3-Handouts_Lec_03_WFig_ver_01

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EE 201 ELECTRIC CIRCUITS

Class Notes

CLASS 3

The material covered in this class will be as follows:

Kirchoffs Voltage Law.

Fundamental Laws of Electric Circuits.

Dependent Voltage Source.

Dependent Current Source.

At the end of this class you should be able to:

Apply Kirchoffs Voltage Law.

Recognize invalid circuits.

Use the fundamental laws to analyze electric circuits.

Recognize the symbol of a dependent source.

Distinguish between the four possible types of dependent sources.

Analyze circuits that contain dependent sources.

Kirchoffs Voltage Law (KVL):

The algebraic sum of voltages around any closedcircuit is equal to zero.

KVL around circuit 1 (CW) 1 2 3 4 5 0v v v v v + + = (1)

KVL around circuit 1 (CCW) 1 2 3 4 5 0v v v v v+ + + = (2) [same as (1)]

CW = clockwise & CCW = counterclockwise

KVL around the outer circuit (CW) 6 8 3 4 5 0v v v v v + + + = (3)

KVL around circuit 2 (CW) 6 7 0v v + = 6v v7= (parallel elements)


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Figure 1

Alternative KVL Statement:

The algebraic sum of voltages between two nodes is independentof the path taken from the first node

to the second node.

KVL1&2path

Node a Node b 2 1 3 4v v v v v5+ + = + + (4) [same as (1)]

KVL 2&3path

Node a Node b 3 4 5 8v v v v v6+ + = + (5) [same as (3)]

Figure 2

Example 1:

Calculate the unknown voltages in the given circuit.

Figure 3


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Solution:

Applying KVL:

Right-hand circuit (CW) 1(7) ( 1) 10 0v + + + = 1 2v V=

Right-hand circuit (CCW) 1(7) (10) ( 1) 0v+ = 1 2v V=

1&2path

Node a Node b 1 (7) (10) ( 1)v+ = + 1 2v V=

Same answer in all cases.

Left-hand circuit (CW) 2(7) ( ) 0v+ = 2 7v V=

3&4path Node a Node c 2 7v+ = + 2 7v V=

Same answer in both cases.

Figure 4

Fundamental Laws of Electric Circuits:

1- Ohms Law, KCL and KVL are the fundamental laws of electric circuits.2- All the fundamental laws of electric circuits must be satisfied.3- If a given circuit violates at least one of the fundamental laws, the circuit is not valid.


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Example 2:

All the given circuits below are invalid. Why?

Figure 5

Solution:

a) KCL at node KCL not satisfieda 4 2 7+ =

b) KVL around left hand circuit 2010 6 4 0+ + + = 0= KVL not satisfied

c) KCL at node KCL not satisfieda 2 0=

KVL around lower part of the circuit (CW) 12 0= KVL not satisfied

Example 3: In the given circuit, calculate the unknown quantities.

Figure 6


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Solution:

10V

Vs

4

2A

67

V3 +-

+- V2 +- V4

V1

+

-

3A

I1

a

Figure 7

KCL at node a 13 2 I= + 1 1I A=

Ohms law 2 17 7 1 7v I= + = = V

KVL around left hand circuit 1 210 0v v+ = 1 10 7 0v + = 1 3v V=

Ohms law 3 3 4 12v V= =

(KVL around right hand circuit)4 1 3 0sv v v v+ + =

(Ohms law)1 3(3 6) 0sv v v+ + =

18 ( 3) ( 12) 0s

v+ + = 27s

v V=

Ideal Dependent Sources.

A voltage source whose voltage depends on another voltage or current is called a dependent voltage

source.

Symbol

Figure 8

A current source whose current depends on another voltage or current is called a dependent current

source.

Symbol

Figure 9


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Example 4:

Circuit (a)s

I depends on 2v sI (voltage-dependentcurrent source)

Circuit (b)s

v depends on 1v sv (voltage-dependent voltage source)

Figure 10

Four possible types of dependent sources:

Voltage-dependent voltage source (it is a voltage source that depends on another voltage)

Current-dependent voltage source (it is a voltage source that depends on another current)

Voltage-dependent current source (it is a current source that depends on another voltage)

Current-dependent current source (it is a current source that depends on another current)

Example 5:

a) Calculate the value of the dependent current source.

b) Show that the power generated is equal to the power dissipated.

Figure 11


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Solution:

a) KCL at node a 2 4i i i + = 0 i2 3i = (1)

(KVL around left hand circuit)2 1 20 0v v + + =

(Ohms law)2(40 ) (20 ) 20 0i i + + =

[Using (1)](40 3 ) 20 20 0i i + + =

120 20 20 0i i + + =20 1

100 5i A= =

1 4

4 45 5

i = = A (Value of the dependent current source)

b)20

1 2020 20 4

5 5Vp i= + = + = = W

20 1

1 1 1 4( )(20 ) ( )(4)5 5 5 5

p iv = + = + = + = W

2

33

5i i= = A & 2 2

340 40 24

5v i= + = + = V

40 2 23 72245 5

p i v = + = + = W

4 2

4 96(4 ) ( ) 24 19.2

5 5iA p i v= = = = W

4 72 20 4 72 964 19.2

5 5 5 5disp W

+ += + + = = =

19.2genp W=

19.2dis gen p p W = =

Figure 12

3-Handouts_Lec_03_WFig_ver_01

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