111

======−

=

−−−

=

=+

++−

6 Ω

i4 12A

4Ω

2 Ω

3 Ω

i3 120V

i1 + _ i2

6 Ω

'4i 4Ω

2 Ω

3 Ω

'3i

120V '1i+

_ '2i

v1

-

+

4

• We continue by finding the branch currents resulting from the current source 12A,

We denote those currents with double-prime we deactivate the ideal voltage

source by replacing it with a short-circuit as shown in figure6.

Figure6

We determine the branch currents in the circuit shown in figure6 by first solving for the

node voltage v3 across 3Ω and the node voltage v4 across 4Ω resistor. The two node-

voltage equations that describe the circuit are

Av

iAvv

i

Av

iAv

i

vandvoftermsiniiiicurrentsbranchtheforressionthewritecanweNow

VvVv

getwevandvforSolving

vvv

vvvv

6424

46

212

2)24(12

2

4312

32

612

6

:exp

2412

,

01242

0263

4''4

43''3

3''2

3''1

43''

4''

3''

2''

1

4

3

43

434

4333

−=−

====−−−

=−

=

−=−

====−

=

−−−

−=−=

=++−

=−

++

To find the branch currents in the original circuit, that is, the currents i1, i2,, i3, and i4 in

figure4, we simply adding the obtained currents :

AiiiAiii

AiiiAiii

1651165

641017215''

4'44

''3

'33

''2

'22

''1

'11

−=−=+==+=+=

=−=+==+=+=

6 Ω

''4i 4Ω 3 Ω

''3i''

2i''

2i

v3 v4

-

+

-

+

12A

5

Superposition with a dependent Source Example: use the principle of superposition to find v0 in the circuit shown in figure7.

Figure7

Solution :

We begin by finding the component v0 resulting from the 10V voltage source. Fgure8 shows the circuit. With 5A source deactivated, '

∆v must equal(-0.4 '∆v ) (10). Hence,

'∆v must be zero, the branch containing the two dependent sources open, and

Vvo 8102520' =×=

Figure8

When the 10V source is deactivated, the circuit reduces to the one shown in figure9.

Figure9

+

-

v∆

+

-

20 Ω

2 i∆

5 Ω

5A

10Ω

0.4 v∆

10V i∆ +

_ v0

- +

+

-

'∆v

+

-

20 Ω

2 '∆i

5 Ω

10Ω

0.4 '∆v

10V

'∆i +

_ 'Ov

- +

+

-

''∆v

+

-

20 Ω

2 ''∆i

5 Ω

5A

10Ω ''Ov

- +

''∆i

a b

c

6

We have added a reference node and the node designations a, b, and c to aid the discussion. - Summing the currents away the node a yields :

085,04.0520

''''''''''

=−=−+ ∆∆ vvorvvv

ooo

- Summing the currents away the node b gives :

024,0510

24.0 ''''

'''' =−+=−

−+ ∆∆

∆∆ ivvor

ivv b

b

- We now use ''''2 ∆∆ += vivb to find the value of ''

∆v . Thus, Vvorv 10,505 '''' == ∆∆ From the node a equation, Vvorv oo 16,805 '''' == The value of v0 is the sum of '

Ov and ''Ov ,

Vvvv OOo 24168''' =+=+=