Chapter 2-Electric Circuits

CGE535

Munawar Zaman Shahruddin

Faculty of Chemical Engineering

Universiti Teknologi MARA, Shah Alam

[email protected]

Tel: 03-5544 8019; 019-249 0416

ELECTRICAL AND INSTRUMENTATION TECHNOLOGY

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Week 2-4

Chapter 2: Electric Circuits

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State and apply basic circuit laws to solve for currents, voltages and powers in electric circuits.

Apply the voltage-division and current-division principles to analyze electric circuit.

Analyze the circuits using the node-voltage and mesh-current technique.

Apply the Thevenins theorem to analyze the electric circuit.

Lesson Outcome

At the end of class, students should be able to:

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Electric Circuit

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The path that the current follows

Consist of a voltage source, load and conductor.

Voltage source: battery,

generator, etc

Load: resistor, capacitor, LED, etc

Conductor

Series

Parallel

Series-Parallel (Combination)

Open Circuit

Closed/Short Circuit

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Types of Electric Circuits/Resistors

TOPIC 1: Circuit/resistor in

parallel and series.

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Series Circuit/Resistors

Parallel Circuit/Resistors

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Series-Parallel Circuit/Resistors

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Open Circuit

Closed Circuit

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Series Circuit/Resistors

In a series circuit, the current flow is the same throughout the circuit

RT=R1+R2+R3+Rn

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Example 1

According to the figure, determine IT

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Example 1

Given VT=12 V, R1=560 , R2=680 , R3=1000 Therefore, RT=R1+R2+R3=560+680+1000=2240

Based on Ohms Law: IT=VT/RT=12/2240= 0.0054 A

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Parallel Circuit/Resistors

In a parallel circuit, the current divides among the branches of the circuit and recombines on returning to the voltage source.

1/RT=1/R1+1/R2+1/R3+1/Rn

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Example 2

According to the figure, determine R2

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Example 2

Given VT=120 V, IT=0.2 A, R1=1000 , R3=5600

Based on Ohms Law: RT=VT/IT=120/0.2= 600

According to parallel circuit, 1/RT=1/R1+1/R2+1/R3

1/600=1/1000+1/R2+1/5600; R2=2048.78

Draw a schematic of the circuit and label all known quantities

Solve for equivalent circuits and redraw the circuit

Solve for the unknown quantities

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Tips on Solving Series and Parallel Circuit

END OF TOPIC 1

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TOPIC 2: Kirchhoffs Law

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Kirchhoffs Law was introduced in 1845 by G. R. Kirchhoff as an extension work of Ohm which solve calculation of currents, voltages and resistances in electrical circuits with multiple loops.

It also in-lines with the first law of thermodynamic that involved the electrical energy.

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Overview

Gustav Robert Kirchhoff (1789-1854)

Branch/Junction any portion of a circuit with two terminals connected to it. A branch may consist of one or more circuit elements.

Node junction of two or more branches.

Loop any closed connection of branches.

Mesh exclusive loop/does not contain other loop

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Basic in term in the electric circuit

Find the branches, nodes, loops and meshes in the following circuit:

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EXAMPLE 3

First Law: Kirchhoffs Current Law (Junction Rule)

Second Law: Kirchhoffs Voltage Law (Loop/Mesh Rule)

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Kirchhoffs Law

The sum of the currents at a node must equal zero (since charge cannot be created but must be conserved)

= 0

=1

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First Law: Kirchhoffs Current Law

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First Law: Kirchhoffs Current Law

Node a: I1 I2 I3 = 0 Node b: I2 + I3 I1 = 0

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Example 4

i2=3A, i3=3A, i6=2A, i5=-2A, i7=1A

Hint: Search all possible nodes one to another

The net voltage around a closed circuit is zero (no energy is lost or created in an electric circuit term-the sum of all voltages associated with the sources must equal the sum of the load voltages

= 0

=1

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Second Law: Kirchhoffs Volatge Law

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Second Law: Kirchhoffs Voltage Law

Mesh ABEF: Vs V1 V2 = 0 Mesh BCDE: V2 V3 V4 = 0 Loop ABCDEF: Vs V1 V3 V4 = 0

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Example 5

Mesh 1: V1 = 8 12 = - 4 V Mesh 2: V2 = 6 12 = -6 V Mesh 3: V3 = 10 6 = 4 V Mesh 4: V4 = 8- 10 = -2 V V2+v3+v4-v1=-6+4-2+4 =0 verified!

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Circuit Analysis

Previously, you have already learned a part of the circuit analysis which are:

Series and Parallel Circuit/Resistors

Kirchhoffs Law

The aim of circuit analysis is to solve the circuit problem with regards to the value of R, I, V or P.

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Circuit Analysis

There are another part of circuit analysis to be learned which are:

Voltage Divider Rule

Current Divider Rule

Wye Delta Transformation

Resistive Circuit/Network Analysis Mesh-Current /Node-Voltage Analysis

Thevenins Theorem

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Voltage Divider Rule

Consider the following series circuit:

1 =

1

.

.

.

=

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Example 6

Find R by using Voltage Divider Rule

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Example 6

First, solve the parallel resistor-load(rl): Rload=3.2/10

-2=320 Rrl=(1/470+1/320)-1=190.4 According to VDR, Vload=Vsource(Rrl/R+Rrl) 3.2=20(190.4)/(R+190.4) So, R=(20(190.4)/3.2)-190.4 =999.61000 or 1k

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Current Divider Rule

Consider the following parallel circuit:

= 1 = 2 = =

I=I1+I2++In

1 = 21 + 2

=11 + 2

2 = 11 + 2

=21 + 2

=

1 + 2 ++

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Example 7

Consider the following parallel circuit:

Calculate I1 and I2 based on Current Divider Rule

30V 5 10

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Example 7

RT=(1/5+1/10)-1=3.33; I = 30/3.33 = 9 A

I1=G1/(G1+G2)x I = ((1/5)/(1/5+1/10))x9 = 6 A

I2=I-I1=9-6=3 A

30V 5 10

END OF TOPIC 2

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TOPIC 3: Wye-Delta Transformation

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Instead of combination between series and parallel resistors, it can be a situation where the circuit does not follow either pattern/network such as: Y or T network and or network

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Wye-Delta-Wye Transformation

Y Network T Network

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Wye-Delta-Wye Transformation

Network Network

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Delta-Wye Transformation

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Delta-Wye Transformation

Each resistance in the network Y is the sum of the product of two adjacent branches of the resistance in the delta network, divided by the sum of the three resistance in delta

=12

1 + 2 + 3

=23

1 + 2 + 3

=13

1 + 2 + 3

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Wye-Delta Transformation

Each resistance in delta network is the product of the total resistance of all the Y taken twice at a time, divided by the resistance against the network Y

1 = + +

2 = + +

3 = + +

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Example 8

By referring to the following circuit, find RT

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Example 8

Should be transformed

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Example 8

=6 18

6 + 18 + 12=108

36= 3

=18 12

36=216

36= 6

=12 6

36=72

36= 2

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Example 8

Parallel Network Based on Parallel Network analysis,

Rn=2+12||6+12 = 14||18 = 7.875 So, RT=20+3+7.875 = 30.875

END OF TOPIC 3

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TOPIC 4: Resistive Circuit Analysis & Thevenins

Theorem

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So far, we have already cover the topic circuit analysis with one voltage or current source, what if there are more than one sources? That is why we have Resistive Network Analysis in solving aforementioned circuit problem.

There are two different methods in Resistive Network Analysis: Mesh-Current@Mesh/Loop analysis

Node-Voltage@Nodal Analysis

Resistive Network Analysis

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Mesh analysis is based on Kirchhoffs Voltage Law. Normally the current direction can be assumed as clockwise.

All positive-negative charge at resistor should be labeled respectively.

In the end, there are 2 or 3 sets of algebraic equations to be solved simultaneously by using Cramer's Rule or other suitable method.

Mesh Analysis

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Mesh Analysis

Mesh 1:

VA-V1-VL1=0

VA=I1R1+RL1(I1-I2)

VA=I1(R1+RL1)-I2RL1

Mesh 2:

VL1-V2-VB=0

VB=RL1(I2-I1)-I2R2

VB=I2(RL1-R2)-I1RL1

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Mesh Analysis

Solve the equation using Linear Algebraic Equation Solution Technique:

VA=I1(R1+RL1)-I2RL1

VB=I2(RL1-R2)-I1RL1 +

Use Cramers Rule to solve it!

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Consider the following circuit:

Example 9

Find I in each mesh

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Example 9

Mesh 1:

20-6I1-10(I1-I3)-4(I1-I2)+10=0

I1(6+10+4)-4I2-10I3=30

20I1-4I2-10I3=30

1 2

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Mesh 2:

-10-4(I2-I1)-11(I2-I3)-8-3I2=0

-4I1+I2(4+11+3)-11I3=-18

-4I1+18I2-11I3=-18

Mesh 3:

8-11(I3-I2)-10(I3-I1)-9I3+12=0

-10I1-11I2+I3(11+10+9)=20

-10I1-11I2+30I3=20

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Example 9

20I1-4I2-10I3=30

-4I1+18I2-11I3=-18

-10I1-11I2+30I3=20

Cramers Rule

I1=2.473 A

I2=0.594 A

I3=1.709 A

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Exercise 1

Find I in each mesh

I1=0.5 A

I2=-0.12 A

I3=0.06 A

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Recap/Additional matters

Source transformation:

AV

An alternative approach on a

mesh/nodal analysis

1.

2.

3.

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Supermesh

A supermesh occurs when a current source is contained between two essential meshes.

The circuit is first treated as if the current source is not there. This leads to one equation that incorporates two mesh currents.

Once this equation is formed, an equation is needed that relates the two mesh currents with the current source.

This will be an equation where the current source is equal to one of the mesh currents minus the other.

The following is a simple example of dealing with a supermesh.

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Supermesh

Mesh 1,2 (assume no current source): Vs-I1R1-I2R2=0 I1R1-I2R2=Vs Current source eq: Is=I2-I1

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Example 10

Find i1 and i2

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Example 10

Mesh 1,2 (assume no current source): Vs-I1R1-I2R2=12-9I1-3I2-6I2=0 9I1+9I2=12 Current source eq: I2-I1=1.5 I2=1.5+I1 So, 9I1+9(I1+1.5)=12 18I1+13.5=12 I1=-1.5/18=-0.083 A I2=1.5-0.083=1.417 A

Write down the equation to solve the following mesh analysis:

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EXERCISE 2

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Nodal Analysis

Nodal analysis is based on Kirchhoffs Current Law. Every point at the junction can be treated as node and one node is assigned as reference node.

The aim is to determine the voltage in the circuit.

Other than reference node, are assumed to have positive sign in front.

Solution will be obtained based on the same method in Mesh Analysis.

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Consider the following circuit:

Nodal Analysis

R5

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Recognize the nodes first

Nodal Analysis

Node 1:

I1+I2+I3=0 1 11

+12+1 2

3= 0

11

1+1

2+1

323=11

R5

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Nodal Analysis

Node 2:

I4+I5+I6=0 2 1

3+24+2 + 25

= 0

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3+1

4+1

513= 25

R5

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Nodal Analysis

To obtain Vnode1 and Vnode2, the equation should be order and solve as Linear Algebraic Equation (Cramers Rule)

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3+1

4+1

513= 25

11

1+1

2+1

323=11

(1

1+1

2+1

3) 1

3

(1

3+1

4+1

5) 1

3

11

25

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Example 11

Based on the following circuit, find Ix and Px

Ix=-2.33A

Px=27.1W

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Supernode

In this circuit, we initially have two unknown voltages, V1 and V2.

The voltage at V3 is already known to be VB because the other terminal of the voltage source is at ground potential.

The current going through voltage source VA cannot be directly calculated. Therefore we can not write the current equations for either V1 or V2.

However, we know that the same current leaving node V2 must enter node V1. Even though the nodes can not be individually solved, we know that the combined current of these two nodes is zero.

This combining of the two nodes is called the supernode technique, and it requires one additional equation:

V1 = V2 + VA.

The complete set of equations for this circuit is:

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Example 12

Based on the following circuit, find Va and Vb

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Example 12

Supernode:

-1.5+Va/6+3.5+Vb/3=0

Va/6+Vb/3=-2

Voltage source: Vb=12+Va

Va/6+(12+Va)/3=-2

Va=-12 V

Vb=0 (surprise?yes its true)

Any two output terminals A&B of an active linear network containing independent sources (voltage/current sources) can be replaced by a simple voltage source of magnitude VTh in series with RTh .

RTh is the equivalent resistance of the network when looking from the output terminals A&B with all sources (voltage/current sources) removed and replaced by their internal resistances.

VTh is equal to the open circuit voltage across the A&B terminal.

This will simplify the complicated circuit network as it will eventually producing Thevenins equivalent circuit as a final result.

Voltage sourcesshort circuit; current sourcesopen circuit

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Thevenin's Theorem

Leon Charles Thevenin (1789-1854)

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Thevenin's Theorem

A

B

Looking from backward

Thevenins Equivalent Circuit

Simplified the following circuit and find Rth

Should the value of voltage source is 10 V, find Vth

Next draw the Thevenins Equivalent Circuit

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Example 13

A

B

Simplified the following circuit and find Rth

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Example 13

*3 12=64=2.4

Final form:

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Example 13

By removing voltage source, determine RTH:

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Example 8

RTH=((((102.4)+1)2.4)+1)5 RTH=1.58

VTH can be obtained by any circuit analysis e.g. mesh analysis

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Example 13

VTH

Mesh 1: 12.4I1-2.4I2=10 Mesh 2: -2.4I1+5.8I2-2.4I3=0 Mesh 3: 2.4I2=8.4I3 By solving the sets of equations, Current @ 5 resistorI3=0.119 A So, VTH=0.119 x 5 = 0.6 V

With the value of RTH and VTH, the Thevenins Equivalent Circuit can be obtained:

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Example 13

RTH=1.58

A resistive network contains independent and dependent sources.

A load is connected to a pair of terminals labeled a b.

What value of load resistance permits maximum power delivery to the load?

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Maximum Power Transfer

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2

V2 Th

p = i R = RL L

R + RTh L

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2(R + R ) - 2R (R + R )dp 2 Th L L Th L

= V = 04Th

dR (R + R )L Th L

2(R + R ) = 2R (R + R )

Th L load Th L

R = RL Th

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2V

2 Thp = I R = Rmax 2L L

(2R )L

2VTh

p =max

4RL

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Example 14

Find the value of RL for maximum power transfer to RL. Next find the maximum power that can be delivered to RL.

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Example 14

RL=RTH=15030=25

VL=VTH=(150/180)x360=300 V

So, P=VTH2/4RTH=300

2/(4x25)

= 900 W

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Conclusion

Kirchhoffs Law Voltage/Current Divider Rule Wye delta transformation Mesh-Nodal Analysis Thevenins Theorem Maximum Power Transfer

Chapter 2-Electric Circuits

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