Electrical Technology Topic 1 Sem 1 Politeknik

DET1013 - ELECTRICAL TECHNOLOGY

Chapter 1:

Introduction to Electric Circuit

AUTHORS:AMINAH BINTI OTHMANJA’AFAR BIN SURADIJUNAIDA BINTI SHAARIZULKURNAIN BIN ABDUL HAMID

COURSE LEARNING OUTCOME

1. Apply the concept and principles of DC electrical circuit using different method and approach. (C3, PLO1)

2. Solve DC circuit problems using appropriate DC electrical laws and theorems. (C3, PLO2)

3. Conduct the laboratory activities of DC electrical circuit using appropriate electrical equipment. (P4, PLO5)

4. Demonstrate ability to work in team to complete assigned task during practical work sessions. (A3, PLO11)

TOPIC TITLE (RTA)CHAPTER TITLE RTA

1.0 INTRODUCTION TO ELECTRIC CIRCUIT

08:12

2.0 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS

08:10

3.0 CAPACITORS AND CAPACITANCE 05:004.0 INDUCTORS AND INDUCTANCE 05:045.0 MAGNETIC CIRCUIT,

ELECTROMAGNETISM AND ELECTROMAGNETIC INDUCTION

04:04

LEARNING OUTCOME (2 Hours)1.1 Know standard symbols for electrical components.

1.1.1 Identify common symbols in electrical circuit diagrams.

1.2 Understand the general features of cells and batteries.1.2.1 Describe the difference between cells and batteries.1.2.2 Show the effects of different cell connections:

a. seriesb. parallelc. series-parallel

1.2.3 Calculate the total voltage of series sourcesa. with the same polarities.b. with opposite polarities.

Standard Symbol for Electrical Components

G V A

conductor /wire switchground/earth

Cell (dc supply) Battery (dc supply) AC supply

resistor inductor capacitor

galvanometer voltmeter ammeter

Cell

• A single unit of a primary or secondary battery that converts chemical energy into electric energy.

Battery

• A battery is a series of two or more connected cells, which changes chemical energy into electrical energy.

Relationship of Cells & Batteries

• A Battery is a combination of cells• Cell combination could be in

SERIES, PARALLEL & SERIES-PARALLEL

• Practically, a cell is also notified as a battery.

Series Connection Cells

Series Connection Cells

Total e.m.f., ET = E1 + E2 + E3 + E4

= 2.0 + 2.0 + 2.0 + 2.0 = 8V

Example 1.1Calculate total e.m.f. of the circuit below

Parallel Connection Cells

Parallel Connection Cells

Total e.m.f., ET = E1 = E2 = E3 = 2.0V


Series-Parallel Connection Cells

Series-Parallel Connection Cells

Total e.m.f. for series cells, ESeries = E1 + E2 + E3 + E4

= 2.0 + 2.0 + 2.0 + 2.0 = 8V Total e.m.f., ET = ESeries = 8V


Series Connection with same Polarities

Total e.m.f., ET = E1 + E2

= 8 + 6 = 14V


Series Connection with opposite Polarities

Total e.m.f., ET = E1 + E2

= 8 - 6 = 2V


i) 5V 11V 4V

B AAnswer: 20V

44V

ii) 44V B A

44VAnswer: 44V

SELF-EXERCISEQUESTION: Calculate total e.m.f. of each cells connection as follow.

ANSWER

ANSWER


iii) 4V 3V

B 2V 5V AAnswer: 7V

6V 1V

iv) 4V 4V 4VB A

Answer: 120V30 cells

ANSWER

ANSWER


v) 14V B A 20 cells 14V

Answer: 14V

14V

5V 5V 5V

vi) B 5V 5V 5V A

5V 5V 5V

Answer: 50V10 cells

ANSWER

ANSWER

LEARNING OUTCOME (1 Hour)1.3 Know electric current and quantity of electricity.

1.3.1 State the definition of electric current. 1.3.2 State the unit of charge. 1.3.3 Indicate charge or quantity of electricity Q from Q=It.

1.4 Know the main effects of electric current. 1.4.1 Identify the three main effects of electric current,

giving practical examples of each.

1.5 Understand resistance and resistivity 1.5.1 Explain that electrical resistance depends on four

factors. 1.5.2 Express that resistance R= ρI/A where ρ is the resistivity.

Electric current, I• Current: - motion of charge

- depends on the rate of flow of charge- electric fluid- unit of current is ampere (A)

• Equation: dq = changing of charge

I = dt = changing of timeI = current (ampere)

• For steady state condition: Q = charge (coulomb)I = , thus Q = It t = time (second)

Example 1.6If a current of 5 A flows for 2 minutes, find the charge transferred.

Electric current, I

Q = It = 5 x 2 x 60 = 600 C

Main Effect of Electric Circuit

1. Heat Effect - Example: soldering iron, water heater, fuse, bulb, cookers, electric fires, furnaces, kettles, iron

2. Magnetic Effect - Example: bells, relays, motors, generators, transformers, telephones, lifting magnets, car ignition

3. Chemical Effect - Example: cell and battery, electroplating

Resistance & resistivity

• Resistance – property of a component which restricts the flow of electric current.

• The value of resistance depends upon 4 factors:

1. Length, l2. Cross-sectional area, A3. resistivity, ρ4. temparature


• Equation:R = [Unit = Ω]

R = resistance [Ω] l = Length [m] A = Cross-sectional area [m2] ρ = resistivity [Ω.m]

• Resistivity is difference for different material

Example 1.7Calculate resistance of a 5m long conductor if it has cross sectional area and resistivity Ω.m


Resistance, R= =

= 1.5Ω

Resistor (R)

• A device that is manufactured to have specific resistance.

• Used to limit current flow and reduce voltage applied to other components.

• Basic unit is ohm (Ω)

Resistor (R)

• Different examples of resistors

SELF-EXERCISEi) In what time would a current of 1 A transfer a charge of 30 C?

Answer: 30sANSWER

ii) What would be the resistivity of 2m length conductor wire if the resistance value is 500Ω and the cross sectional area 0.5

Answer: 125µΩmANSWER

LEARNING OUTCOME (1 Hour)

1.6 Understand Ohm’s Law. 1.6.1 Explain Ohm’s Law. 1.6.2 Outline the procedure adopted when using Ohm’s Law

1.7 Apply Ohm’s Law in circuit. 1.7.1 Construct circuit to explain Ohm’s Law. 1.7.2 Use Ohm’s Law to find current, voltage and resistance in a circuit.

.

Ohm’s Law

• Ohm’s Law states that the current (I) through a conductor between two points is directly proportional to the potential difference or voltage (V) across the two points, and inversely proportional to the resistance (R) between them.

I =

Ohm’s Law Triangle

V = IR

I =

R =

V

RI

Simple Circuit

E = E.M.F. (Electromotive force)

- Generates from voltage source

- Example: cells / batteries

I =

Current =

From Ohm’s Law:

Simple Circuit

V drop = Voltage drop--------------------------- - appears when current, I flows through resistor,R.- Inverse polarity from E

V drop = IR

From Ohm’s Law:

Voltage = Current x Resistance

Simple Circuit

• A battery possess e.m.f. that produces DC current.

• A complete circuit should consist of at least 1 electricity source (battery) and 1 load (resistor)

Source Load

E R

+

-• Current will only produce when the

source (battery) is connected to the load (resistor) in close loop connection.

I

Simple Circuit



Source Load

E R

+

- • Current will only produce when the source (battery) is connected to the load (resistor) in close loop connection.

I

• When current flows across resistor, R, voltage drop, Vd will be produced across R

+

-

Vd

Simple Circuit



Source Load

E R

+

- • Current will only produce when the source (battery) is connected to the load (resistor) in close loop connection.

I

• When current flows across resistor, R, voltage drop, Vd will be produced across R

+

-

Vd

Simple Circuit (Example)

15V 10Ω

+

-

I

+

-

Vdrop

Example 1.8QUESTION: By referring to the circuit below, calculate:i) Current, Iii) Voltage drop across resistor 10Ω, Vdrop

Simple Circuit (Example)

15V 10Ω

+

-

I

+

-

Vdrop

i) Current, I = = = 1.5A

ii) Voltage drop, = IR = 1.5 x 10 = 15V

SELF-EXERCISEA 100 V battery is connected across a resistor and causes a current of 5 mA to flow. Determine the resistance of the resistor. If the voltage is now reduced to 25 V, what will be the new value of the current flowing?

R = 20kΩ

I = 1.25mA

ANSWER

ANSWER

LEARNING OUTCOME (2 Hours)1.8 Understand series, parallel and series-parallel connections.

1.8.1 Identify a series circuit.1.8.2 Explain the flow of current and voltage division in the series circuit.1.8.3 Identify a parallel circuit.1.8.4 Explain the voltage drop and the current division in the parallel circuit.1.8.5 Explain the equivalent resistance in series and parallel

circuits.1.8.6 Identify a combination of series and parallel circuit.1.8.7 Explain the total resistance for the combination of series and parallel circuit.

LEARNING OUTCOME (2 Hours)1.9 Apply series, parallel and series-parallel connections to dc circuit.

1.9.1 Construct a series connection circuit1.9.2 Calculate the flow of current and voltage division in the series circuit.1.9.3 Construct a parallel circuit.1.9.4 Calculate the voltage drop and the current division in the parallel circuit.1.9.5 Construct a series-parallel connection circuit.1.9.6 Calculate the equivalent resistance in series and parallel circuits.1.9.7 Calculate the total resistance for the combination of series and parallel circuit.1.9.8 Use of voltage divider in series circuit and use of current divider in parallel circuit.1.9.9 Solve problems related to series, parallel and combination of series and parallel circuits.

Series Circuit

• Is formed when any number of devices are connected end-to-end so that there is only one path for current to flow.

Series Circuit Characteristics

Series Circuit Characteristics

1. Resistances are additive RT = R1 + R2 + R3

2. The current flows throughout the circuit is same.I = IR1 = IR2 = IR3

3. Different resistors have their individual voltage drop

VR1 ≠ VR2 ≠ VR3

4. Total e.m.f equals to the sum of voltage drops across each resistorE = VR1 + VR2 + VR3

Equivalent resistance in series

RT = R1 + R2 + R3Equ. 1

• Applicable to any means of resistors.• Standard equation of series connection

resistors.

Equivalent resistance in series (resistors with same value)

r = resistance valuen = amount of resistors

Equ. 2 RT = r x n

• Applicable for any means of resistors with same value.

Voltage Divider Rule

VR1 = x E

Series Circuit (Example)

15V

4Ω 6Ω

8Ω

Example 1.9By referring to the circuit above, calculate:i) Total resistance of the circuit, ii) Current, Iiii) Voltage drop across resistor 6Ω,

Series Circuit (Example)

15V

4Ω 6Ω

8Ω

i) Rtotal = 4 + 6 + 8 = 18Ω

ii) I = = = 0.833A

iii) VR2 = IR2 = 0.833 x 6 = 5V or

VR2 = x E = x 15 = 5V

use VDR

Parallel Circuit

• Is formed when two or more devices are arranged in a circuit side by side so that current can flow through more than one path

Parallel Circuit Characteristic

Parallel Circuit Characteristic1. Total resistance can be determined from:

RT =

2. Different resistors have their individual current.IR1 ≠ IR2 ≠ IR3

3. Same voltage acts across all parts of the circuit E = VR1 = VR2 = VR3

4. Supplied current equals to the sum of different current flows through each resistor.

I = IR1 + IR2 + IR3

Equivalent resistance in parallel

RT = Equ. 1

• Applicable to any means of resistors.• Standard equation of parallel connection

resistors.

Equivalent resistance in parallel (2 resistors case)

RT = Equ. 2

• Applicable for 2 resistors connection only.

Equivalent resistance in parallel (same value case)

RT = r = resistance valuen = amount of resistors

Equ. 3

• Applicable for any means of resistors with same value.

Current Divider Rule (CDR)

IR1 = x IEqu. 1

• Applicable to any means of resistors.• Standard equation of current divider rule

Current Divider Rule (2 Resistors case)

IR1 = x IEqu. 2

• Applicable for 2 resistors connection only.

Parallel Circuit (Example)

6Ω

Example 1.10QUESTION: By referring to the circuit above, calculate:i) Total resistance of the circuit, ii) Current, Iiii) Voltage drop across resistor 8Ω, iv) Current through resistor 4Ω,

20V4Ω 8Ω

Parallel Circuit (Example)

20V6Ω4Ω 8Ω

i) Rtotal = = 1.846Ω

ii) I = = = 10.83A

iii) VR3 = E = 20V

iv) IR1 = = = 5A or

IR1 = x I = x 10.83 = 5A

CDR

Series-Parallel Circuit

For this diagram:• R1 is parallel with R2.• Ra is series with equivalent resistance of R1

and R2.

Total Resistance of Series-Parallel Circuit

• RT is the equivalent resistance of Ra, R1 and R2

referencepoint

• Start solving by calculating the total resistance of parts located farthest away from the reference point.

• Exception: if there are any series/parallel connection resistors at any part of circuit which is not farthest from the reference point, solve the total resistance of the series/parallel connection first. Then you can use the tips mentioned above to solve your problem.

Total Resistance of Series-Parallel Circuit (Example )

Example 1.11Calculate equivalent resistance, of the circuit below.


Rb =

RT = Ra + Rb

Total Resistance of Series-Parallel Circuit (Example)

A

BRT

10Ω 10Ω 5Ω

6Ω3Ω

4Ω

8Ω

9Ω

Example 1.12Calculate the total resistance, RT of the circuit below.


A

BRT

10Ω 10Ω 5Ω

6Ω3Ω

4Ω

8Ω

9Ω

Step 1: Identify any series/parallel connection (in between) and calculate the total resistance.

Ra

Ra = 4 + 8 = 12Ω


A

BRT

10Ω 10Ω 5Ω

6Ω3Ω

4Ω

8Ω

9Ω

Step 1: Identify any series connection (in between) and calculate the total resistance.

Ra

Ra = 4 + 8 = 12Ω

12Ω


A

BRT

10Ω 10Ω 5Ω

6Ω3Ω

9Ω

Step 2: Identify the farthest part from ref. point and calculate the total resistance.

Ra

Rb = 5 + 6 = 11Ω

12Ω

Rb


A

BRT

10Ω 10Ω 5Ω

6Ω3Ω

9Ω

Step 2: Identify the farthest part from ref. point and calculate the total resistance.

Ra

Rb = 5 + 6 = 11Ω

12Ω Rb 11Ω


A

BRT

10Ω 10Ω

3Ω

9Ω

Step 3: calculate the total resistance of next portion until reach ref. point.

Ra

Rc = = 2.36Ω

12Ω Rb 11Ω

Rc


A

BRT

10Ω 10Ω

3Ω

9Ω


Ra

Rc = = 2.36Ω

12Ω Rb 11ΩRc

2.36Ω


A

BRT

10Ω 10Ω

9Ω


Ra

Rd = 10 + 2.36= 12.36Ω

12ΩRc

2.36Ω

Rd


A

BRT

10Ω 10Ω

9Ω


Ra

Rd = 10 + 2.36= 12.36Ω

12ΩRc

2.36ΩRd

12.36Ω


A

BRT

10Ω

9Ω


Ra

Re = = 6.09Ω

12ΩRd

12.36Ω

Re


A

BRT

10Ω

9Ω


Ra

Re = = 6.09Ω

12ΩRd

12.36Ω

Re6.09Ω


A

BRT

10Ω

9Ω

Step 4: Finally, calculate the total resistance, RT of the circuit.

RT = 10 + 6.09 + 9 = 25.09Ω

Re 6.09Ω


A B10kΩ 9kΩ3kΩ 3kΩ

6kΩ

6kΩ

Example 1.13Calculate the total resistance across point A - B



6kΩ

6kΩ

RT

Reference point

Ra

Ra = 3k + 3k = 6kΩ



6kΩ

6kΩ

RT

Reference point

Ra

Ra = 3k + 3k = 6kΩ

6kΩ


A B10kΩ 9kΩ

6kΩ

6kΩ

RT

Reference point

Rb = = 2kΩ

6kΩ

Rb


A B10kΩ 9kΩ

6kΩ

6kΩ

RT

Reference point

Rb = = 2kΩ

6kΩ

Rb

2kΩ


A B10kΩ 9kΩ

RT

Reference point

RT = 10k + 2k + 9k = 21kΩ

2kΩ

∴


Series-Parallel Circuit (Example)

+- 20V

2kΩ 4kΩ

8kΩ

6kΩ

18kΩ

20kΩ

Is

Example 1.14QUESTION: By referring to the circuit above, calculate:i) Equivalent resistance of the circuit, Rtotalii) Current from supply, Is iii) Current through resistor 18kΩiv)Voltage drop across resistor 8kΩ,

Series-Parallel Circuit (Example)2kΩ 4kΩ

8kΩ

6kΩ

18kΩ

20kΩ

+-

20V

Isi) Rtotal Calculation

• Temporarily, remove voltage source from the circuit.

• The open nodes leaved by your voltage source would be your reference point

Rtotal Ra

Ra = 4k + 8k + 6k = 18kΩ

Series-Parallel Circuit (Example)2kΩ 4kΩ

8kΩ

6kΩ

18kΩ

20kΩ

i) Rtotal Calculation



Rtotal

Ra

Ra = 4k + 8k + 6k = 18kΩ

18kΩ

Series-Parallel Circuit (Example)2kΩ

18kΩ18kΩ

20kΩ




Rtotal

Ra

Rb = = 9kΩ

Rb


18kΩ18kΩ

20kΩ




Rtotal

Ra

Rb = = 9kΩ

Rb 9kΩ


20kΩ




Rtotal

Rtotal = 2k + 9k + 20k= 31kΩ

Rb 9kΩ31kΩ

Series-Parallel Circuit (Example)ii) Is Calculation

• Place voltage source back to the circuit.

• Your current from source is calculated using Ohm’s Law

Rtotal

Is = = = 645.16 μA

31kΩ+-

20V

Is


+-

20V

2kΩ 4kΩ

8kΩ

6kΩ

18kΩ

20kΩ

645.16μA ii) I18 Calculation

• Use current divider rules (CDR) or any other relevant methods

I18


+-

20V

2kΩ 4kΩ

8kΩ

6kΩ

18kΩ

20kΩ

645.16μA ii) I18 Calculation

• Use current divider rules or any other methods relevantI18

18kΩRa

If Use CDR: I18 = x 645.16μ = 322.58 μA

Other method: I18 = = 322.58 μA


+-

20V

2kΩ 4kΩ

8kΩ

6kΩ

18kΩ

20kΩ

645.16μA ii) V8 Calculation

• Calculate the current flows through 8kΩ resistor first

• Use Ohm’s Law to calculate the Voltage drop

• Other method as Voltage Divider Rule (VDR) also could be used here if you understand well the technique

322.58μA 322.58μA

I8 = 645.16μ – 322.58μ= 322.58 μA


+-

20V

2kΩ 4kΩ

8kΩ

6kΩ

18kΩ

20kΩ

645.16μA ii) V8 Calculation

• Calculate the current flows through 8kΩ resistor first

• Use Ohm’s Law to calculate the Voltage drop

• Other method as Voltage Divider Rule (VDR) also could be used here if you understand well the technique

322.58μA 322.58μA

I8 = 645.16μ – 322.58μ= 322.58 μA

+

V8

-

V8 = IR = 322.58μ x 8k = 2.58V

SELF-EXERCISEFind the value of the total resistance, current from supply and voltage drop across resistor 90Ω in the diagram as below

= 24.5Ω

I = 2.041A

= 45.92V

ANSWER

ANSWER

50V

2Ω 4Ω

90Ω

22Ω

8Ω 8Ω

ANSWER

LEARNING OUTCOME1.10 Understand Delta–Star transformation. 1.10.1 Express formula required to transform from Delta to Star

and Star to Delta1.10.2 Illustrate circuits to show star and delta connections.1.10.3 Explain steps to solve problems involving Star-Delta transformation.

1.11 Apply the concept of Delta–Star transformation.1.11.1 Construct circuits to show star and delta connections.1.11.2 Solve problems involving Star-Delta transformation.

1.12 Understand electrical power and energy.1.12.1 Explain electrical power and energy.1.12.2 Express electrical power formula from Ohm’s Law and the unit.1.12.3 Calculate the electrical power and energy in a circuit.

Delta-Star Transformation

• Standard 3-phase circuits or networks take on two major forms with names that represent the way in which the resistances are connected, a Star connected network which has the symbol of the letter, Υ (wye) and a Delta connected network which has the symbol of a triangle, Δ (delta).

Delta-Star Transformation

R1 R3

R2

a

b

c

Ra

RcRb

Ra = Rb =

Rc =

Star-Delta Transformation

R1 R3

R2

a

b

c

Ra

RcRb

R1 =

R2 =

R3 =

Delta-Star (Example)

x y

4Ω

6Ω

8Ω

12Ω

10Ω

Example 1.15Calculate the total resistance, Rxy of the circuit below.

Rxy


x y

4Ω

6Ω

8Ω

12Ω

10Ω

Ra

Rb

Rc

Ra = = 1.78Ω Rb = = 2.67Ω

Rc = = 1.33Ω

Convert --- Y


x y

12Ω

10Ω

1.78Ω

2.67Ω

1.33Ω

Rd

Re

Rd = 1.33 + 12= 13.33 Ω

Re = 2.67 + 10= 12.67 Ω


x y1.78Ω

10Ω

2.67Ω

12Ω1.33Ω

Rd

Re

Rd = 1.33 + 12= 13.33 Ω

Re = 2.67 + 10= 12.67 Ω

13.33Ω

12.67Ω


x y1.78Ω

Rf = = 6.5Ω

13.33Ω

12.67Ω

Rf


x y1.78Ω

Rf = = 6.5Ω

13.33Ω

12.67Ω

6.5Ω

Rxy = 1.78 + 6.5 = 8.28Ω

Electrical Power & Energy

• ELECTRICAL POWER is defined as the rate at which electrical energy is transferred by an electric circuit.

• The SI unit of power is Watt.• Equation:

Power, P = VI Equ. 1

V – voltage measured in Volts (V)

I – current measured in Ampere (A)

http://en.wikipedia.org/wiki/Watt_(unit)

Electrical Power & Energy• From Ohm’s Law;

I = V/R and V = I*R

Hence Power, P = Equ.2

P = I2R Equ.3


• ENERGY can be defined as capacity to do work

• The unit of energy is Joule• Equation :

Energy/Work Done, W = Pt

P – power measured in Watt (W)

t – time measured in seconds (s)


20V

15Ω

25Ω

Example 1.16By referring to the circuit below, calculate:i) Power that’s supplied by the batteryii) Power that’s absorbed by 25Ω resistoriii) Energy supplied by the battery after 30siv) Energy absorbed by the 15Ω resistor after 2 hours


20V

15Ω

25Ω

i) Power that’s supplied by the battery, PsRT

RT = 15 + 25 = 40ΩIT

IT = = = 0.5A

Use Equ. 1:

Power, Ps = V*I = 20 x 0.5 = 10W


20V

15Ω

25Ω

ii) Power that’s absorbed by 25Ω resistor PL

0.5A

Use Equ. 3:

Power, PL = I2*R = 0.52 x 25 = 6.25W


20V

15Ω

25Ω

iii) Energy supplied by the battery after 30s

0.5A

Energy, W = P*t = 10 x 30 = 300 J


20V

15Ω

25Ω

iv) Energy absorbed by the 15Ω resistor after 2 hours

0.5A

Energy,

W = P*t = I2*R*t = 0.52 x 15 x 2 x 60 x 60 = 27 kJ

SELF-EXERCISEi) Diagrams below show a delta connection circuit

with its equivalent star connection circuit. If R1=20kΩ, R2=40kΩ and R3 =80kΩ, calculate Ra, Rb and Rc

11.43kΩ

5.71kΩ 22.86kΩ

ANSWER

ANSWERANSWER

SELF-EXERCISEii) With refer to the diagram as below, calculate power that supplied by the battery and power dissipation at resistor 40kΩ.

= 1.6mW

Ps = 4.8mWANSWER

ANSWER

RECAP• Cell and battery are sources of DC type of

electricity.• Voltage, current and resistance are recognized

as three basic elements of electrical circuit which contribute in Ohm’s Law.

• Electrical circuit can be constructed in series, parallel and combination of series-parallel connection.

• Star-Delta transformation technique is required to analyze network that involve Star/Delta connection.

• Power and Energy is the product of voltage and current elements of a circuit.

REFERENCES

Main: John Bird (2010). Electrical Circuit Theory & Technology. Fourth

Edition. Newness. (ISBN: 978-0-08-089056-2)

Additional: 1. Allan R. Hambley (2011). Electrical Engineering, Principles

and Applications, Fifth Edition. Prentice Hall. (ISBN-13: 978-0-13-213006-6)

2. B.L. Theraja (2010).Textbook of Electrical Technology .S Chand & Co Ltd. (ISBN: 978-8121924900)

REFERENCES

3. Darren Ashby (2011). Electrical Engineering 101, (3rd Ed ) [Paperback] Elsevier Inc. (ISBN: 978-0123860019)

4. John Bird. (2010). Electrical And Electronic Principles And Technology. Fourth Edition. Newness. (ISBN: 978-1-85617-770-2)

5. Meizhong Wang. (2010). Understandable Electric Circuits First edition © 2005 Higher Education Press, China, English translation ©2010 The Institution of Engineering and Technology. (ISBN 978-0-86341-952-2)

6. V. K. Mehta (2010). Principles of Electrical Engineering and Electrical [Paperback] S Chand & Co Ltd. (ISBN: 978-8121927291)

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