Basic Electrical 1

105
Basics of Electrical Circuits Dr. Sameir Abd Alkhalik Aziez University of Technology Department of Electromechanical Engineering Resources :- Introductory circuit Analysis; by Robert L. Boylestad . Basic Electrical Engineering science; by Mckenzie smith and K.T. Hosie Electrical Technology. د ترجمة، لكھربائية ا الھندسة علم. زكيحمد م& د. أنور مظفرDefinitions :- Electric charge : Fundamental property of sub-atomic particles . Positive charge proton negative charge electron no charge neutron Like charges repel & opposite charges attract. Unit of charge coulomb Symbol and Units:- Difference between a symbol and unit. . ) ( 180 ) ( Unit C Symbol = Η Current :- The movement of electrons is the current which results in work begin done in an electric circuit. Hence, the Electric Current is the rate of flow of charge. We have a current when there is a few of charges; S C A t q I = Δ Δ = The direction of motion of positive charges is opposite to direction of motion of negative charges. Amper ( ) I t q t Sec Coul = Δ Δ = Δ × × = = 6 19 10 10 6 . 1 . . q = Charge that flows through section in time Δ t. Δ Uniform flow of charges direct current ( dc ) - 1 -

Transcript of Basic Electrical 1

Page 1: Basic Electrical 1

Basics of Electrical Circuits

Dr. Sameir Abd Alkhalik AziezUniversity of TechnologyDepartment of Electromechanical Engineering

Resources :-

• Introductory circuit Analysis; by Robert L. Boylestad .

• Basic Electrical Engineering science; by Mckenzie smith and K.T. Hosie

• Electrical Technology.

مظفر أنور. د& محمد زكي . علم الھندسة الكھربائية ، ترجمة د •

Definitions :-

Electric charge : Fundamental property of sub-atomic particles .

Positive charge proton

negative charge electron

no charge neutron

Like charges repel & opposite charges attract.

Unit of charge coulomb

Symbol and Units:- Difference between a symbol and unit.

. )(180)( UnitCSymbol =Η

Current :- The movement of electrons is the current which results in work

begin done in an electric circuit. Hence, the Electric Current is the rate of flow

of charge.

We have a current when there is a few of charges;

SCA

tqI =⇒ΔΔ

=

The direction of motion of positive charges is opposite to direction of motion of

negative charges.

Amper ( ) Itq

tSecCoul

=ΔΔ

××==

− 619 10106.1..

q = Charge that flows through section in time Δ t. Δ

Uniform flow of charges ⇒ direct current ( dc )

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Example: In a copper wire a flow of charge is 0.12 C in a time of 58 ms . Find

the current in this wire ?

Solution: AtQ

tq

I 06.21058

12.03=

×==

Δ

Δ=

Work & Energy:

dFw Δ=Δ .

Where:

wΔ dΔ: Is the work done by a force ( F ) acting over a displacement ( ).

As charges move, subject to various forces they may gain or lose energy.

= work done = charge in energy wΔ

External force work done on charges charges gain energy.

( Like Sources )

Work done by charges charges lose energy

(Like loads )

= Joules = N.m wΔ

Potential difference:

Potential difference between two points is the energy gained or lost by a unit

charge as it moves from on point to the other.

coulombJoulevolt =⇒

ΔΔ

=qwv

wΔ q= work done for transporting a total charge Δ .

Potential difference

Voltage

Potential

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Power :- Time rate for doing work .

IVtq

qw

qq

twP

twP

.=ΔΔ

×ΔΔ

=ΔΔ

×ΔΔ

=

ΔΔ

=

Watt (w) = sJ = V.A

Systems of unit ( S.I. ) :-

Quantity Unit Symbol

1) Charge (q) Coulombs C

2) Current (I) Amper SCA =

3) Force (F) Newton N

4) Work , energy (w) Joule J=N.m

5) Voltage (V) Volt cJV =

6) Power (P) watt AVsJw .==

7) Length ( l ) meter m

8) Temperature (T) Kelvien K

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Electrical Circuits :-

Circuit element :- is a two – terminal electrical component .

Electrical circuit :- Interconnected group of elements .

Source = current and voltage in the same direction.

load = current and voltage in the opposite direction .

Resistive element, is always load and always dissipates power

+ VA

I

V

- VB

I

+ VA

I

V

- VB

I

- VA

I

V

+ VB

I

+ VA

I

V

- VB

I

- VA

I

V

+ VB

I

(1) Load

issipates power) P = V.I

(2) Source

(supplies power)

(3) Source

(supplies power , charging)

(4) Load

(dissipates power) (d

+I

V

-

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Source of emf ( electro motive force ) s

Source ( supplies power )

Load ( consumes power )

Ex. Charging of batteries.

Resistance :-

The unit of resistance ( R ) is ohm ( Ω )

The resistance of any material is depends on four factors:

1. The length of conductor ( l ) lαR⇒

2. Cross – section area of conductor ( A ) AR 1α⇒

3. The nature of material conductor.

4. The temperature of conductor.

AR lα

AR lρ=

ρ = constant is called resistivity or specific resistance unit of resistivity.

-

-

V I

I

-

l

A

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mmmRA

AR .Ω===⇒=

lρρ or

.Ωl

2

cm.Ω

22

2⎟⎠

⎜⎝

==A π ⎞⎛ dr π

r = radians of section.

d = diameter of section.

Con ctance ( G ):-du

ll ρρA

AR

===11

(Siemens (S)) G

Prefix :-

pico 10 P -12

nano 10-9 n

micro 10-6 μ

milli 10-3 m

centi 10-2 c

deci 10-1 d

Kilo 103 K

Mega 106 M

Giga 109 G

Tera 1012 T

Example: What is the resistance of 3 Km length of wire with cross section area

6 mm2 and resistivity 1.8 μΩcm .

Solution:

( ) ( )( ) =×

== − 9106 6A

R ×××× −− 10310108.1 326lρΩ

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Example: What is the resistance of 100 m length of copper wire with a

diameter of (1 mm) and resistivity 0.0159 μΩm .

Solution:

( ) ( )Ω=

⎟⎟⎠

×−

02.21001023

6

π

⎜⎝⎠⎝ 22⎜⎛ ×⎟

⎞⎜⎛ 101

2

πdA×

=××

==− 0159.0100100159.0 6ρR l

Effect of Temperature on a resistance :-

slop = TΔRΔ = constant =

1

1

212 TTTT212

TTRRRRRR

−−

=−−

=−−

Example: The resistance of material is 300 Ω at 10Co, and 400 Ω at 60Co. Find

its resistance at 50 Co ?

Solution:

oCTTRR /2

1060300400

12

12 Ω=−−

=−−

= slop

Ω=⇒+=⇒=−

−=

−−

=−−

=

380300808030040

3001050

30021

1

RRR

RRTTRR

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Also from

the above figure we can sea

01

12

TR

TTR

−=

− 02

01

1

02

2 00

T

TTR

T −−

=−−

RT

01

02

1

2

TTTT

RR

−−

=∴ , hence 01

022022 TTTTA −

=⇒−

ρ

ρ

l

1011

TTTTA

−− ρ

l

ple: Aluminum conductor has resistance 0.25 Ω at 10 Co . Find its

resistance at 65 Co ?

Solution:

Exam

( )( )

.31.024630125.0

236102366525.02

01

0212

01

02

1

2

Ω=⎥⎦⎤

⎢⎣⎡=

⎥⎦

⎤⎢⎣

⎡−−−−

=∴

⎥⎦

⎤⎢⎣

⎡−−

=⇒−−

=

R

TTTTRR

TTTTR

The following table illustrate the value of resistivity (

R

ρ ) for different materials

t 20 Co temperature.

a

Material ρ at 20 Co , Ω.m

Silver

Paper

Mica

1.63*10-8

2.83*10-8

20*10-8

10*1010

1011-1015

*Copper

Aluminum

(1.72 - 1.77)*10-8

Iron

Material To , Co

Silver

Aluminum

Iron

-234

.5

-236

-180

Copper -234

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Example: Aluminum conductor with length of 75 cm and 1.5 mm2 cross

section area. Find its resistance at 90 Co ?

Solution:

( ) ( )

Ω=×=××= −− m15.14105.141105083.2 44

××××

== −

−−

AR

105.110751083.2

6

28lρ

فان الحل ينته20Co اذا كانت المقاومة مطلوبة في درجة حرارة : . الى هنا يمالحظة

( )( )

.236202369015.

209015

1

2

Ω⎥⎦⎤

⎢⎣⎡

++

⎢⎣

⎡−−−−

− TTR

r method

18= m14=

236 ⎦

236⎥⎤ .142 =R

01

02

−=

TTR

Anothe

⎟⎠⎞

⎜⎝⎛

++

=⇒−−

=2362023690

209001

02

1

2 ρρρρ

TTTT

( )

Ω=∴

×=

×

××⎟⎠⎞

⎜⎝⎛

++

==

−0 8

m

AR

15.14

183.2

105.1

10752362023690

90

20

6

220

90

ρ

ρ

ρρ l

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Driving :

( )

( )⎥⎥⎦

⎢⎢⎣

⎡−⎟⎟

⎞⎜⎜⎝

⎛−

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−−−+=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−−

+=

⎟⎟⎠

⎞⎜⎝ − 01

12 TT⎜⎛

−+−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=

−−

=

1201

12

01

1212

01

010212

01

0212

02

01

0212

01

02

1

2

11

1

1

11

11

TTTT

RR

TTTTRR

TTTTTTRR

TTTTR

TTRR

TTTTRR

TTTT

RR

Let

R

011

1TT −

=α temperature coefficient of resistance at a temperature T1

( )[ ]12112 1 TTRR −+= α ∴

Where T0 for copper = -234.5

ource, T0 take an absolute value, which means In some res 0T = 234.5, hence

e can sea w

& 10

20

1

2

TTTT

RR

++

= 1

11

TT +=α

Example:

a) Find the value of 1α at (T1 = 40 Co) for copper wire.

b) Using the result of (a), find the resistance of a copper wire at 75 Co if its

resistance is 30 Ω at 40 Co ?

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-11-

Solution:

a) ( ) 00364.05.274

15.23440

11

011 ==

−−=

−=

TTα 1/K

Or 00364.05.274

1405.234

11

11 ==

+=

+=

TTα 1/K

b) ( )[ ]12112 1 TTRR −+= α

( )[ ] Ω=−+= 8.33407500364.0130

.الحرارة المقاومة ازدادت عندما زادت درجة إنالحظ

Example: If the resistance of a copper wire at freezing ( 0 Co ) is 30 Ω , Find its

sistance at -40 Co ?

Solution:

re

( )( )

Ω=⎟⎞

⎜⎛= 88.245.19430

⎠⎝

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=

5.234

5.23405.2344030

01

0212 TT

TTR

Or

R

⎟⎟⎞

⎜⎛ + 20 TT

⎠⎜⎝ +

=101

2

TTRR

Ω=⎟⎠

⎜⎝

= 88.245.234

30

⎞⎛

5.194

40⎟⎠

⎜⎝ +

=05.234

30⎛ −5.234

Page 12: Basic Electrical 1

Ohm's Law :- Ohm's law states that the voltages ( V ) across a resistor ( R ) is

directly proportional to the current ( I ) flowing through the resistor .

Slop = RV

I 1=

ΔΔ

V = constant = R I

IVR = Ω ; V = I . R ; ⇒

RVI =

The resistance of short circuit element is approaching to zero.

The resistance of open circuit is approaching to infinity.

S.C.∞==

=

RG

R10R

Hence VI

RG ==

1 Siemens ( S ) or mhos ( υ ) .

O.C.01

==

∞=

GR

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Electrical Energy ( W ) :-

tPWt

WP .=⇒=Q KWh

( )( )

tR

V

tRItIVtPW

.

.....

2

2

⎟⎟⎠

⎞⎜⎜⎝

⎛=

=

==

Energy in KWh ( W ) = ( ) ( )1000

ttimePPower ×

Example : For the following circuit diagram , calculate the conductance and

the power ?

Solution :

mARVI 6

10530

3 =×

==

υmR

G 2.010511

3 =×

==

( ) mWVIP 18030106. 3 =××== −

or ( ) ( ) mWRIP 180105106. 332 =×××== −

or ( ) ( ) mWGVP 180102.030. 322 =××== −

or ( )( ) mW

RVP 180

102.030

3

22

== −

-13-

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Efficiency ( η ) :-

Wi/p = Wo/p + Wloss

ttt+=

WWW losspopi //

Pi/p = Po/p + Ploss

%100powerInput powerOutput

× Efficiency ( η ) =

%100×=i

o

PPη

%100×=i

o

WWη

nT ηηηηη ××××= ......321

Example: A 2 hp motor operates at an efficiency of 75 %, what is the power

input in Watt, if the input current is (9.05) A, calculate also the input

voltage?

Solution:

1 hours power (hp) = 746 Watt

%100×=i

o

PPη

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WPP i

i

33.198975.0

1492746275.0 ==⇒×

=

VIPEIEP 22082.219

05.933.1989. ≅===⇒=

Example: What is the energy in KWh of using the following loads:-

a) 1200 W toaster for 30 min.

b) Six 50 W bulbs for 4 h.

c) 400 W washing machines for 45 min.

d) 4800 W electric clothes dryer for 20 min.

Solution : ( ) ( )

KWh

W

htWPW

7.310003700

100016003001200600

100060204800

60454004506

60301200

1000

==+++

=

⎟⎠⎞

⎜⎝⎛×+⎟

⎠⎞

⎜⎝⎛×+××+⎟

⎠⎞

⎜⎝⎛×

=

×=

D.C. Sources:-

The d.c. sources can be classified to:-

1- Batteries . Voltage

Amper - hours

2- generators .

3- Photo cells .

4- Rectifiers .

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V

E

E

V

I

V = E = constant voltage element

V

IIo

IIo

I = Io = constant current element

.مصدر الفولتية يولد تيار و فولتية و لكن الفولتية تكون ثابتة

.مصدر التيار يولد تيار و فولتية و لكن التيار يكون ثابت

Series Circuit :-

V1 = I.R1

V2 = I.R2

V3 = I.R3

E – V1 – V2 – V3 = 0 E = V1 + V2 + V3 ⇒

E = I.R1 + I.R2 + I.R3 ∴

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E = I.[R1 + R2 + R3] = I.RT

The current in the series circuit is the same through each series element &

RT = R1 + R2 + R3 + -------- + RN

3

3

2

2

1

1

RV

RV

RV

REI

T

====

Pt = P1 + P2 + P3 = E.I

Voltage Source in Series:-

-17-

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Example: Find the current for the following circuit diagram?

Solution:

ET = 10 + 7 + 6 – 3 = 20 V

RT = 2 + 3 = 5 Ω

AREII

T

TT 4

520

====

Kirchoff's voltage law ( K.V.L. ):-

The algebraic sum of all voltages around any closed path is zero.

∑=

=m

mmV

1

0

Where m is the number of voltages in the path ( loop ) , and Vm is the mth

voltage .

E – V = 0

E = V

RE

RVI ==

-18-

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E – V1 – V2 = 0

E = V1 + V2 ; RT = R1 + R2

TT RVV

REI 21 +==

Example: Use K.V.L. to find the current in the following circuit diagram?

E1 E2

R1

V1

R2

V2

I

Solution: From K.V.L. ∑ = 0V

∴ E1 – V1 – E2 – V2 = 0

E1 – E2 = V1 + V2

E1 – E2 = IR1 + IR2

E1 – E2 = I ( R1 + R2 )

21

21

RREEI

+−

=∴

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Page 20: Basic Electrical 1

Example: For the following circuit diagram, Find I using:-

a) Ohm's law.

b) K.V.L.

10Ω

40V

10V

20V10V

17Ω

Solution:

a ) By applying ohm's law :-

-20-

AREI T 1

4040

17671010104020

==+++

−+==

b ) By applying K.V.L. :-

10 + 6I + 7I - 40 + 10I – 20 + 10 + 17I = 0

10 – 40 – 20 + 10 + I ( 6 + 7 + 10 + 17 ) = 0

-40 = -I ( 40 ) AI 14040

==⇒

Page 21: Basic Electrical 1

Example :- For the following circuit diagram , find the potential difference

between Node ( A & D ) , and Node ( A & F ) ?

5V

12V8V

6V

C F

DA

B E

Solution : To find the potential difference between Node A & D , we will apply

K.V.L. on the closed loop BADEB

5V

12V8V

6V

C F

DA

B E

+6 + V – 12 – 8 = 0 ∴V2

V1

VV = 20 – 6 = 14 volt

or

+6 – V1 – 12 – 8 = 0

–14 – V1 = 0

V1 = –14 volt ∴

أعلـى مـن جهـد Dمعنى ذلك ان جهـد نقطـة *

. فولت 14 بـ Aنقطة

Take the loop FEDAF to find the potential difference between Node C & F .

–5 + 12 – V – V2 = 0 ∴

–5 + 12 – 14 – V2 = 0 –7–V2 = 0 ⇒ V2 = –7 volt . ⇒

Or Take the loop FEBAF –5 –8 +6 – V2 = 0 V2 = –7 volt . ⇒ ⇒

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Example :- For the following circuit diagram , find the current ?

Solution :

3Ω15V

8V

2Ω2Ω

2Ω2Ω2Ω

IV

A B C

F E D Take the loop FABCDEF

+8 + V – 15 = 0 ⇒ +V – 7 = 0 V = +7 volt ⇒

V = IR ⇒ AI 5.327==

Definitions :-

Node :- Meeting point of 3 or more branches .

Branch :- Series of elements carrying the same current .

Loop :- Is any closed path in a circuit .

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Hence for the loop circuit, we can find :-

A4 nodes and 6 branches

E1 E2

V3 V4

and we can find : V1 , V2 , V3 and V4

as follows :- BD

C

E3

V1

V2

Take the loop BACB ; to find V1

E2 – V1 – E3 = 0 V1 = E2 – E3 ⇒

Or, if we take BCAB ;

E3 – V1 – E2 = 0 V1 = E2 – E3 ⇒

Take the loop ADBA ; to find V2

–E1 + V2 + E2 = 0 V2 = E1 – E2⇒

Take the loop ABCDA ; to find V3

–E2 + E3 + V3 + E1 = 0

V3 = E2 – E3 – E1

V4 = –V3

Or;

–E2 + E3 – V4 + E1 = 0

V4 = E1 + E3 – E2

Example :- For the following circuit diagram , find ; RT , I , V1 , V2 , P4Ω , P6Ω

, PE , verify by K.V.L. ?

R1 R2

V1 V2

I

E=20V

4Ω 6Ω

Solution :-

RT = R1 + R2 = 4 + 6 = 10

AREI

T

21020

===

-23-

Page 24: Basic Electrical 1

VIRV 84211 =×==

VIRV 126222 =×==

( ) WRIP 1642 21

24 =×==Ω ; or ( ) w

RVP 16

48 2

1

21

4 ===Ω

( ) WRIP 2462 22

26 =×==Ω ; or ( ) w

RVP 24

612 2

2

22

6 ===Ω

WIEPE 40202 =×== ; or WPPPE 40241664 =+=+= ΩΩ

To verify results by using K.V.L. ; then

01

=∑=

N

iiV

E – V1 – V2 = 0

E = V1 + V2

20 = 8 + 12

20 = 20 checks

Internal Resistance :-

Every practical voltage or current source has an internal resistance that

adversely affects the operation of the source.

In a practical voltage source the internal resistance represent as a resistor in series

with an ideal voltage source.

In a practical current source the internal resistance represent as a resistor in

parallel with an ideal current source, as shown in the following figures.

RL

E

I

VVo

Practical voltage source

RoRoI

Practical current source

RL

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Where

Ro = Internal resistance

RL = load resistance

According to K.V.L.

E – Vo – V = 0

E – IRo – V = 0

V = E – IRo ∴

Note that an ideal sources have Ro = 0

We can representing a load as a group of parallel resistances.

Hence as the load will increase the current will be increase ( because the

resistance will decrease ) and the voltage will decrease .

This is because the drop voltage due to the internal resistance , as shown in the

following figure :-

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Page 26: Basic Electrical 1

As seen from the above figure , if Ro2 > Ro1 , then V2 < V1 and the drop voltage

will be ( E – V2 ) , which is greater than ( E –V1 ) .

Example :- For the following circuit diagram , calculate I and VL for the

following cases :-

a) Ro = 0 Ω

b) Ro = 8 Ω

c) Ro = 16 Ω

Solution :-

a.) By apply K.V.L.

E – Vo – V = 0

120 – IRo – IRL = 0 120 – 0 – 22I = 0 ⇒

120 = 22I ⇒ I = 5.46 A

VL = I RL = 5.46 * 22 = 120 V

b.) E – Vo – V = 0

120 – 8I – 22I = 0 ⇒ 120 – 30I = 0

120 = 30I ⇒ I = 4 A

VL = I RL = 4 * 22 = 88 V

c.) E – Vo – V = 0

120 – 16I – 22I = 0 ⇒ 120 – 38I = 0

120 = 38I ⇒ I = 3.16 A

VL = I RL = 3.16 * 22 = 69.5 V

Then we can conclude that as Ro increase the total current and load voltage will

decrease.

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Example :- A circuit have load one with 20 Ω and 4A , and load two with 10 Ω

& 6A . Find the current for load three which have 30 Ω ?

Solution :-

1) 20 Ω & 4A

2) 10 Ω & 6A

3) 30 Ω & I = ?

From K.V.L. , then

E – Vo – VL = 0

VL = E – IRo

IRL = E – IRo

4 * 20 = E – 4Ro

80 = E – 4Ro

----------------- ( 1 )

Also

6 * 10 = E – 6 Ro ----------------- ( 2 ) 60 = E – 6Ro

From eq. (1) & (2) , we have

20 = ( 6 – 4 ) Ro ⇒ ∴ Ro = 10 Ω ; sub. this result it in eq. (1) , then

80 = E – 4 * 10 ⇒ ∴ E = 120 V

Now , we Apply K.V.L. for load 3 ;

V3 = E – IRo 30I = 120 – 10I ⇒

40I = 120 ⇒ ∴ I = 3 A for load three.

See from this example that the current will increase as the load will decrease

with constant E & Ro .

-27-

Page 28: Basic Electrical 1

Example :- A circuit have Voc = 25 v and Isc = 50 A , find its current and RL

when VL = 15 V ?

Solution :-

E = Voc = 25 V

Ω=== 5.05025

sco I

ER

From K.V.L.

E – Vo – VL = 0

E – 0.5I – 15 = 0

25 – 0.5I – 15 = 0

0.5I = 25 – 15

-28-

AI 205.0

10==∴

Ω=== 75.02015

IVR L

L

Page 29: Basic Electrical 1

Voltage divider Rule :-

RT = R1 + R2

TREI =

TT RRER

RERIV 1

111... =⎟⎟

⎞⎜⎜⎝

⎛==

TRRER

RERIV 2

22

22... =⎟⎟

⎞⎜⎜⎝

⎛==

Voltage divider rule

Vn = Voltage across Rn

E = The ( emf ) voltage across the series elements .

RT = The total resistance of the series circuits .

Example :- Using voltage divider rule , determine the voltage V1 , V2 , V3 and

V4 for the series circuit in figure below , given that ; R1 = 2KΩ , R2 = 5KΩ ,

R3 = 8KΩ , E = 45 V ?

Solution :-

R1 R2

V1 V2

I

E

R3

V3

V4

VR

ERVT

1510*15

45*10*53

32

2 ===

T

nn R

ERV =

VR

ERVT

610*15

45*10*23

31

1 ===

-29-

Page 30: Basic Electrical 1

VR

ERVT

2410*15

45*10*83

33

3 ===

( ) VR

ERRVT

2110*15

45*10*73

321

4 ==+

= or V4 = V1 + V2 = 21V

To check: E – V1 – V2 – V3 = 0

E = V1 + V2 + V3 45 = 6 + 15 + 24 ⇒

45 = 45

Active Potential :-

رموز االرضي

a

b

Va = 14 V

Vb = 8 V

Vab = 6 V

Vab is the voltage difference between the

point a and point b

Vab = Va – Vb = 14 – 8 = 6V

Vba = Vb – Va = 8 – 14 = - 6V

Vab = - Vba

a

-7 V

a

10 V

a

Va = -7 V

Va = 10 V

Va = 0 V

-30-

Page 31: Basic Electrical 1

R1

R2

R1

R2

20 V

E = 20 V

R1

R2

-12 V

E = -12 V

R1

R2

-31-

Page 32: Basic Electrical 1

Example :- Find Va , Vb , Vc , Vab , Vac and Vbc for the following diagram .

Solution :-

RT = R1 + R2 + R3

= 2 + 5 + 3 = 10 Ω

AREI

T

11010

===

E – V2 – Va = 0

Va = E – V2 = 10 – (2 * 1) = 8 V

Vb = V5 = (1 * 5) = 5 V = Vbc ; Vc = 0 V

or E – V2 – V3 – Vb = 0 Vb = E – V2 – V3 = 10 – 2 – 3 = 5 V ⇒

Vab = Va – Vb = 8 – 5 = 3 V

Vac = Va – Vc = 8 – 0 = 8 V

Vbc = Vb – Vc = 5 – 0 = 5 V

-32-

Page 33: Basic Electrical 1

Equivalence of actual sources :-

Voltage

Source

Current

Source

Open

Circuit

Voc = E

I = 0 oooc G

IV 1=

Short

circuit osc R

EI =

V = 0

Isc = Io

Kirchoff's Current Law ( K.C.L. ) :-

The algebraic sum of ingoing currents is equal to the out going currents

at any point .

∑ ∑= outin II

Or , At any point , the algebraic sum of entering and leaving current is zero .

∑ = 0I

I1

I5

I2

I3

I4

I1 + I2 + I4 = I3 + I5

Or I1 + I2 + I4 - I3 - I5 = 0

-33-

Page 34: Basic Electrical 1

At a

I1 = I2 + I3 13 + 5 – I = 0

Or I1 - I2 - I3 = 0 18 – I = 0

At b ∴ I = 18 A

-I1 + I2 + I3 = 0

Example :- Find the current in each section in the cct. Shown ?

a

b

cd

e 3A

4A8A

2A

3A

Iab

Icd

Ibc

Ide

1A

Solution :-

At node a

3 – 1 – Iab = 0

2 – Iab = 0 ⇒ Iab = 2 A

-34-

Page 35: Basic Electrical 1

At node b

Iab + 3 – Ibc = 0

2 + 3 – Ibc = 0 ⇒ Ibc = 5 A

At node c

Ibc + 4 – Icd = 0

5 + 4 – Icd = 0 ⇒ Icb = 9 A

At node d

Icd – 8 – Ide = 0

9 – 8 – Ide = 0 ⇒ Ide = 1 A

At node e

2 – 3 + Ide = 0

2 – 3 + 1 = 0

0 = 0 check .

Example :- Find the magnitude and direction of the currents I3 , I4 , I6 , I7 in the

following cct. Diagram?

a

b

c

dI1 = 10A

I 2= 12

AI5 = 8A

I7

I6I3

I4

Solution :-

∑∑ = leaveenter II

∴ I1 = I7 = 10 A

-35-

Page 36: Basic Electrical 1

At node a ; suppose I3 is entering

I1 + I3 – I2 = 0

10 + I3 – 12 = 0 I3 = 2 A ⇒

At node b;

I2 enter , I5 leave , ∴ I4 must be leaving

I2 = I5 + I4

12 = 8 + I4 ⇒ I4 = 12 – 8 = 4 A

At node c;

I4 enter , I3 leave , ∴ I6 leave

I4 = I3 + I6

4 = 2 + I6 I6 = 2 A ⇒

At node d;

I5 and I6 enter , I7 leave

I7 = I5 + I6

10 = 8 + 2

10 = 10 Ok.

Resisters in Parallel :-

R1 R2

I2I1

V1 V2

I

V

A

B

From K.V.L. V = V1 = V2

From K.C.L. I = I1 + I2

2

2

1

1

RV

RVI +=From Ω.L.

= V1G1 + V2G2 = V1 ( G1 + G2 )

or I = V ( G1 + G2 )

I = VGT

-36-

Page 37: Basic Electrical 1

Where GT = G1 + G2

Hence 21

21

21 .111

RRRR

RRR T

+=+=

or 21

21.RR

RRRT +=

In the same minner , if we have three resistors in parallel , then:

321

1111RRRR T

++=

321

213132

.....1

RRRRRRRRR

RT

++=

213132

321

.....

RRRRRRRRRR T ++

=

And , if we have N of parallel

resistance , then

NT RRRRR11111

321

−−−−−+++=

Also

PT = P1 + P2 + P3

1

21

12

1111 RVRIIVP ===

Source power T

TTTTs R

ERIEIP2

2 ===

-37-

Page 38: Basic Electrical 1

Example :- For the following cct. Find RT , PT , IT , Ib?

8Ω16 V

IT

8Ω 8Ω 8Ω

Solution :-

===Ω في حالة كون قيم المقاومات متساوية 248

NRRT

AREI

TT 8

216

===

AREIbranch 2

816

1

===

( ) ( ) WRIP TTT 1282.8 22 ===

or PT = E.IT = 16 * 8 = 128W

or PT = P1 + P2 + P3 + P4

( ) ( ) ( ) ( )

W12832323232

8*28*28*28*2 2222

=+++=

+++=

-38-

Page 39: Basic Electrical 1

Example :- For the parallel network in fig. below , find :-

a) R3 , b) E , c) IT , I2 , d) P2 ; given that RT = 4 Ω ?

Solution :-

a)

Ω==⇒=

=−−

++=

++=

++=

101.0

111.0

105.01.025.0

105.01.025.0

1201

101

41

1111

33

3

3

3

321

RR

R

R

R

RRRRT

b) E = V1 = I1R1 = 4 * 10 = 40 V

c)

ARE

RVI

AREI

TT

22040

104

40

22

22 ====

===

d) ( ) ( ) WRIP 8020.2 22

222 ===

or 2

22

2 RVP = , or P2 = I2V2

-39-

Page 40: Basic Electrical 1

Current division Rule :-

21

21.RR

RRIV+

=

1

21

21.

RRR

RRI

RVI

T

+==

21

21 RR

RII+

=∴

In the same miner

21

12 RR

RII+

=

Also 1

2

21

1

21

2

2

1

RR

RRRI

RRRI

II

=

+

+=

2

1

1

2

2

1

GG

RR

II

== ∴

Example :- For the following circut. , find V , I1 and I2?

Solution :-

Ω==+

=+

= 0999.01.100

101.01001.0*100.

21

21

RRRRRT

V = I . RT = 5 * 0.0999 = 0.4995 V

-40-

Page 41: Basic Electrical 1

AVI 004995.01001 ==

AVI 995.41.02 ==

To check I = I1 + I2

5 = 0.004995 + 4.995

5 = 5 Ok.

Example :- Determine the resistance R1 in the figure below?

Solution :-

I = I1 + I2

or I2 = I – I1 = 27 – 21 = 6 mA

V2 = I2R2 = 6 * 10-3 * 7 = 42 mV

V1 = V2 = 42 mV

Ω=== −

210*2110*42

3

3

1

11 I

VR

or

Ω=

+=⇒

+=

−−

2

77*10*2710*21

1

1

33

21

21

R

RRRRII

-41-

Page 42: Basic Electrical 1

Voltage Regulation :-

Voltage Regulation %100% ×−

=FL

FLNLR V

VVV

Where

VNL = No load voltage

VFL = Full load voltage

Also we can write

%100% .int ×=L

R RRV

Where

Rint. = Internal resistor .

RL = load resistor .

Example :- Find the voltage VL and power lost to the internal resistance , if the

applied load is 13 Ω , also find the voltage regulation ?

Rint. = 2Ω

RL = 13ΩVLE = 30V

Solution :-

ARR

EIL

L 2132

30

int

=+

=+

=

VRIEV LL 262*230.int =−=−=

( ) ( ) WRIP Lloss 82.2 2.int

2 ===

%385.15%10026

2630%100% =×−

=×−

=FL

FLNLR V

VVV

or %385.15%100132%100% .int =×=×=

LR R

RV

-42-

Page 43: Basic Electrical 1

Example :- Find the current I , for the network shown:

R1 = 6Ω

I1

I = 42 mA

R2 =24Ω R3 =24Ω

Solution :- All resistance in parallel , so if we define that R = R2 // R3 then :-

Ω=+

=+

= 12242424*24

32

32

RRRRR

Hence

( ) mARR

RII 28612

1210*42 3

11 =

+=

+= −

Example :- Calculate I & V for the network shown

Solution :- We have a short circuit on R2 resistance , hence no current through

R2 , hence the above cct. Can redrawn as fellows:

-43-

Page 44: Basic Electrical 1

mARE

REI

T

6.310*5

183

1

====

VERIV 18. 1 ===

Example :- For the following cct. Network , find RT , IA , IB , IC , VA , VB , I1 ,

I2 ?

R1 = 9Ω

R2 = 6Ω

R5 = 3ΩR4 = 6Ω

R3 = 4Ω

R6 = 3Ω16.8 V

IA

I1

I2 IB IC

Solution :-

RC = 3Ω

IC

IB

IA

RB = 6Ω

RA = 3.6ΩΩ=

+=

+= 6.3

696*9

21

21

RRRRRA

Ω=+

+= 6363*94RB = R3 + R4 // R5

RC = 3 Ω

RT = RA + RB // RC

Ω=+

+= 6.5363*66.3

ARE

TA 3

6.58.16=== I∴

-44-

Page 45: Basic Electrical 1

Apply C.D.R.

ARR

RIICB

CAB 1

633*3=

+=

+=

By K.C.L.

IC = IA – IB = 3 – 1 = 2 A

VA = IARA = 3 * 3.6 = 10.8 V

VB = IBRB = 1 * 6 = 6 V = VC

ARR

RII A 2.1963*6

21

21 =

+=

+=

I2 = IA – I1 = 3 – 1.2 = 1.8 A

To check

E – VA – VB = 0

16.8 – 10.8 – 6 = 0

0 = 0 Ok.

-45-

Page 46: Basic Electrical 1

Example :- Find the resistor required to connect in parallel with the ammeter to

flow 1.2 A , if you know that the fsd ( full scale deflection ) of ammeter is 120

mA , and the resistance of ammeter is 2.7 Ω ?

Solution :-

From K.C.L.

Ish = 1.2 – 0.12 = 1.08 A

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

+=

sh

shA

Ash

R

RRRII

7.27.22.108.1

( )

Ω=

∴=

+=+=

+=

3.008.1324.0

08.1916.224.37.208.124.3

7.224.308.1

sh

sh

sh

sh

sh

RR

RR

R

-46-

Page 47: Basic Electrical 1

Example :- for the following cct. Network , Given that (V= 24 v), Find E ?

E

24Ω

16Ω12Ω

Solution :-

AI 22.1

241 == V

The same voltage ( V = 24 V ) on the

resistor Ra = 4 +16 + 4 = 24 Ω

Hence

AI 12424

2 ==

AIII 3213 =+=∴

Also from K.C.L. I4 = I3 = 3A

Take the closed loop ABCDA , from

K.V.L.

V1 – 6I4 – V – 6I3 = 0

E

24Ω

16Ω12Ω

V

V1

V2

I5B

V3C

I6

I4

D

I2AI3

I1V1 = 6 * 3 + 24 + 6 * 3

VV 601 =∴

AI 5.22460

5 ==∴

AIII 5.535.2456 =+=+=∴

Take the closed loop CBC

– V1 – V2 + E – V3 = 0

E = V1 + V2 + V3

E = 60 + 25.5 * 8 + 5.5 * 8

= 60 + 44 + 44

E = 148 V

-47-

Page 48: Basic Electrical 1

Current Source :-

Example :- Find the voltage ( Vs ) for the circuit below:

10 A

Solution :-

I RL (2-5)ΩVs = IRL = 10 * 2 = 20 V if RL = 2 Ω Vs

Vs = IRL = 10 * 5 = 50 V if RL = 5 Ω

Example :- Calculate V1 , V2 , Vs for the following cct.:

Solution :-

R1

R2

VsI = 5A

V1

V2

V1 = IR1 = 5 * 2 = 20 V

V2 = IR2 = 5 * 3 = 15 V

Vs = V1 + V2 = 10 + 15 = 25 V

Source Conversions :-

A voltage source with voltage E and series resistor Rs can be replaced by

a current source with a current I and parallel resistor Rs as shown :-

sREI =

Current source to voltage source

Voltage source to current source

-48-

Page 49: Basic Electrical 1

Example :- Convert the voltage source in the cct. Below to a current source ,

then calculate the current through the load for each source:

IL

Solution :-

Rs = 2Ω

RL = 4ΩA

RREI

LsL 1

426

=+

=+

= E = 6V

• For the current source cct. ⇓

ARR

RIILs

sL 1

4223 =⎟

⎠⎞

⎜⎝⎛

+=

+=

RL = 4Ω

IL3A

Rs = 2Ω

A

REI

s

326==

=

Example :- Convert the current source in the cct. Shown below to a voltage

source and determine IL for each cct.:

Solution :-

متساوي في الحالتين و هذا ILالحظ ان

.صحيح

• For the current cct.

⇓( )mAI

RRRII

L

Ls

sL

310*610*3

10*310*9 33

33

=∴

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=+

= −

• For the voltage source cct.

( )mAI

RRE

REI

L

LsTL

310*63

273

=+

=+

==

-49-

Page 50: Basic Electrical 1

Current source in parallel :-

Is = 10 – 6 = 4 A & Rs = 3 Ω // 6 Ω = 2 Ω

Example :-

Is = 7 – 3 + 4 = 8 A

Example :- Find the load current in the following cct.:

Solution :-

-50-

Page 51: Basic Electrical 1

AREI 4

832

11 ===

Is = I1 + I2 = 4 + 6 = 10 A

Rs = R1 // R2 ARR

RIILs

ssL 3

1466*106

24824*8

=+

=+

=⇒∴Ω=+

=

-51-

Page 52: Basic Electrical 1

Matrices :-

Second order determinate

Col. 1 Col. 2

-

122122

1 babaa

baD −== 1

b

Col. 1 Col. 2 Col. 3

a1x b1y = c1

a2x b2y = c2

⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

2

1

22

11

cc

yx

baba

2121

2121

22

11

22

11

1

abbacbbc

bababcbc

DDx

−−

=

⎥⎦

⎤⎢⎣

⎥⎦

⎤⎢⎣

==

2121

2121

22

11

22

11

2

abbaacca

babacaca

DDy

−−

=

⎥⎦

⎤⎢⎣

⎥⎦

⎤⎢⎣

==

Example :- Find the value of D

⎥⎦

⎤⎢⎣

⎡ −=

2614

D

Solution :-

( ) 14686*12*42614

=+=−−=⎥⎦

⎤⎢⎣

⎡ −=D

-51-

Page 53: Basic Electrical 1

Example :- Solving the equations below by determinates

4I1 – 6 I2 = 8

2I1 + 4 I2 = 20

Solution :-

⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −208

4264

2

1

II

( )( ) A

DDI 428.5

121612032

2*64*420*64*8

426442068

11 =

++

=−−−−

=

⎥⎦

⎤⎢⎣

⎡ −

⎥⎦

⎤⎢⎣

⎡ −

==

ADDI 28.2

282*820*4

2820284

11 =

−=

⎥⎦

⎤⎢⎣

==

Third – order determinant :-

333

222

111

cbacbacba

3

2

1

aaa

3

2

1

bbb

D =

[ ] [ ]321321321321321321 cabbcaabcbacacacbaD ++= − + +

Example :- Find the value of D

⎥⎥⎥

⎢⎢⎢

⎡−=

240012321

D

-52-

Page 54: Basic Electrical 1

Solution :-

412

02

1

240012321−⎥⎥⎥

⎢⎢⎢

⎡−=D

( )[ ] [ ]

[ ] [ ] 148228002402

2*2*21*0*43*1*04*2*30*0*22*1*1

−=+−=++−−+=

D

D −++= − + + −

Example :- Find V1 , V2 , V3 from the following equations :-

2V1 + 4V2 +2V3 = 8

5V1 – 2V2 – 10V3 = 18

V1 + 8V2 – 20V3 = -8

Solution :-

⎥⎥⎥

⎢⎢⎢

−=

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

−−−

8188

20811025242

3

2

1

VVV

82

4

152

20811025242

82

4

8188

208810218248

11

−⎥⎥⎥

⎢⎢⎢

−−−

−−⎥⎥⎥

⎢⎢⎢

−−−−

==DDV

( ) ( ) ( ) ( )[ ] ([ ) ( ) ( ) ( ) ]

( ) ( ) ( )[ ] ( ) ( ) ( )[ ]4*5*202*10*82*2*18*5*21*10*420*2*24*18*208*10*82*2*88*18*28*10*420*2*8

1 −+−+−−+−+−−−+−−

=V − + − − − + +− −

VV 35.46842976

1 ==

-53-

Page 55: Basic Electrical 1

6848

188

152

208110185282

22

−⎥⎥⎥

⎢⎢⎢

−−−

==DDV

68482

4

152

8811825842

33

−⎥⎥⎥

⎢⎢⎢

−−

==DDV

Star – Delta ( Δ→Υ ) and Delta – Star ( Υ→Δ ) transformation :-

1. ) Delta – Star ( ) transformation :- Υ→Δ

If the value of RAB , RCA , RBC are

known, and we need to get the values

of RA , RB , RC ; then :-

If RAb = RBC = RCA = R∆ , in this case ΥΔ ==== RRRRR CBA 3

or 3Δ

Υ =RR

BCCAAB

CAABA RRR

RRR++

=

BCCAAB

BCABB RRR

RRR++

=

BCCAAB

BCCAC RRR

RRR++

=

-54-

Page 56: Basic Electrical 1

B

A

C30Ω

30Ω30Ω ⇒

B

A

C

10Ω

10Ω 10Ω

2. ) Star – Delta ( ) transformation :- Δ→Υ

If the value of RA , RB , RC are known ,

and we need to get the values of RAB ,

RCA , RBC ; as follows :-

If Υ=== RRRR CBA , in this case RAB = RCA = RBC = R∆ = 3 RY

B

A

C

4Ω 4Ω

B

A

C12Ω

12Ω12Ω

C

CACBBAAB R

RRRRRRR ++=

A

CACBBABC R

RRRRRRR ++=

B

CACBBACA R

RRRRRRR ++=

-55-

Page 57: Basic Electrical 1

Examples of star and delta connections and transformation :-

Delta connection Star connection

Example :- Find the current flow in the 25 V source for the following circuit :-

Solution :-

Ω=++

= 5.215105

15*5aR

Ω=++

= 515105

15*10bR

Ω=++

= 67.115105

10*5cR

25 V

Rc Rb Ra

5Ω 8Ω 10Ω

( ) ( ) ( ) ( ) ( ) ( )

AREI

R

R

RR

RRRR

T

T

T

T

T

cbaT

92.104.13

2504.1367.637.6

67.6135.1213*5.12

67.613//5.12567.185//105.2

58//10

===

Ω=+=

+⎟⎠⎞

⎜⎝⎛

+=

+=++++=++++=

-56-

Page 58: Basic Electrical 1

Example :- For the following network , find I ?

Solution :- The resistances ( 6 , 3 , 3 ) are delta , can convert to star connection

as follows:

Ω=++

= 5.1336

3*6aR , Ω=

++= 5.1

3363*6

bR , Ω=++

= 75.0336

3*3cR

( ) ( )[ ]( ) ( )[ ]

75.05.35.55.3*5.5

5.3//5.52//4

+⎥⎦⎤

⎢⎣⎡

+=

+=+++=

c

cbaT

RRRRR

AREI

T

077.2889.26

===

-57-

Page 59: Basic Electrical 1

Example :- Find I for the following cct. network :-

12 V

20Ω15Ω

12.5Ω 10Ω

30Ω

I

Solution :- The resistors ( 5Ω , 10Ω , 20Ω ) are star convert to delta

I

15Ω

12.5Ω

30Ω

b

a

c

Rbc

Rab

Rac

12 V

Ω=++

=

Ω=++

=

Ω=++

=

3510

20*55*1020*10

5.1720

20*55*1020*10

705

20*55*1020*10

bc

ac

ab

R

R

R

It is clear that ( 12.5 Ω // Rac ) and (15 Ω // Rbc ) and (30 Ω // Rab ) , hence the

circuit can be reduce to the following network :-

-58-

Page 60: Basic Electrical 1

12 VR3

I

R1

R2

( )

-59-

Ω=+

== 3.75.175.125.17*5.12//5.121 acRR

( ) Ω=+

== 5.10351535*15//152 bcRR

( ) Ω=+

== 21703070*30//303 abRR

( )

( )

Ω=+

==

+=

+=

634.9218.1721*8.1721//8.17

21//5.103.7

// 321 RRRRT∴

AREI

T

246.1634.912

===

Page 61: Basic Electrical 1

Network Solution :-

To solve a circuit is to find the current and voltage in all branches.

1) Loop ( Mesh ) current method :-

Example( 1 ):- Find the current through the 10 Ω resistor of the network

shown:

Solution :-

الفولتيـة m المشترك Loop تيار الـ *المقاومة المشتركة + Loop تيار الـ *لمقاومات مجموع ا -

.صفر ) = Loopتيار الـ حسب اتجاهها مع (

The loop equations are :-

Loop 1 :-

- ( 8+3 )I1 + 3I2 +8I3 + 15 = 0

Loop 2 :-

- ( 3+5+2 )I2 + 3I1 +5I3 = 0

Loop 3 :-

- ( 10+8+5 )I3 + 8I1 +5I2 = 0

Rearrange the equations , then :-

-11I1 + 3I2 +8I3 = -15

3I1 - 10I2 +5I3 = 0

8I1 + 5I2 - 23I3 = 0

-60-

Page 62: Basic Electrical 1

AII

ADDI

22.1

22.1

235851038311058010315311

103

33

==∴

=

⎥⎥⎥

⎢⎢⎢

−−

⎥⎥⎥

⎢⎢⎢

⎡−

==

Ω

Example( 2 ):- Solve following circuit diagram;

Solution :-

-I1 ( 5+7 ) + 7I2 + 20 – 5 = 0

-I2 ( 7+2+6 ) + 7I1 + 6I3 + 5 + 5 + 5 = 0

-I3 ( 6+8 ) + 6I2 – 5 – 30 = 0

Rearrange;

-12I1 + 7I2 +0 = -15

7I1 - 15I2 +6I3 = -15

0 + 6I2 - 14I3 = 35

------------------- ( 1 )

------------------- ( 2 )

------------------- ( 3 )

-61-

Page 63: Basic Electrical 1

ADDI 862.1

14022610

14606157071214635615150715

11 ==

⎥⎥⎥

⎢⎢⎢

−−

−⎥⎥⎥

⎢⎢⎢

−−−

==

ADDI 049.1

14021470

140214350615701512

22 ==

⎥⎥⎥

⎢⎢⎢

−−−−

==

ADDI 05.2

14022875

140235601515715712

33 −=

−=

⎥⎥⎥

⎢⎢⎢

⎡−−−−

==

Example( 3 ):- Find the current in the 10V source , for the following network;

10V

4Ω 3Ω

6Ω I2I1

5A

Solution :-

I2 = -5 A

Hence , we need only one equation to solve this circuit

-I1 ( 4+6 ) + 6 * ( -5 ) + 10 = 0

-10I1 – 20 = 0 ⇒ -10I1 = 20

AI 210

201 −=

−=∴

-62-

Page 64: Basic Electrical 1

Example( 4 ):- Solve the following circuit diagram, also find the voltage across

15 Ω resistance?

Solution:-

I4 = 2 A

-I1 ( 4+6+5 ) + 4I2 + 12 – 18 = 0

-I2 ( 8+3+4+7 ) + 4I1 + 8I3 - 12 = 0

-I3 ( 15+2+8+9 ) + 8I2 + 15 * 2 = 0

Rearrange:-

-15I1 + 4I2 +0 = 6

4I1 - 22I2 +8I3 = 12

0 + 8I2 - 34I3 = -30

6O

5O

8O

3O

18V

4O12V

7O

15O

2O 9O

2A

I1

I2

I3

I4

------------------- ( 1 )

------------------- ( 2 )

------------------- ( 3 )

-63-

Page 65: Basic Electrical 1

DDI 1

1 =

DDI 2

2 =

DDI 3

3 =

V15 = I15 * R15

= ( I3 – I4 ) * 15

= ( I3 – 2 ) * 15

Example( 5 ):- Solve the following circuit diagram .

3Ω 15Ω

40Ω

60Ω20A

10A

5A

Solution:- The above diagram can be reduced to the following diagram;

أكمل الحل

-64-

Page 66: Basic Electrical 1

3Ω 15Ω

40Ω

60Ω

50V

75V60V

I1

I2

-I1 ( 5+15+60 ) + 60I2 - 50 – 75 = 0

-I2 ( 3+60+40 ) + 60I1 + 60 = 0

Rearrange:-

-80I1 + 60I2 = 125

60I1 - 103I2 = -60

------------------- ( 1 )

------------------- ( 2 )

DDI 1

1 =

DDI 2

2 =

Example( 6 ):- Solve the following circuit diagram:

10Ω

4Ω2Ω

20V

I1 I2

6AVo

أكمل الحل

-65-

Page 67: Basic Electrical 1

Solution:-

-( 6+2 )I1 + 2I2 +20 - Vo = 0

-( 10+2+4 )I2 + 2I1 + Vo = 0

I2 - I1 = 0

Add eq. 1 & eq. 2

-8I1 + 2I2 +20 -16 I2 + 2I1 = 0

-6I1 - 14I2 = -20

From eq. 3

I1 – I2 = -6

ADDI 2.3

1468420

11146

161420

11 −=

+−

=

⎥⎦

⎤⎢⎣

⎡−−−

⎥⎦

⎤⎢⎣

⎡−−−−

==

ADDI 8.2

202036

2061206

22 =

+=

⎥⎦

⎤⎢⎣

⎡−−−

==

------------------- ( 1 )

------------------- ( 2 )

------------------- ( 3 )

------------------- ( 1 )

------------------- ( 2 )

-66-

Page 68: Basic Electrical 1

Example ( 7 ) : Solve the following circuit , using loop current method :-

12V

Ib

Ic

3A Vo

5Ω Ia

Solution:

Loop a :- ------------------- ( 1 ) -( 5+8 )Ia + 8Ic - Vo = 0

Loop b :- ------------------- ( 2 ) -( 6+7 )Ib + 7Ic + Vo – 12 = 0

Loop c :- ------------------- ( 3 ) -( 8+7+9 )Ic + 8Ia + 7Ib = 0

------------------- ( 4 ) Ib – Ia = 3

ـــل هـــذه االســـئلة التـــي تحتـــوي علـــى -:حظـــة مال • ، نجـــري عمليـــات جمـــع او طـــرح Vo فـــي مث

. للتخلص منها و نبسط الحل Voالمعادالت التي تحتوي على

عــــن طريــــق جمــــع او طــــرح ( نهــــا اوال م، فايــــضا نقــــوم بــــالتخلص Voأمــــا اذا كــــان المطلــــوب ايجــــاد

Voرات نعوضها في المعادلـة التـي تحتـوي علـى ثم بعد ايجاد التيا ) Voالتي تحتوي على المعادالت

.ونجدها

-67-

Page 69: Basic Electrical 1

Loop a+b :

-13Ia - 13Ib + 15Ic – 12 = 0

Loop c :-

-24Ic + 8Ia + 7Ib = 0

Ib – Ia = 3

Rearrange Eq.s :-

-13Ia - 13Ib + 15Ic = 12

8Ia + 7Ib -24Ic = 0

Ia – Ib = 3

ADDIa 862.1

14022610

3112478

1513133102470

151312

1 ==

⎥⎥⎥

⎢⎢⎢

⎡−

−−⎥⎥⎥

⎢⎢⎢

⎡−

==

DDIb

2=

DDIc

3=

------------------- (1 ) ′

) ------------------- ( 2′

------------------- ( ) 3′

------------------- (1 ) ′′

------------------- ( ) 2 ′′

) ------------------- ( 3 ′′

أكمل الحل

-68-

Page 70: Basic Electrical 1

Example, (Sheet 4 – Q. 25): Solve the following circuit diagram using loop

current:

30Ω

15Ω

13V

9V

12Ω

1.2A

0.8A

20Ω

6V

Solution:-

15 Ω // 30 Ω = Ω=+

10301530*15 ; 12 + 3 = 15 Ω

13V

15Ω

1.2A

10Ω

6V

9V

16V

20Ω

Ib

Ic

Ia

Loop a:-

-( 15+7 )Ia + 6 + 9 + 7Ib - 13 = 0

------------------- ( 1 )

-69-

Page 71: Basic Electrical 1

Loop b:-

-( 7+10 )Ib - Vo + 13 + 7Ia = 0

Loop c:-

-20Ic – 16 – 6 + Vo = 0

Ic – Ib = 1.2

Loop b+c:

-17Ib - 20Ic + 7Ia – 9 = 0

Rearrange Eq.s :-

Loop a: -22Ia + 7Ib = -2

Loop b+c: 7Ia - 17Ib -20Ic = 9

Ib – Ic = -1.2

------------------- ( 2 )

------------------- ( 3 )

------------------- ( 4 )

------------------- ( 1 ) ------------------- ( 2 )

------------------- ( 3 )

أكمل الحل

Example, (Sheet 4 – Q. 7): Solve the following circuit diagram:

-70-

Page 72: Basic Electrical 1

Solution:

120Ω

80Ω

2Ω 10V

10Ω 2Ω

8V20V

4V10V

Ib

Ia

Loop a :-

-( 10+120+2+80 )Ia + 80Ib - 8 + 20 = 0

Loop b :-

-( 2+5+80+1 )Ib + 80Ia - 4 - 10 + 10 = 0

Rearrange Eq.s :-

-212Ia + 80Ib = -12

80Ia - 88Ib = 4

------------------- ( 1 )

------------------- ( 2 )

------------------- ( 1 )

------------------- ( 2 )

أكمل الحل

-71-

Page 73: Basic Electrical 1

Nodal voltage:-

Example 1 :- Solve the following circuit using the nodal voltage method:

R4

R1

E2

R2R3

R5 R6

E1

VB - VCVA - VB

VC

VA

VB

A C

D

B

VC - VA

I1

I4

I6I5

I3

I2

Solution :-

Choose reference point N = 4

Let D be a reference point IN = N-1 = 3

Kcl at B:

I5 – I2 – I6 = 0

(VA – VB ) G5 – ( VB – E2 ) G2 – ( VB – VC ) G6 = 0

Kcl at A:

I3 – I5 – I1 = 0

-VA G3 – ( VA – VB ) G5 – [( VA – VC )- E1] G1 = 0

نقاط التفرع

عدد المعادالت

-72-

Page 74: Basic Electrical 1

Kcl at C:

I6 + I1 – I4 = 0

( VB – VC ) G6 + [( VA – VC )- E1] G1 – VC G4 = 0

Rearrange:

A : ( VB – VA ) G5 -VA G3 + ( VC – VA ) G1 - E1G1 = 0

B : (VA – VB ) G5 + ( VC – VB ) G6 - VB G2 + E2 G2 = 0

C : ( VB – VC ) G6 – VC G4 + ( VA – VC ) G1 - E1G1 = 0

Hence, we can arrange the above equations in the following form:-

A : - VA ( G1 + G3 + G5 ) + VBG5 + VCG1 + E1G1 = 0

B : - VB ( G2 + G5 + G6 ) + VAG5 + VCG6 + E2G2 = 0

C : - VC ( G1 + G4 + G6 ) + VAG1 + VBG6 - E1G1 = 0

Then, we can find VA , VB , VC by the determinate method .

Example 2 :- Solve the following circuit diagram using nodal voltage .

Solution:

First we simplify the circuit and make a less nodal point.

----------- ( 1 )

----------- ( 2 )

----------- ( 3 )

-73-

Page 75: Basic Electrical 1

N = 4 ; IN = 4 – 1 = 3

Let D be a reference point

A : 01935

830

191

81

71

71

81

191

=++⎟⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ ++− CBA VVV

B : 033351

71

331

71

=−+⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ +− AB VV

C : 01935

830

81

191

191

81

61

=−−⎟⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛ ++− AC VV

Rearrange

-0.321 VA + 0.143 VB + 0.178 VC = -5.592

0.143 VA - 0.174 VB = 1.455

----------- ( 1 ) ----------- ( 2 )

----------- ( 3 ) 0.178 VA - 0.344 VC = 5.592

VA , VB , VCاكمل الحل جد

-74-

Page 76: Basic Electrical 1

Example 3 :- Solve the following circuit using nodal voltage method:

9Ω 50Ω

30Ω

15Ω

20Ω

15V

A C

D

B

E

Solution:

Let D reference

∴VA = 15 V

B : 0201

301

615

201

301

91

61

=⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ +++− ECB VVV

C : 0501

315

301

151

501

31

301

=⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ +++− EBC VVV

D : 0501

201

51

501

201

=⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ ++− CBE VVV

Then rearrange the above equations and find VB , VC , VE .

-75-

Page 77: Basic Electrical 1

Example, (Sheet 4 – Q. 24): For the following circuit diagram, find I & I1,

using nodal voltage method:

15V

25Ω

15Ω

35Ω

17V

15Ω

25Ω

12V

0.8A

0.3A

20Ω A

C

D

B

DD

A I1

IB

Solution:

First; let D reference:-

A : 03.02515

4012

4025201

251

401

=+++++⎟⎠⎞

⎜⎝⎛ ++− CB

AVV

V

B : 02515

25351

151

251

=−−+⎟⎠⎞

⎜⎝⎛ ++− IVV A

B

C : 08.04040

12401

=+++−⎟⎠⎞

⎜⎝⎛− I

VV A

C

VB + 17 = VC

Then the above equations can be minimized to:-

B + C:

( ) 0401178.0

4012

401

251

2515

351

151

251

=⎟⎠⎞

⎜⎝⎛+−+−⎟

⎠⎞

⎜⎝⎛ ++−⎟

⎠⎞

⎜⎝⎛ ++− BAB VVV

----------- ( 1 )

----------- ( 2 )

----------- ( 3 )

----------- ( 4 )

----------- ( 1 )

-76-

Page 78: Basic Electrical 1

A :

( )03.0

2515

4012

4017

25201

251

401

=++++

++⎟⎠⎞

⎜⎝⎛ ++− BB

AVVV

Then solve to find VA & VB; hence we can find VC from eq. ( 4 ); to find I sub.

VA & VB in eq. ( 2 ); and to find I1; then I1 = I + 0.8

Second solution; let B reference

Hence VC = 17 V

A : 03.02515

204017

4012

201

251

401

=+++++⎟⎠⎞

⎜⎝⎛ ++− D

AVV

D : 08.03.02020

1351

151

=−−+⎟⎠⎞

⎜⎝⎛ ++− A

DVV

Then find VA & VD from the above equations; hence to find I;

C : 040128.0

404017

=+−++− IVA

And to find I1; I1 = I + 0.8

----------- ( 2 )

----------- ( 1 )

----------- ( 2 )

-77-

Page 79: Basic Electrical 1

):9(المحاضرة غيداء كائن صالح. د سمير عبد الخالق عزيز . د

قسم الھندسة الكھروميكانيكية/ المرحلة االولى

Network Theorems:-

1- Superposition Theorem:-

In any circuit network contain more than one sources ( voltage or

current ) to find the current ( or voltage ) in a certain part of a network , remove

the sources of the network and find the current ( or voltage ) in the existence of

only one source each time. The resultant current ( or voltage ) will be the

algebraic sum of current ( or voltage ) due to all sources when acting

independently once a time .

(Removing the sources means:- Short circuiting the voltage source and open

circuiting the current source) .

Example 1:- In the following circuit diagram, find all branch current's using

superposition theorem:-

25V

3A4Ω

I5

I2

I1I4

I3

-79-

Page 80: Basic Electrical 1

Solution :-

1.) Effect of 25 V source :-

AI 5.237

251 =

+=′

AI 5.264

252 =

+=′

AIII 5213 =′+′=′

2.) Effect of 3 A source :-

3A4Ω

I"2

I"1I"4

I"3

I"5

AI 1.237

7*31 =+

=′′

AI 9.037

3*34 =+

=′′

AI 2.164

4*32 =+

=′′

AI 8.164

6*35 =+

=′′

AIIIII 9.02.11.245213 =−=′′−′′=′′−′′=′′

3.) Superpose :-

AIII 6.41.25.2111 =+=′′+′=

AIII 3.15.22.1222 −=−=′−′′=

AIII 3.48.15.2525 =+=′′+′=

AIII 6.19.05.2414 =−=′′−′=

AIII 9.59.05333 =+=′′+′=

-80-

Page 81: Basic Electrical 1

Example 2:- For the following circuit network, find the current in all branches,

using superposition theorem:-

30Ω

60Ω40Ω

150V 270V

I1

I2

I3

Solution:-

1.) Effect of 150 V source:-

30Ω

60Ω40Ω

150V

I'1

I'2

I'3

Ω=+

+= 60306030*6040TR

AI 5.260

1501 ==′

AI 83.03060

30*5.22 =+

=′

AI 67.13060

60*5.23 =+

=′

-81-

Page 82: Basic Electrical 1

2.) Effect of 270 V source:-

Ω=+

+= 54604060*4030TR

AI 554270

1 ==′′

AI 26040

40*52 =+

=′′

AI 36040

60*53 =+

=′′

3.) Superpose :-

AIII 5.035.2311 −=−=′′−′=

AIII 83.2283.0222 =+=′′+′=

AIII 33.3567.1133 −=−=′′−′=

Example 3:- Find the current in all branch in the following circuit diagram:-

15V

2A

12V

I4

I2

I3

I1

I5

-82-

Page 83: Basic Electrical 1

Solution:-

1.) Effect of 15 V source:-

AI 12.2

858*54

151 =

++

=′

AI 3.185

8*12.22 =+

=′

AI 82.085

5*12.23 =+

=′

2.) Effect of 12 V source:-

12V

I"2

I"1

I"3

AI 57.1

484*85

121 =

++

=′′

AI 52.048

4*57.12 =+

=′′

AI 04.148

8*57.13 =+

=″

-83-

Page 84: Basic Electrical 1

3.) Effect of 2 A source:-

4Ω8Ω

2A

I'"3

I'"2

I'"5

I'"4

I'"1

Vo

Ω=⇒∴++= 74.151

41

811

TT

RR

VRV To 5.3*2 ==

AVI o 88.041 ==′′′

AVI o 7.052 ==′′′

AVI o 44.083 ==′′′

AII 12.12 14 =′′′−=′′′

AII 3.12 25 =′′′−=′′′

3.) Superpose :-

AIIII 04.012.104.112.24311 −=−−=′′′−′′−′=

AIIII 97.07.057.13.12122 −=−−=′′′−′′−′=

AIIII 03.13.157.13.15123 =+−=′′′+′′−′=

AIIII 9.044.052.082.03234 =−+=′′′−′′+′=

AIIII 96.188.004.112.21315 =+−=′′′+′′−′=

-84-

Page 85: Basic Electrical 1

):10(المحاضرة غيداء كائن صالح. د سمير عبد الخالق عزيز . د

قسم الھندسة الكھروميكانيكية/ المرحلة االولى

2-) Thevenin's Theorems:-

.في اغلب االحيان اذا كان المطلوب ايجاد التيار او الفولتية في مقاومة محددة في الدائرة تستخدم

Any two terminal linear network can be replaced by an equivalent circuit

of a voltage source ( Eth ) and a series resistor ( Rth ); as shown in figure below:-

Hence; Lth

th

RREI+

=

Steps to find Eth & Rth :-

1. Remove that portion of the network across which the Thevenins

equivalent circuit is to be find.

2. Mark the terminals of the remaining two – terminal network.

3. Calculate Rth by first setting all sources to zero ( voltage sources are

replaced by short circuits and current sources are replaced by open

circuit ), and finding the resultant resistance between the two marked

terminals.

4. Calculate Eth by first returning all sources to their origin positions and

finding the open circuit voltage between the marked terminals.

5. Draw the Thevenins equivalent circuit with the portion of the circuit

previously removed replaced between the terminals of the equivalent

circuit.

-85-

Page 86: Basic Electrical 1

Example 1:- For the following circuit diagram, find the current in ( 6Ω )

resistor?

5Ω4Ω

25V

6Ω3Ω

7A

Solution:-

5Ω4Ω

25V

7A

A B

1.) Find Rth :

5Ω4Ω

3Ω A B

Rth

( )

Ω=+=+⎭⎬⎫

⎩⎨⎧

+=

++=

22.759205

454*5

54//32thR

-86-

Page 87: Basic Electrical 1

2.) Find Eth :

A925

( )

VE

VV

V

VVV

th

oc

oc

oc

89.23

89.23359

100

05*7925*4

021

=∴

−=−=

=+⎟⎠⎞

⎜⎝⎛−

=+−

Rth = 7.22 Ω

Eth = 23.89 V

A

B

I

A

RREI

Lth

th

8.1622.7

89.23=

+=

+=

-87-

Page 88: Basic Electrical 1

Example 2:- Find the current in the 25Ω resistor for the following circuit

network?

10Ω 40Ω

25Ω

20Ω 20Ω

2V

Solution:-

1.) Find Rth :

10Ω 40Ω

20Ω 20Ω

A B

( ) ( )

Ω=+

++

=∴

+=

20204020*40

201020*10

20//4020//10

th

th

R

R

2.) Find Eth :

10Ω 40Ω

20Ω 20Ω

A B2V

Voc

A602

A302

VE

VV

V

th

oc

oc

67.0

67.06040

3020

6080

0602*40

302*10

=∴

==−=

=⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛+

-88-

Page 89: Basic Electrical 1

AI

RREI

Lth

th

4567.0

252067.0

=+

=∴

+=

Example 3:- Find I in the ( 9Ω ) resistor for the following cct. diagram?

6A

4A

10Ω

25V

Solution :-

Ω=++= 258107thR

-89-

Page 90: Basic Electrical 1

6A

4A

10Ω

25V

A B

Voc

6A

4A

10A

( ) ( ) ( )

VV

V

oc

oc

201

06*82510*107*4

=∴

=−−−−

A

RREI

Lth

th

91.5925

201=

+=

+=

-90-

Page 91: Basic Electrical 1

Norton's Theorems:-

Any two terminal linear network can be replaced by an equivalent circuit

consisting of a current source and a parallel resistor.

≡ ≡th

th

RE

≡ ≡

RN = Rth as before .

IN = Isc = short circuit current between the two terminals of the active network.

Example 1:- Find the current in 25Ω resistor for the following circuit network

using Norton's Theorem?

10Ω 40Ω

25Ω

20Ω 20Ω

2V

-91-

Page 92: Basic Electrical 1

Solution:-

First find RN :-

10Ω 40Ω

20Ω 20Ω

A B

Second find IN :-

10Ω 40Ω

20Ω 20Ω

A B2VIN

I1I2

I4I3

I

)( ) (

Ω=+

++

=

+=

20204020*40

201020*10

20//4020//10NR

AI91

1082

202020*20

401040*10

2=

+=

++

+

=

4020*

91&

4020*

91

5010*

91&

5040*

91

43

21

==

==

II

II

KC

I1 – IN – I3 = 0

L at A

∴ IN = I1 – I3

A033.04020*

91

5040*

91

=⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=

AIL 0147.02520

20*033.0 =+

=∴

-92-

Page 93: Basic Electrical 1

Example 2:- Find I in 50v voltage source, for the following circuit using

eorem? Norton's Th

25Ω

12Ω

20Ω

17Ω

30Ω

65V 50V45V

Solution:-

1.) Find RN :-

12Ω

20Ω

17Ω

30Ω

25Ω

A

B

R1 R2 R3

30Ω 20ΩA

B

-93-

Page 94: Basic Electrical 1

Ω== 8.754

25*171R , Ω== 78.3

5417*12

2R , Ω== 56.554

12*253R

2.) Find IN :-

( ) ( )[ ]

[ ] Ω=+=

+++=

1978.356.25//8.37

20//30 231 RRRRN

12Ω

20Ω

17Ω

30Ω

65V

25Ω

45V

A

BIN

Ib

Ic

Ia

-47Ia + 17Ic + 65 = 0

-32Ib + 12Ic - 45 = 0

-54Ic + 17Ia + 12Ib = 0

After find Ia , Ib , Ic

IN = Ia – Ib

195050 I -I I

0 I-I-I

aNL

LaN

−=−==

=

NN

N IR

I

-94-

Page 95: Basic Electrical 1

Maximum Power Transfer:-

A load will receive maximum power from a d.c. network when its total

resistive value is exactly equal to the Thevenin resistance of the network.

For Thevenin cct. Nortan cct.

For Max. power

LLth

thLLL R

RRERIP *

22

.max ⎟⎟⎠

⎞⎜⎜⎝

⎛+

==

thth

th RRE *

4 2

2

=

RNIN RL

LLN

NNLLL R

RRRIRIP *

22

.max ⎟⎟⎠

⎞⎜⎜⎝

⎛+

==

NN

NN R

REI *

4 2

22=

RL = Rth RL = RN

Under Max. Power transfer conditions, the efficiency is:-

%100*i

o

PPη % =

%100*Lth

LL

IEIV

= %100*th

L

EV

=

th

thL R

EP4

2

.max= ∴

4

2

.max

NNL

RIP =∴

-95-

Page 96: Basic Electrical 1

Eth = VL + Rth Ith

max. power transfer )

th = VL + RL IL

= VL + VL = 2VL

Q Rth = RL ( for

E

%50%100100*=thE

η *2

% ==L

LL

VVV

h efficiency will always be 50% under max. power transfer conditions .

Practical example:-

T e *

RL

Eth

Rth

I

et Rth = 3Ω & RL = 1Ω & Eth = 15 V L

WPL 141*13

15 2

≈⎟⎠⎞

⎜⎝⎛

+=∴

For RL = 2Ω WPL 182*5

15 2

=⎟⎠⎞

⎜⎝⎛=⇒

For RL = 3Ω WPL 75.183*6

15 2

=⎟⎠⎞

⎜⎝⎛=⇒

LL RIP 2=

LLth

RRR

E *⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

2

-96-

Page 97: Basic Electrical 1

For RL = 4Ω WP 36.184*15 2

=⎟⎞

⎜⎛=⇒ L 7 ⎠⎝

WPL 57.175*8

15 2

=⎟⎠⎞

⎜⎝⎛=⇒For RL = 5Ω

L = Rth , we get the max. power of PL .

ence

Note that when R

H EIPP inL

1122

==

r Pin = 2 PL Example 1:- Find the value of RL for maximum power transfer to RL , and

determine the power delivered under these conditions ?

o

Solution:-

L , and find the equivalent resistance ( Rth ) First remove R

-97-

Page 98: Basic Electrical 1

(

For Max. power RL = Rth

Ω=∴ 14LR

12Ω

18V

Vab

a

b

Find the value of R for the following cct. for max. power transfer,

L

Example 2:- L

and find P ?

Solution:-

6ΩRth

a

b

)

Ω=+

+=∴

+

141263

12

th

R = 6//3th

6*3R

VVE abth 12366*18=

+==

( ) WR

EPth

th 57.214*4

124

22

.max ===∴

-98-

Page 99: Basic Electrical 1

( ) ( )

V

EVV thaboc

7215*25

120

78*7846

120

=

++++

===

=

( ) WRIP LL 2166*66

722

2 =⎥⎦

⎤⎢⎡

+==

or

RL= 6Ω

Eth = 72 V

Rth = 6Ω

a

b

( ) ( )

L

theq

R

RR

=Ω=+

=

+==

6101510*15

46//7//8.

WR

EPth

thL 216

6*47

4

22

===

-99-

Page 100: Basic Electrical 1

Example 3 (sheet 5, fig. 20):- Find the maximum power in ( R ), for the

following cct. diagram?

1Ω 2Ω

6Ω 6Ω

2A4Ω

10V

6V

2Ω 4Ω

R

6V

Solution:-

1-) Find Rth.:-

-100-

Page 101: Basic Electrical 1

RRR theq =Ω=+++==∴ 102314.

2.) Find Eth :-

-101-

Page 102: Basic Electrical 1

1Ω 2Ω

6Ω 6Ω

2A4Ω

10V

6V

2Ω 4Ω

6V

A

B

IY

IX

IZ

2A

Voc

( )

( )

( ) AII

AII

AII

zz

yy

xx

25.102*268

33.106*261012

202*1104

=⇒=++−

=⇒=+−+−

−=⇒=+−−

From KVL

( ) ( ) ( )

VV

V

oc

oc

154856

02*233.1*625.1*46

=−++=

=−++−∴

AI 75.01010

15=

+=

-102-

Page 103: Basic Electrical 1

Rth = 10Ω RL= 10Ω

Eth = 15 V

B

AI

Example 4 (sheet 5, fig. 21):- Find the maximum power in ( R ), for the

following cct. diagram?

( ) W

RIP

625.510*75.0 2

2.max

==

=

80mA18Ω

12Ω

R

15Ω

0.86V

0.3V

20mA

Solution:-

First find Req.:- // )515( +)1812(. +=eqR //8 = 4.8Ω

-103-

Page 104: Basic Electrical 1

Second find IN:-

18Ω

12Ω

15Ω

0.86V

0.3V

0.36V

1.2V

IscA

B

30Ω

20Ω

0.5V

1.5V

A

B

Isc

IY

IX

mAII xysc 205.1−=−=II

II

II

N

yy

xx

3.5830

5.0

205.105.120

305.005.030

==

=⇒=+−

=⇒=+−

-104-

Page 105: Basic Electrical 1

A2

3.58

mWP 1.48.4*10*2

3.58 23

.max =⎟⎠⎞

⎜⎝⎛=∴ −

-105-