chapter 1 techniques of dc circuit analysis

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Chapter 1 : Techniques of DC Circuit

Analysis

Engr. Mohd Riduwan bin Ghazali (Grad. IEM)

1.1 Review Circuit Analysis I:

Most Important thing that should remember:

Ohm Laws, V = IR

(Kirchoff’s) KCL/KVL

Independent/Dependent Source

Nodal/SuperNode

Mesh/SuperMesh Mesh/SuperMesh

Source Transformation

OHM’S LAW

Ohm’s Law state that the voltage, v across a resistor is directly proportional to the current, i flowing through the resistor.

Current and voltage are linearly proportional

V=iR

KIRCHHOFF’S LAW

• States that the sum of currents entering a node (or closed boundary) is zero.

Kirchhoff current law (KCL)

1 2 3 4 5( ) ( ) 0i i i i i+ − + + + − =

1 3 4 2 5i i i i i+ + = +

Total current in = Total current out

KIRCHHOFF’S LAW

• States that the sum of voltages around a closed path (or loop) is zero

Kirchhoff voltage law (KVL)

1 2 3 4 5 0v v v v v− + + − + =

2 3 5 1 4v v v v v+ + = +

CIRCUIT ELEMENTS

Active Elements Passive Elements

IndependentSources

(round shape)

DependantSources

(diamond shape)

• voltage source comes with polarities (+-) in its symbol• current source comes with an arrow

NODAL ANALYSIS

Provides a general procedure for analyzing circuits using node voltagesas the circuit variables.• Steps:

i) select a node as a reference node (ground)ii) assign voltage designations to nonreference nodesiii) redraw the circuit to avoid too much information on the

same circuit

iv) apply KCL for each nonreference node- at node 1: I1 = I2 + i1 + i2 -at node 2: I2 = i3 – i2

v) apply Ohm’s law (current flows from a higher potential to a lower potential)

R

vvi lowerhigher −

=1

11

0

R

vi

−=2

212 R

vvi

−=3

23

0

R

vi

−=

Nodal Analysis With Voltage Sources

We now consider how voltage sources affect nodal analysis:

CASE 1: If a voltage source is connected between the reference node, we simply set the voltage at the nonreference node equal to the voltage of the voltage source

E.g.: v1 = 10V

CASE 2: If the voltage source (dependent or independent) is connected CASE 2: If the voltage source (dependent or independent) is connected between two nonreference nodes, the two nonreference nodes form a generalized node or supernode; apply both KCL and KVL to determine the node voltagesE.g.: nodes 2 and 3 form a supernode

MESH ANALYSIS

Steps to determine mesh currents

• assign mesh currents i1, i2,,…..,in to the nmeshes

• apply KVL to each of the n meshes. Use Ohm’s law to express the

voltages in terms of the mesh currents

• solve the resulting simultaneous equations to get the mesh

currents1,A p p ly in g K V L to m e s h

1 1 1 3 1 2

1 3 1 3 2 1

1,

( ) 0

( )

A p p ly in g K V L to m e s h

V R i R i i

R R i R i V

− + + − =+ − =

2 2 2 3 2 1

3 1 2 3 2 2

2 ,

( ) 0

( )

A p p ly in g K V L to m e s h

R i V R i i

R i R R i V

+ + − =− + + = −

1 1

2 2

1 3 3

3 2 3

i VR R R

R R R i V

+ − = − + −

CASE 1

When a current source exists only in one mesh, we set i2 = -5A and write a mesh equation for the other mesh in the usual way:

-10 + 4i1 + 6(i1 – i2) = 0 i1 = -2A

MESH ANALYSIS WITH CURRENT SOURCES

CASE 2

When a current source exists between two meshes, we create a supermeshby excluding the current source and any elements connected in series with it

MESH ANALYSIS WITH CURRENT SOURCES

Applying KVL in (b),

-20 + 6i1 + 10i2 + 4i2 = 0 6i1 + 14i2 = 20 ………..(i)

Applying KCL to a node where the two meshes intersect,

i2 = i1 + 6 ………… (ii)

Solve (i) and (ii), i1 = -3.2 A, i2 = 2.8A

SOURCE TRANSFORMATION

• series parallel combination and wye-delta transformation help simplifycircuits• source transformation is another tool for simplifying circuits• is the process of replacing a voltage source, Vs in series with a resistor by acurrent source in parallel with a resistor, or vice versa• basic to these tools is the concept of equivalence

Riv ss =

R

vi s

s =

Superposition theorem

For circuit network have more than one independent source

voltage or current produced by a source acting in isolation can be determined by assuming other sources do not work, where the resources should be switched off in the following manner: -be switched off in the following manner: -

Independent voltage source – short circuit (0 V) or internal resistance if have

Independent current source – open circuit (0 A)

Superposition theorem cont.

Example for superposition theorem.

Solution.

IB

Solution.1. Current source IB work ( voltage source VB off- short circuit)

Get the value of I’I’

Superposition theorem cont.

2. Voltage source VB work (current source IB off – open circuit)

I”

Get the value of I”

3. So get the value of current flowing at resistance R2 with I = I’ +I”

Record:- Various methods can be used to obtain the value of I 'and I “, such as current divider or mesh analysis or nodal analysis or node.

Superposition theorem cont.

Example 1.10

For the circuit in the figure below, find the value of I, the voltage across

the resistor 2Ω and the power absorbed by the resistor

0.5 Ω 0.5 Ω

I

0.5 Ω2 Ω

5 V

0.5 Ω

10 VI

V2Ω

Superposition theorem cont.

1. Voltage source 5V work ( voltage source 10 V off – short circuit)

Solution

Find the value of I’ ( )1 2 //1 1.67

53

1.67

1' 1

2 1

T

T

T

R

I A

I xI A

= + = Ω

= =

= =+

Superposition theorem cont.2. Voltage source 10 V work ( voltage source 5 V off – short circuit)

Find the value of I” ( )1 2 //1 1.67R = + = ΩFind the value of I” ( )1 2 //1 1.67

106

1.67

1" 2

2 1

T

T

T

R

I A

I xI A

= + = Ω

= =

= =+

3. So the value of current I (make sure the direction of I’ and I”)

2

2 22

' ( ") 1

( 1)(2) 2

( ) ( 1) (2) 2

I I I A

V IR V

P I R W

Ω

Ω

= + − = −= = − = −

= = − =

Superposition theorem cont.

Example 1.11Refer to the figure below, calculate current flowing and voltage across resistance 4Ω.

8 Ω1 Ω

I

5 A2 Ω4 Ω

1 Ω

Superposition theorem cont.

Example 1.12

Refer to the figure below, determine voltage across resistance 4 Ω.

5 Ω 2 Ω 4 Ω

1 Ω3 Ω 3 A 4 Vx

-+Vx

Thevenin theorem

This theorem is in use to facilitate a complex circuit network to a simple circuit called the Thevenin equivalent circuit.

The equivalent circuit contains a voltage source Vth in series with a resistor Rth

Complex Circuit

a

b

Thevenin theorem cont.

The steps to get the Thevenin equivalent circuit:-a. Remove section of the network where to find the thevenin

equivalent circuit and mark clearly the two terminals as a-b

Complex Circuit

I=0 A

VTH

a

b

OFF Circuit

a

bRTH

equivalent circuit and mark clearly the two terminals as a-b

b. Determine the Thevenin equivalent resistance seen from the terminal a-b with independent sources is turn off

c. Get the values of Thevenin voltage on the voltage across the terminal a-b when the terminal at open circuit. (various method can be used to obtain Vth, whether to used loop analysis/nod analysis)

d. Draw the Thevenin equivalent circuit and connect the back portion removed from the (a) above

Thevenin theorem cont.

Example 1.13For circuit below, sketch the Thevenin equivalent circuit at terminals a-b.

6 Ω 2 Ωa

20 V

4 Ω

5 A RL

b

Thevenin theorem cont.

Solutiona. Remove RL from circuit

b. Determine RTH seen from terminal a-b with all independent sources are turn off.

Thevenin theorem cont.

c. Get VTH at terminal a-b

5

205

1070

70

X

X

TH A X

V

V V

V V V

− =

== = =

VTH

I=0 A

d. Draw the Thevenin equivalent circuit and connect the back portion removed from the (a) above

VTH=70 V RL

RTH=12Ω

a

b

Thevenin theorem cont.

Example 1.14

Refer to the circuit below, sketch Thevenin equivalent circuit at terminal

a-b, next calculate the current flowing, I3Ω and voltage across, V3Ω the resistor 3Ω,

5 Ω 1 Ωa

4 Ω 3 ΩV3Ω

a

b

28 V

I3Ω

Thevenin theorem cont.

Example 1.16Refer to the circuit below, get the value of V1/3Ω

½ Ω

½ Ω3 V

¼ Ω 1/3 Ω V1/3Ω2 A

Norton theorem

This theorem is in use to facilitate a complex circuit network to a simple circuit called the Norton equivalent circuit.

This equivalent circuit consists of a current source IN connected in parallel with a resistor RN.

Complex Circuit

a

b

Norton theorem cont.

The steps to get the Norton equivalent circuit:-a. Remove section of the network where to find the Norton

equivalent circuit and mark clearly the two terminals as a-b

Complex Circuit

IN

a

b

OFF Circuit

a

bRN

equivalent circuit and mark clearly the two terminals as a-b

b. Determine the Norton equivalent resistance seen from the terminal a-b with independent sources is turn off

c. Get the Norton current value of current flowing through the terminals a-b when a short circuit in the terminal. (various method can be used to obtain IN, whether to used loop analysis/nod analysis)

d. Draw the Norton equivalent circuit and connect the back portion removed from the (a) above

Norton theorem cont.

Examples

Determine Norton equivalent circuit at terminals a-b for circuit below.

Next calculate current flowing and voltage across resistance 3Ω

5 Ω 1 Ωa

I

4 Ω 3 ΩV3Ω

b

28 V

I3Ω

Norton theorem cont.

Solution.a. Remove RL from circuit

b. Determine RN seen from terminal a-b with all independent sources are turn off.

RN=(1+(5//4))=3.22 Ω

Norton theorem cont.

c. Get IN at terminal a-b (short a-b)

IN

IT

4 28 43.86

5 (5 (4 //1)) 5N TI I A = = = +

d. Draw the Norton equivalent circuit and connect the back portion removed from the (a) above

IN=3.86 A RN=3.22Ω 3Ω

Current flowing 3Ω,

( )3

3.223.86 2

(3 (3.22))I AΩ = =

+

Voltage across 3Ω

( )3 3 3 6V I VΩ Ω= =

Norton theorem cont.

Example 1.19

½ Ω

½ Ω3 V

Get the value of V1/3Ω

¼ Ω 1/3 Ω V1/3Ω2 A

Thevenin and Norton theorem with dependent sources

To analyze circuits with independent sources, (IN) and (VTH) may be obtained by using the analysis as before.

However, the Thevenin and Norton resistance can not be obtained directly from the network because of dependent sources can not be turned off as an independent source.independent source.

therefore, to solve the circuit dependent sources, two ways:

1. Determine the value of VTH and IN, so

THN TH

N

VR R

I= =

Thevenin and Norton theorem with dependent

sources CONT.

2. Introduce an independent voltage source,VT or an independent current source, IT at the root a-b. VT

and IT value is any value. However, free resources available on the network must be turned off prior circuit.

I

Off Circuit

a

b

I

VTOff Circuit

a

bIT

TTH N

VR R

I= = Ω ab

TH NT

VR R

I= = Ω

Vab

Thevenin and Norton theorem with dependent

sources CONT.

Example 1.20Sketch the Thevenin equivalent circuit at terminal a-b, next calculate value Iab

Thevenin and Norton theorem with dependent

sources CONT.

Solution.

a) Remove resistance 3 Ω from circuit

b) Get the value VTH- +

Write equation every loop

Thevenin and Norton theorem with

dependent sources CONT.

Solve the equation above to get value I2, next find value VTH

c) Get resistance value of Thevenin equivalent, RTH

RTH can be solve in two way.

i) Get value Ii) Get value IN- +

Thevenin and Norton theorem with

dependent sources CONT.

ii) Introduce an independent source.

1. introduce independent voltage source

- +

Thevenin and Norton theorem with

dependent sources CONT.

II. introduce independent current source

Thevenin and Norton theorem with

dependent sources CONT.

So, sketch the Thevenin equivalent circuit

VTH=48.2 V

Relationship between the Thevenin and Norton theorem

Thevenin equivalent circuit can be converted into the Norton equivalent circuit or vice versa by changing the concept (super transformation)of supply where: -

a) Thevenin resistance (RTH) value is equal to the Norton resistance (RN)

Maximum power transfer

a circuit will supply maximum power to the load if the load resistance RL is equal to the equivalent resistance seen by the load

Maximum power transfer can be obtained by replace a complex circuit with the Thevenin equivalent circuit or Norton equivalent circuitcircuit or Norton equivalent circuit

Maximum power transfer cont.

Power to the load RL,

Condition maximum power transfer RL = RTH

Therefore, maximum power supplied to the load is:

2

2 THRL L L

TH L

VP I R R

R R

= = +

2 2V V2 2

4 4TH TH

RLmakL TH

V VP

R R= =

Maximum power transfer cont.

Th

ThThL R

VpRR

4

2

max ==

Maximum power transfer cont.

ExamplesFrom circuit below, calculate:-

a) Value RL when maximum output power

b) Maximum power absorb by load RL

Maximum power transfer cont.

Solution.a) Remove RL from circuit and off all source from circuit to find RTH

b) Get the thevenin voltage at terminal a-b when RL removed from circuit

Maximum power transfer cont.

So, draw Thevenin equivalent circuit when maximum power transfer happen.

Maximum power transfer cont.

Examples 1.22

From circuit below, calculate:-

a) Value RL when maximum output power

b) Maximum power absorb by load RL