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---

Dr
...
Boylestad
...
T
...

• ‫علم الھندسة الكھربائية ، ترجمة د
...

Positive charge

proton

negative charge

electron

no charge

neutron

Like charges repel & opposite charges attract
...


Η ( Symbol ) = 180C (Unit )
...
Hence, the Electric Current is the rate of flow
of charge
...

Amper =

(

)

Coul
...
6 × 10−19 × 106 Δq
=
=
=I
Δt
Sec
...


Uniform flow of charges ⇒ direct current ( dc )

-1-

Example: In a copper wire a flow of charge is 0
...
Find
the current in this wire ?

Solution: I =

Δq
Δt

Q

=

t

=

0
...
06 A

Work & Energy:
Δw = F
...


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

( Like Sources )

Work done by charges

charges lose energy
(Like loads )

Δw = Joules = N
...

v=

Δw
Joule
⇒ volt =
Δq
coulomb

Δw = work done for transporting a total charge Δq
...

P=

Δw
Δt

P=

Δw Δq Δw Δq
×
=
×
= V
...
A
s

Systems of unit ( S
...
) :Quantity

Unit

Symbol

1) Charge (q)

Coulombs

C

2) Current (I)

Amper

A=

3) Force (F)

Newton

N

4) Work , energy (w)

Joule

J=N
...
A
s

7) Length ( l )

meter

m

8) Temperature (T)

Kelvien

K

-3-

C
S

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

+

+
-

VA

VA
VA

+
I

I
V

V

VB

(1)
Load
(dissipates power)
P = V
...

load = current and voltage in the opposite direction
...
Charging of batteries
...
The length of conductor ( l ) ⇒ R α l
2
...
The nature of material conductor
...
The temperature of conductor
...


-5-

RA Ω
...
m or Ω
...

d = diameter of section
...
8 μΩcm
...
8 × 10
=

−6

) (
)

)

× 10 −2 × 3 × 10 3
= 9Ω
6 × 10 − 6

(

-6-

Example: What is the resistance of 100 m length of copper wire with a
diameter of (1 mm) and resistivity 0
...

Solution:
R =

ρl
A

=

(0
...
0159

× 10

⎛ 1 × 10
⎜
⎜
2
⎝

−3

−6

)× 100
2

⎞
⎟ π
⎟
⎠

= 2
...
Find
its resistance at 50 Co ?
Solution:
slop =
2=

R2 − R1 400 − 300
=
= 2Ω / C o
T2 − T1
60 − 10

R − R1 R − 300 R − 300
=
=
50 − 10
40
T − T1

R − 300 = 80 ⇒ R = 80 + 300 ⇒ R = 380Ω
-7-

Also from the above figure we can sea
R2 − 0 R1 − 0
=
T2 − T0 T1 − T0
R2
R1
=
T2 − T0 T1 − T0

∴

R2 T2 − T0
=
R1 T1 − T0

l
A = T2 − T0 ⇒ ρ 2 = T2 − T0
, hence
l
T1 − T0
ρ 1 T1 − T0
ρ1
A

ρ2

Example: Aluminum conductor has resistance 0
...
Find its
resistance at 65 Co ?
Solution:
⎡ T − T0 ⎤
R 2 T2 − T0
=
⇒ R 2 = R1 ⎢ 2
⎥
T1 − T0
R1
⎣ T1 − T0 ⎦
⎡ 65 − (− 236 ) ⎤
∴ R 2 = 0
...
25 ⎢
⎥ = 0
...

⎣ 246 ⎦

The following table illustrate the value of resistivity ( ρ ) for different materials
at 20 Co temperature
...
m

Material

To , Co

Silver

1
...
72 - 1
...
5

Aluminum

2
...
5 mm2 cross
section area
...
83 × 10 )× (75 × 10 )
=
−8

−2

1
...
83 × 50 × 10− 4 = 141
...
15mΩ

...
15 ⎢
⎥
⎣ 20 − (− 236 ) ⎦
⎡ 90 + 236 ⎤
= 14
...

⎣ 20 + 236 ⎦

Another method

ρ 2 T2 − T0
⎛ 90 + 236 ⎞
=
⇒ ρ90 = ρ 20 ⎜
⎟
ρ1 T1 − T0
⎝ 20 + 236 ⎠

R=

ρ90 l
A

=

⎛ 90 + 236 ⎞
−2
⎟ × 75 ×10
⎝ 20 + 236 ⎠
1
...
83 ×10 −8
∴ ρ 90 = 14
...
5
In some resource, T0 take an absolute value, which means T0 = 234
...

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 ?
-10-

Solution:
a)

Or

α1 =

α1 =

1
1
1
=
=
= 0
...
5) 274
...
00364
T + T1 234
...
5

1/K

1/K

R2 = R1[1 + α1 (T2 − T1 )]

b)

= 30[1 + 0
...
8Ω


...
5) ⎞
⎟ = 30⎜
⎟
⎜ 0 − (− 234
...
5 ⎞
= 30⎜
⎟ = 24
...
5 ⎠

Or

R 2 ⎛ T0 + T 2
=⎜
R1 ⎜ T0 + T1
⎝

⎞
⎟
⎟
⎠

⎛ 234
...
5 + 0 ⎠
⎛ 194
...
88Ω
⎝ 234
...


Slop =

ΔI
1
=
ΔV R

V
= constant = R
I

R=

V
I

Ω ; ⇒ V=I
...

The resistance of open circuit is approaching to infinity
...
C
...
C
...


-12-

R=0
1
G= =∞
R

Electrical Energy ( W ) :Q P =

W
t

⇒ W

= P
...
t
= (V
...
t

(

)

⎛V 2
=⎜
⎜ R
⎝

⎞
⎟
...
R
...
2mυ
R 5 × 103

(

)

P = I
...
R = (6 × 10−3 )× (5 × 103 ) = 180mW
or P = V 2
...
2 × 10−3 ) = 180mW

(30) = 180mW
V2
=
R
0
...
× η

n

Example: A 2 hp motor operates at an efficiency of 75 %, what is the power
input in Watt, if the input current is (9
...
75 =

2 × 746
1492
⇒ Pi =
= 1989
...
75

P = E
...
33
=
= 219
...
05
I

Example: What is the energy in KWh of using the following loads:a) 1200 W toaster for 30 min
...

c) 400 W washing machines for 45 min
...

Solution :
W =

P(W ) × t (h )
1000

⎛ 30 ⎞
⎛ 45 ⎞
⎛ 20 ⎞
1200 × ⎜ ⎟ + 6 × 50 × 4 + 400 × ⎜ ⎟ + 4800 × ⎜ ⎟
⎝ 60 ⎠
⎝ 60 ⎠
⎝ 60 ⎠
W =
1000
=

600 + 1200 + 300 + 1600 3700
=
= 3
...
C
...
c
...


Voltage
Amper - hours

2- generators
...

4- Rectifiers
...
‫ﻣﺻدر اﻟﻔوﻟﺗﻳﺔ ﻳوﻟد ﺗﻳﺎر و ﻓوﻟﺗﻳﺔ و ﻟﻛن اﻟﻔوﻟﺗﻳﺔ ﺗﻛون ﺛﺎﺑﺗﺔ‬

...
R1
V2 = I
...
R3
E – V1 – V2 – V3 = 0 ⇒ E = V1 + V2 + V3
∴ E = I
...
R2 + I
...
[R1 + R2 + R3] = I
...
I
Voltage Source in Series:-

-17-

Example: Find the current for the following circuit diagram?

Solution:
ET = 10 + 7 + 6 – 3 = 20 V
RT = 2 + 3 = 5 Ω
I = IT =

ET 20
=
= 4A
RT
5

Kirchoff's voltage law ( K
...
L
...

m

∑

m =1

Vm = 0

Where m is the number of voltages in the path ( loop ) , and Vm is the mth
voltage
...
V
...
to find the current in the following circuit diagram?
I

R1
V1
E2

E1
V2
R2

Solution:

From K
...
L
...

b) K
...
L
...
V
...
:10 + 6I + 7I - 40 + 10I – 20 + 10 + 17I = 0
10 – 40 – 20 + 10 + I ( 6 + 7 + 10 + 17 ) = 0
-40 = -I ( 40 ) ⇒ I =

40
= 1A
40

-20-

10V

Example :- For the following circuit diagram , find the potential difference
between Node ( A & D ) , and Node ( A & F ) ?

A

D
12V

6V

8V
B

E
5V

C

F

Solution : To find the potential difference between Node A & D , we will apply
K
...
L
...
‫ ﺑـ 41 ﻓوﻟت‬A ‫ﻧﻘطﺔ‬
C

F

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


Or Take the loop FEBAF ⇒ –5 –8 +6 – V2 = 0 ⇒ V2 = –7 volt
...
5 A
2

Definitions :Node :- Meeting point of 3 or more branches
...

Loop :- Is any closed path in a circuit
...
V
...
?
R1
I
E=20V

Solution :RT = R1 + R2 = 4 + 6 = 10
I=

E
20
=
= 2A
RT 10
-23-

4Ω
V1

R2
6Ω
V2

V1 = IR1 = 2 × 4 = 8V
V2 = IR2 = 2 × 6 = 12V

; or

V12 (8)
=
= 16w
P4 Ω =
R1
4

P6Ω = I R2 = (2 ) × 6 = 24W

; or

V 2 (12 )
= 24w
P6 Ω = 2 =
R2
6

PE = IE = 2 × 20 = 40W

; or PE = P4Ω + P6Ω = 16 + 24 = 40W

P4Ω = I 2 R1 = (2) × 4 = 16W
2

2

2

2

2

To verify results by using K
...
L
...

In a practical voltage source the internal resistance represent as a resistor in series
with an ideal voltage source
...


I
Ro

Vo

V

RL

I

Ro

E

Practical voltage source

Practical current source

-24-

RL

Where
Ro = Internal resistance
RL = load resistance
According to K
...
L
...

Hence as the load will increase the current will be increase ( because the
resistance will decrease ) and the voltage will decrease
...

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
...
V
...

E – Vo – V = 0
120 – IRo – IRL = 0

⇒ 120 – 0 – 22I = 0

120 = 22I ⇒ I = 5
...
46 * 22 = 120 V
b
...
) E – Vo – V = 0
120 – 16I – 22I = 0

⇒ 120 – 38I = 0

120 = 38I ⇒ I = 3
...
16 * 22 = 69
...

-26-

Example :- A circuit have load one with 20 Ω and 4A , and load two with 10 Ω
& 6A
...
V
...
, then
E – Vo – VL = 0
VL = E – IRo
IRL = E – IRo
4 * 20 = E – 4Ro
80 = E – 4Ro

-----------------

(1)

-----------------

(2)

Also
6 * 10 = E – 6 Ro
60 = E – 6Ro
From eq
...
this result it in eq
...
V
...
for load 3 ;
V3 = E – IRo ⇒ 30I = 120 – 10I
40I = 120 ⇒ ∴ I = 3 A for load three
...


-27-

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
Ro =

E 25
=
= 0
...
V
...

E – Vo – VL = 0
E – 0
...
5I – 15 = 0
0
...
5

VL 15
=
= 0
...
R1
V1 = I
...
R1 =
⎜R ⎟
RT
⎝ T⎠
⎛E⎞
E
...
R2 = ⎜ ⎟
...

RT = The total resistance of the series circuits
...


Solution :RT = R1 + R2 + R3
= 2 + 5 + 3 = 10 Ω
I=

E 10
=
= 1A
RT 10

E – V2 – Va = 0
Va = E – V2 = 10 – (2 * 1) = 8 V
Vb = V5 = (1 * 5) = 5 V = Vbc
or

E – V 2 – V3 – Vb = 0

; Vc = 0 V
⇒ 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-

Equivalence of actual sources :-

Voltage
Source
Open

Source

Voc = E

Circuit

Current

I=0

Short

I sc =

circuit

Voc = I o

E
Ro

1
Go

Isc = Io

V=0

Kirchoff's Current Law ( K
...
L
...


∑I

in

= ∑ I out

Or , At any point , the algebraic sum of entering and leaving current is zero
...
Shown ?
1A
a
Iab
3A

b 3A

e
2A

Ibc
Ide

d

c
4A

8A
Icd

Solution :At node a
3 – 1 – Iab = 0
2 – Iab = 0 ⇒ Iab = 2 A

-34-

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
...
Diagram?
b

I2

=

8A

12

=

A

I5

I1 = 10A a

d

I4

I7
I6

I3
c

Solution :-

∑I

enter

= ∑ I leave

∴ I1 = I7 = 10 A

-35-

At node a ; suppose I3 is entering
I1 + I3 – I 2 = 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
...
V
...


V = V1 = V2

From K
...
L
...
L
...
R 2

Hence

or

RT =

R1
...
R + R1
...
R 2
= 2 3
RT
R1
...
R 3
RT =

R1
...
R 3
R 2
...
R 3 + R1
...
Find RT , PT , IT , Ib?
IT

8Ω

8Ω

8Ω

8Ω

16 V

Solution :RT =
IT =

R 8
= = 2Ω
N 4

‫ﻓﻲ ﺣﺎﻟﺔ ﻛون ﻗﻳم اﻟﻣﻘﺎوﻣﺎت ﻣﺗﺳﺎوﻳﺔ‬

E 16
=
= 8A
RT
2

I branch =

E 16
=
= 2A
R1
8

PT = IT2 RT = (8)
...
IT = 16 * 8 = 128W

or

P T = P 1 + P 2 + P 3 + P4
= (2 ) * 8 + (2 ) * 8 + (2 ) * 8 + (2 ) * 8
2

2

2

2

= 32 + 32 + 32 + 32 = 128W

-38-

Example :- For the parallel network in fig
...
25 = 0
...
05 +

1
R3

0
...
1 − 0
...
1 =

b)
c)

1
1
⇒ R3 =
= 10Ω
0
...
(20) = 80W

or

P2 =

2

V22
R2

, or P2 = I2V2

-39-

Current division Rule :V =I

I=

R1
...
R2
R1 + R2
R1

R2
R1 + R2

In the same miner
I2 = I

R1
R1 + R2

R2
I
R
R +R
Also 1 = 1 2 = 2
I 2 I R1
R1
R1 + R2
I

∴

I1 R2 G1
=
=
I 2 R1 G2

Example :- For the following circut
...
1
10
R1
...
0999Ω
R1 + R2 100 + 0
...
1

V = I
...
0999 = 0
...
004995 A
100

I2 =

V
= 4
...
1

To check

I = I 1 + I2
5 = 0
...
995
5 = 5 Ok
...

× 100 %
RL

Where
Rint
...

RL = load resistor
...
= 2Ω

VL

E = 30V

RL = 13Ω

Solution :IL =

E
30
=
= 2A
Rint + RL 2 + 13

VL = E − I L Rint
...
= (2 )
...
385%
26
VFL

VR % =

2
Rint
...
385%
13
RL

-42-

Example :- Find the current I , for the network shown:
I = 42 mA

I1
R2 =24Ω

R1 = 6Ω

R3 =24Ω

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

24 * 24
R2 R3
=
= 12Ω
R2 + R3 24 + 24

Hence
I1 = I

(

)

R
12
= 42 * 10− 3
= 28mA
R + R1
12 + 6

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
...
6mA
RT R1 5 * 103

V = I
...
Network , find RT , IA , IB , IC , VA , VB , I1 ,
I2 ?
I1 R1 = 9Ω
IA
I2

IB
R2 = 6Ω

IC
R3 = 4Ω
R6 = 3Ω

16
...
6Ω
R1 + R2 9 + 6

RB = R3 + R4 // R5 = 4 +

IA

RB = 6Ω

RT = RA + RB // RC

∴ IA =

IC
IB

9*3
= 6Ω
6+3

RC = 3 Ω

= 3
...
6Ω

6*3
= 5
...
8
=
= 3A
RT
5
...
D
...

IB =

3*3
I A RC
=
= 1A
RB + RC 3 + 6

By K
...
L
...
6 = 10
...
2 A
R1 + R2 6 + 9

I2 = IA – I1 = 3 – 1
...
8 A
To check
E – VA – VB = 0
16
...
8 – 6 = 0
0=0

Ok
...
2 A , if you know that the fsd ( full scale deflection ) of ammeter is 120
mA , and the resistance of ammeter is 2
...
C
...

Ish = 1
...
12 = 1
...
7 ⎞
1
...
2⎜
⎜ 2
...
08 =

3
...
7 + Rsh

3
...
08(2
...
24 = 2
...
08Rsh
0
...
08Rsh
∴ Rsh = 0
...
Network , Given that (V= 24 v), Find E ?
4Ω

6Ω

Solution :I1 =

24
= 2A
1
...
C
...
I4 = I3 = 3A
Take the closed loop ABCDA , from
8Ω

K
...
L
...
5 A
24

D
I4

∴ I 6 = I 5 + I 4 = 2
...
5 A

4Ω

Take the closed loop CBC

6Ω

V1

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

4Ω

A

∴ V1 = 60V

∴ I5 =

E

B

24Ω

8Ω

I5
C

E = 60 + 25
...
5 * 8
V2

= 60 + 44 + 44
E = 148 V

8Ω

-47-

V3

E

I6

Current Source :Example :- Find the voltage ( Vs ) for the circuit below:
10 A

Solution :Vs = IRL = 10 * 2 = 20 V

if RL = 2 Ω

Vs = IRL = 10 * 5 = 50 V

if RL = 5 Ω

I

RL (2-5)Ω

Vs

Example :- Calculate V1 , V2 , Vs for the following cct
...
Below to a current source ,
then calculate the current through the load for each source:
IL

Solution :Rs = 2Ω

E
6
IL =
=
= 1A
Rs + RL 2 + 4

RL = 4Ω

E = 6V

• For the current source cct
...
‫ﺻﺣﻳﺢ‬

E
Rs

IL

Rs = 2Ω

6
= = 3A
2

RL = 4Ω

Example :- Convert the current source in the cct
...
:

Solution :• For the current cct
...

IL =

E
E
27
=
=
RT Rs + R L (3 + 6) *10 3

∴ I L = 3mA

-49-

⇓

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
...
1

D=

Col
...
1

Col
...
3

b1 ⎤ ⎡ x ⎤ ⎡ c1 ⎤
=
b2 ⎥ ⎢ y ⎥ ⎢c2 ⎥
⎦⎣ ⎦ ⎣ ⎦

⎡ c1 b1 ⎤
⎢
⎥
D1 ⎣c2 b2 ⎦ c1b2 − b1c2
x=
=
=
D ⎡ a1 b1 ⎤ a1b2 − b1a2
⎢a b ⎥
⎣ 2 2⎦
⎡ a1
⎢
D2 ⎣a2
y=
=
D ⎡ a1
⎢a
⎣ 2

c1 ⎤
c2 ⎥ a1c2 − c1a2
⎦=
b1 ⎤ a1b2 − b1a2
b2 ⎥
⎦

Example :- Find the value of D
⎡4 − 1⎤
D=⎢
⎥
⎣6 2 ⎦

Solution :⎡4 − 1⎤
D=⎢
⎥ = 4 * 2 − (− 1) * 6 = 8 + 6 = 14
⎣6 2 ⎦

-51-

Example :- Solving the equations below by determinates
4I1 – 6 I2 = 8
2I1 + 4 I2 = 20

Solution :⎡ 4 − 6 ⎤ ⎡ I1 ⎤ ⎡ 8 ⎤
⎢2 4 ⎥ ⎢ I ⎥ = ⎢20⎥
⎣
⎦⎣ 2 ⎦ ⎣ ⎦

⎡ 8 − 6⎤
⎢
⎥
D1 ⎣20 4 ⎦ 8 * 4 − (− 6 ) * 20 32 + 120
I1 =
=
=
=
= 5
...
28 A
I1 = 1 = ⎣
D
28
28

Third – order determinant :-

a1
D = a2

b1
b2

c1 a1 b1
c2 a2 b2

a3

b3

c3 a3 b3

D = [a1b2 c 3 + a1c 2 a 3 + c1 a 2 b3 ] − [c1b2 a 3 + a1c 2 b3 + b1 a 2 c 3 ]

Example :- Find the value of D
⎡ 1 2 3⎤
D = ⎢− 2 1 0⎥
⎥
⎢
⎢ 0 4 2⎥
⎦
⎣

-52-

Solution :⎡ 1 2 3⎤ 1 2
D = ⎢− 2 1 0⎥ − 2 1
⎥
⎢
⎢ 0 4 2⎥ 0 4
⎦
⎣
∴ D = [1 * 1 * 2 + 2 * 0 * 0 + 3 * (− 2 ) * 4] − [0 * 1 * 3 + 4 * 0 *1 + 2 * −2 * 2]
D = [2 + 0 − 24] − [0 + 0 + 8] = −22 + 8 = −14

Example :- Find V1 , V2 , V3 from the following equations :2V1 + 4V2 +2V3 = 8
5V1 – 2V2 – 10V3 = 18
V1 + 8V2 – 20V3 = -8
Solution :2 ⎤ ⎡V1 ⎤ ⎡ 8 ⎤
⎡2 4
⎢5 − 2 − 10 ⎥ ⎢V ⎥ = ⎢ 18 ⎥
⎥⎢ 2 ⎥ ⎢ ⎥
⎢
⎢1 8 − 20⎥ ⎢V3 ⎥ ⎢− 8⎥
⎦⎣ ⎦ ⎣ ⎦
⎣

4
2 ⎤ 8 4
⎡8
⎢ 18 − 2 − 10 ⎥ 18 − 2
⎢
⎥
D1 ⎢− 8 8 − 20⎥ − 8 8
⎦
V1 =
=⎣
2 ⎤2 4
D
⎡2 4
⎢5 − 2 − 10 ⎥ 5 − 2
⎢
⎥
⎢1 8 − 20⎥ 1 8
⎣
⎦

V1 =

[8 * (− 2) * (− 20) + 4 * (− 10) * (− 8) + 2 *18 * 8] − [(− 8) * (− 2) * 2 + 8 * (− 10) * 8 + (− 20) *18 * 4]
[2 * (− 2) * (− 20) + 4 * (− 10) *1 + 2 * 5 * 8] − [1 * (− 2) * 2 + 8 * (− 10) * 2 + (− 20) * 5 * 4]

V1 =

2976
= 4
...
) Delta – Star ( Δ → Υ ) transformation :If the value of RAB , RCA , RBC are
known, and we need to get the values
of RA , RB , RC ; then :RA =

RAB

RAB RCA
+ RCA + RBC

RB =

RAB RBC
RAB + RCA + RBC

RC =

RCA RBC
RAB + RCA + RBC

If RAb = RBC = RCA = R∆ , in this case RA = RB = RC =
or RΥ =

RΔ
3

-54-

RΔ
= RΥ
3

A

30Ω

A

10Ω

⇒

30Ω

10Ω
C

B

10Ω
C

B

30Ω

2
...
5Ω
5 + 10 + 15

Rb =

10 * 15
= 5Ω
5 + 10 + 15

Rc =

5 *10
= 1
...
5 + 10 ) // (5 + 8) } + (1
...
5 // 13 } + 6
...
92 A
RT 13
...
5 * 13 ⎞
RT = ⎜
⎟ + 6
...
5 + 13 ⎠
RT = 6
...
67 = 13
...
5Ω , Rb =
= 1
...
5 ) // (3
...
5 * 3
...
75
⎣ 5
...
5 ⎥
⎦

I=

6
E
=
= 2
...
889

-57-

, Rc =

3*3
= 0
...
network :I
12
...
5Ω

Rac

c

12 V

Rab

30Ω

15Ω
Rbc
b

Rab =

10 * 20 + 10 * 5 + 5 * 20
= 70Ω
5

Rac =

10 * 20 + 10 * 5 + 5 * 20
= 17
...
5 Ω // Rac ) and (15 Ω // Rbc ) and (30 Ω // Rab ) , hence the
circuit can be reduce to the following network :-

-58-

I

R1
R3

12 V
R2

R1 = (12
...
5 *17
...
3Ω
12
...
5

R2 = (15 // Rbc ) =

15 * 35
= 10
...
3 + 10
...
8 // 21 =

I=

17
...
634Ω
17
...
246 A
RT 9
...

1) Loop ( Mesh ) current method :Example( 1 ):- Find the current through the 10 Ω resistor of the network
shown:

Solution :‫ اﻟﻔوﻟﺗﻳـﺔ‬m ‫ اﻟﻣﺷﺗرك‬Loop ‫ + اﻟﻣﻘﺎوﻣﺔ اﻟﻣﺷﺗرﻛﺔ * ﺗﻳﺎر اﻟـ‬Loop ‫ع اﻟﻣﻘﺎوﻣﺎت * ﺗﻳﺎر اﻟـ‬
‫ ﻣﺟﻣو‬
...
22 A
=
I3 =
8 ⎤
D ⎡11 3
⎢ 3 − 10
5 ⎥
⎢
⎥
⎢8
− 23⎥
5
⎣
⎦
∴ I 3 = I10 Ω = 1
...
862 A
0 ⎤ 1402
D ⎡− 12 7
⎢ 7
− 15 6 ⎥
⎢
⎥
⎢ 0
− 14⎥
6
⎣
⎦

I2 =

D2
=
D

⎡− 12 − 15 0 ⎤
⎢ 7
− 15 6 ⎥
⎢
⎥
⎢ 0
35 − 14⎥
⎣
⎦

1402

=

1470
= 1
...
05 A
D
1402
1402

Example( 3 ):- Find the current in the 10V source , for the following network;
4Ω

10V

I1

3Ω

6Ω

I2
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
∴ I1 =

20
= −2 A
− 10

-62-

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

5O

18V

I1
6O
12V

4O

I2

3O

7O

8O

I3

2O

9O

15O

I4

2A

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

-------------------

(1)

4I1 - 22I2 +8I3 = 12

-------------------

(2)

0 + 8I2 - 34I3 = -30

-------------------

(3)

-63-

I1 =

D1
D

I2 =

D2
D

I3 =

D3
D

‫أﻛﻣﻝ اﻟﺣﻝ‬

V15 = I15 * R15
= ( I3 – I4 ) * 15
= ( I3 – 2 ) * 15
Example( 5 ):- Solve the following circuit diagram
...
1 & eq
...
3
I1 – I2 = -6
⎡− 20 − 14⎤
⎢
⎥
D1 ⎣ − 6 − 1 ⎦ 20 − 84
I1 =
=
= −3
...
8 A
=
D
20
20

-66-

‫: ‪Example ( 7 ) : Solve the following circuit , using loop current method‬‬‫‪6Ω‬‬

‫‪12V‬‬

‫‪Ib‬‬

‫‪Vo‬‬

‫‪3A‬‬

‫‪7Ω‬‬

‫‪9Ω‬‬

‫‪Ia‬‬

‫‪Ic‬‬

‫‪5Ω‬‬

‫‪8Ω‬‬

‫:‪Solution‬‬
‫: ‪Loop a‬‬‫)1(‬

‫-------------------‬

‫)2(‬

‫-------------------‬

‫)3(‬

‫-------------------‬

‫0 = ‪-( 8+7+9 )Ic + 8Ia + 7Ib‬‬

‫)4(‬

‫-------------------‬

‫3 = ‪Ib – I a‬‬

‫0 = ‪-( 5+8 )Ia + 8Ic - Vo‬‬
‫: ‪Loop b‬‬‫0 = 21 – ‪-( 6+7 )Ib + 7Ic + Vo‬‬
‫-: ‪Loop c‬‬

‫• ﻣﻼﺣظ ــﺔ :- ﻓ ــﻲ ﻣﺛ ــﻝ ﻫ ــذﻩ اﻻﺳ ــﺋﻠﺔ اﻟﺗ ــﻲ ﺗﺣﺗ ــوي ﻋﻠ ــﻰ ‪ ، Vo‬ﻧﺟ ـ ي ﻋﻣﻠﻳ ــﺎت ﺟﻣ ــﻊ او ط ـ ح‬
‫ـر‬
‫ـر‬
‫اﻟﻣﻌﺎدﻻت اﻟﺗﻲ ﺗﺣﺗوي ﻋﻠﻰ ‪ Vo‬ﻟﻠﺗﺧﻠص ﻣﻧﻬﺎ و ﻧﺑﺳط اﻟﺣﻝ
...
‬
‫-76-‬

Loop a+b :
-------------------

( 1′ )

-24Ic + 8Ia + 7Ib = 0

-------------------

( 2′ )

Ib – I a = 3

-------------------

( 3′ )

-13Ia - 13Ib + 15Ic = 12

-------------------

( 1′′ )

8Ia + 7Ib -24Ic = 0

-------------------

( 2′′ )

Ia – I b

-------------------

( 3′′ )

-13Ia - 13Ib + 15Ic – 12 = 0
Loop c :-

Rearrange Eq
...
862 A
Ia =
= ⎣
D ⎡− 13 − 13 15 ⎤ 1402
⎢ 8
7
− 24⎥
⎢
⎥
⎢ 1
1
3 ⎥
⎣
⎦

Ib =

D2
D

Ic =

D3
D

‫أﻛﻣﻝ اﻟﺣﻝ‬

-68-

Example, (Sheet 4 – Q
...
2A

0
...
2A

Ic
20Ω
16V

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

-69-

-------------------

(1)

Loop b:-------------------

(2)

-20Ic – 16 – 6 + Vo = 0

-------------------

(3)

Ic – Ib = 1
...
s :Loop a:

-22Ia + 7Ib = -2

-------------------

(1)

Loop b+c:

7Ia - 17Ib -20Ic = 9

-------------------

(2)

Ib – Ic = -1
...
7): Solve the following circuit diagram:

-70-

Solution:
120Ω

10Ω

2Ω

Ia

20V

8V

80Ω

5Ω

1Ω
Ib

10V

4V

2Ω

10V

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

-------------------

(1)

-------------------

(2)

Loop b :-( 2+5+80+1 )Ib + 80Ia - 4 - 10 + 10 = 0
Rearrange Eq
...

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


-73-

N = 4 ; IN = 4 – 1 = 3
Let D be a reference point
A:

⎛ 1 1 1⎞
⎛1⎞
⎛ 1 1 ⎞ 30 35
− VA ⎜ + + ⎟ + VB ⎜ ⎟ + VC ⎜ + ⎟ +
+
=0
⎝ 19 8 7 ⎠
⎝7⎠
⎝ 8 19 ⎠ 8 19

B:

⎛1 1 ⎞
⎛ 1 ⎞ 51
− VB ⎜ + ⎟ + VA ⎜ ⎟ + − 3 = 0
⎝ 7 33 ⎠
⎝ 7 ⎠ 33

C:

⎛1 1 1 ⎞
⎛ 1 1 ⎞ 30 35
− VC ⎜ + + ⎟ + VA ⎜ + ⎟ − −
=0
⎝ 6 8 19 ⎠
⎝ 19 8 ⎠ 8 19

Rearrange
-0
...
143 VB + 0
...
592

-----------

(1)

0
...
174 VB = 1
...
178 VA - 0
...
592

-----------

(3)

VA , VB , VC ‫اكمل الحل جد‬

-74-

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

A

6Ω

B

30Ω
C
15Ω

9Ω

50Ω

15V
20Ω

D

5Ω

Solution:
Let D reference
∴ VA = 15 V

B:

1 ⎞ ⎛ 15 ⎞
⎛ 1 ⎞
⎛ 1 ⎞
⎛1 1 1
− VB ⎜ + +
+ ⎟ + ⎜ ⎟ + VC ⎜ ⎟ + VE ⎜ ⎟ = 0
⎝ 20 ⎠
⎝ 30 ⎠
⎝ 6 9 30 20 ⎠ ⎝ 6 ⎠

C:

1⎞
⎛ 1 ⎞
⎛ 1 ⎞ ⎛ 15 ⎞
⎛ 1 1 1
− VC ⎜ + +
+ ⎟ + VB ⎜ ⎟ + ⎜ ⎟ + VE ⎜ ⎟ = 0
⎝ 50 ⎠
⎝ 30 ⎠ ⎝ 3 ⎠
⎝ 30 3 50 15 ⎠

D:

1 1⎞
⎛ 1 ⎞
⎛ 1 ⎞
⎛ 1
− VE ⎜ +
+ ⎟ + VB ⎜ ⎟ + VC ⎜ ⎟ = 0
⎝ 50 ⎠
⎝ 20 ⎠
⎝ 20 50 5 ⎠

Then rearrange the above equations and find VB , VC , VE
...
24): For the following circuit diagram, find I & I1,
using nodal voltage method:
12V
20Ω

A

15Ω

A

I1

25Ω
25Ω
0
...
8A

D

D

Solution:
First; let D reference:V
1
1
1 ⎞ V
12 15
+
+ 0
...
8 + I = 0
⎟−
⎝ 40 ⎠ 40 40

-----------

(3)

VB + 17 = VC

-----------

(4)

A : − VA ⎛
⎜

B : − VB ⎛
⎜

Then the above equations can be minimized to:B + C:
1
1 ⎞ 15
1 ⎞ 12
⎛ 1 ⎞
⎛ 1
⎛ 1
+ 0
...
3 = 0
1
1 ⎞ V
⎛ 1
− VA ⎜ +
+ ⎟+ B + B
40
40 25
⎝ 40 25 20 ⎠ 25

-----------

(2)

Then solve to find VA & VB; hence we can find VC from eq
...

VA & VB in eq
...
8
Second solution; let B reference
Hence VC = 17 V
1
1
1 ⎞ 12 17 VD 15
+
+
+
+ 0
...
3 − 0
...
8 −
+I =0
40 40
40

And to find I1; I1 = I + 0
...
غيداء كائن صالح‬

‫د
...
The resultant current ( or voltage ) will be the
algebraic sum of current ( or voltage ) due to all sources when acting
independently once a time
...

Example 1:- In the following circuit diagram, find all branch current's using
superposition theorem:25V

I3

7Ω

I4

I1

3Ω
I2

I5
4Ω

6Ω
3A

-79-

Solution :1
...
5 A
7+3

′
I2 =

25
= 2
...
) Effect of 3 A source :I"3

I1′′ = 3 *

7
= 2
...
9 A
7+3

′
I 2′ = 3 *

4
= 1
...
8 A
4+6

7Ω

I"4

I"1

3Ω

6Ω
3A

4Ω

I"5
′
′
′ ′
I 3′ = I1′′ − I 2′ = I 5′ − I 4′ = 2
...
2 = 0
...
) Superpose :I1 = I1′ + I1′′ = 2
...
1 = 4
...
2 − 2
...
3 A

′
′
I 5 = I 2 + I 5′ = 2
...
8 = 4
...
5 − 0
...
6 A

′ ′
I 3 = I 3 + I 3′ = 5 + 0
...
9 A

-80-

I"2

Example 2:- For the following circuit network, find the current in all branches,
using superposition theorem:I3
I2
30Ω

40Ω
60Ω

I1

270V

150V

Solution:1
...
5 A
60

′
I 2 = 2
...
83 A
60 + 30

′
I 3 = 2
...
67 A
60 + 30

-81-

2
...
) Superpose :′
I1 = I1′ − I 3′ = 2
...
5 A
′
′
I 2 = I 2 + I 2′ = 0
...
83 A

′
I 3 = I 3 − I1′′ = 1
...
33 A

Example 3:- Find the current in all branch in the following circuit diagram:2A

5Ω

15V

I1

I2
I4

I3

8Ω

4Ω

I5
12V

-82-

Solution:1
...
12 A
5*8
4+
5+8

′
I 2 = 2
...
3 A
5+8

′
I 3 = 2
...
82 A
5+8

2
...
57 A
8*4
5+
8+4

′
I 2′ = 1
...
52 A
8+4

8
″
I 3 = 1
...
04 A
8+4

-83-

4Ω

3
...
74Ω
RT 8 4 5
Vo = 2 * RT = 3
...
88 A
4

′
I 2′′ =

Vo
= 0
...
44 A
8

′
I 4′′ = 2 − I1′′′= 1
...
3 A

3
...
12 − 1
...
12 = −0
...
3 − 1
...
7 = −0
...
3 − 1
...
3 = 1
...
82 + 0
...
44 = 0
...
12 − 1
...
88 = 1
...
غيداء كائن صالح‬

‫د
...
‫ﺗﺳﺗﺧدم ﻓﻲ اﻏﻠب اﻻﺣﻳﺎن اذا ﻛﺎن اﻟﻣطﻠوب اﻳﺟﺎد اﻟﺗﻳﺎر او اﻟﻔوﻟﺗﻳﺔ ﻓﻲ ﻣﻘﺎوﻣﺔ ﻣﺣددة ﻓﻲ اﻟداﺋ ة‬
‫ر‬
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; I =

Eth
Rth + RL

Steps to find Eth & Rth :1
...

2
...

3
...

4
...

5
...


-85-

Example 1:- For the following circuit diagram, find the current in ( 6Ω )
resistor?
3Ω

6Ω

2Ω

5Ω

7A

5Ω

4Ω

7A

25V

Solution:3Ω

A

B

2Ω
4Ω
25V

1
...
22Ω
⎬+5 =
9
⎩5 + 4 ⎭

-86-

2
...
89V
Voc =
9
∴ Eth = 23
...
22 Ω
6Ω
Eth = 23
...
89
= 1
...
22 + 6

-87-

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

40Ω

10Ω

2V

25Ω

20Ω

20Ω

Solution:1
...
) Find Eth :

2 ⎞ ⎛
2 ⎞
⎛
Voc + ⎜10 * ⎟ − ⎜ 40 * ⎟ = 0
30 ⎠ ⎝
60 ⎠
⎝
80 20 40
−
=
= 0
...
67V

20Ω

-88-

20Ω

I=

Eth
Rth + RL

∴I =

0
...
67
=
A
20 + 25
45

Example 3:- Find I in the ( 9Ω ) resistor for the following cct
...
91 A
25 + 9

-90-

Norton's Theorems:Any two terminal linear network can be replaced by an equivalent circuit
consisting of a current source and a parallel resistor
...

IN = Isc = short circuit current between the two terminals of the active network
...
033 A
⎝ 9 50 ⎠ ⎝ 9 40 ⎠

∴ I L = 0
...
0147 A
20 + 25

-92-

20Ω
I3

I4

Example 2:- Find I in 50v voltage source, for the following circuit using
Norton's Theorem?
25Ω

12Ω

17Ω

30Ω

20Ω

65V

50V
45V

Solution:1
...
8Ω , R2 =
= 3
...
56Ω
54

RN = [(R1 + 30 ) // (R3 + 20 )] + R2
= [37
...
56] + 3
...
) Find IN :25Ω

Ic
12Ω

17Ω
A

30Ω
Ia
65V

IN
B

20Ω
Ib
45V

-47Ia + 17Ic + 65 = 0
-32Ib + 12Ic - 45 = 0
-54Ic + 17Ia + 12Ib = 0
After find Ia , Ib , Ic
IN = Ia – I b

I N - Ia - IL = 0
IL = I N - Ia = I N −

50
50
= IN −
19
RN

-94-

Maximum Power Transfer:A load will receive maximum power from a d
...
network when its total
resistive value is exactly equal to the Thevenin resistance of the network
...


Nortan cct
...
power
RL = Rth

RL = RN
2

2

PLmax
...


⎛
RN ⎞
= I RL = ⎜ I N
⎟
⎜ R + R ⎟ * RL
N
L ⎠
⎝
2
L

2
Eth
=
* Rth
2
4 Rth

∴ PLmax
...
=

2
EN
* RN
2
4 RN

2
I N RN
4

Under Max
...
power transfer )

Q Rth = RL

Eth = VL + RL IL
= VL + VL = 2VL

η=

VL
V
* 100% = L * 100% = 50%
2VL
Eth

The efficiency will always be 50% under max
...

* Practical example:I
Rth

PL = I 2 RL

RL
Eth

2

⎛ E ⎞
=⎜
⎜ R + R ⎟ * RL
⎟
L ⎠
⎝ th

Let

Rth = 3Ω

&

RL = 1Ω &

Eth = 15 V

2

⎛ 15 ⎞
∴ PL = ⎜
⎟ *1 ≈ 14W
⎝ 3 +1⎠
2

For RL = 2Ω

⎛ 15 ⎞
⇒ PL = ⎜ ⎟ * 2 = 18W
⎝5⎠

For RL = 3Ω

⎛ 15 ⎞
⇒ PL = ⎜ ⎟ * 3 = 18
...
36W
⎝7⎠

For RL = 5Ω

⎛ 15 ⎞
⇒ PL = ⎜ ⎟ * 5 = 17
...
power of PL
...
power RL = Rth
∴ RL = 14Ω

Eth = Vab =

12Ω

3Ω

18 * 6
= 12V
6+3

a

6Ω
Vab

E2
(12 ) = 2
...


18V
b

Example 2:- Find the value of RL for the following cct
...
power transfer,
and find PL?

Solution:8Ω

4Ω
a

6Ω

Rth
b

-98-

7Ω

Req
...
20):- Find the maximum power in ( R ), for the
following cct
...
:-

-100-

R

⇒

∴ Req
...
) Find Eth :-

-101-

1Ω

1Ω

IX

6V

2Ω

10V

2A

4Ω
2A

6Ω

IY

Voc
6Ω

B

6V

2Ω

IZ

2Ω

− 4 I x − 10 + (1 * 2 ) = 0 ⇒ I x = −2 A
− 12 I y + 10 − 6 + (2 * 6 ) = 0 ⇒ I y = 1
...
25 A

From KVL
∴ 6 − Voc + (4 *1
...
33) − (2 * 2 ) = 0
Voc = 6 + 5 + 8 − 4 = 15V

I=

15
= 0
...
= I 2 R

Rth = 10Ω

= (0
...
625W
2

RL= 10Ω

Eth = 15 V
B

Example 4 (sheet 5, fig
...
diagram?
0
...
3V

Solution:First find Req
...
= (12 + 18) // (15 + 5) //8 = 4
...
86V
12Ω
A

Isc

0
...
2V
5Ω
0
...
5V
30Ω
Isc
IX

A

20Ω

B
IY
1
...
5 = 0 ⇒ I x =

0
...
5 = 0 ⇒ I y =

1
...
5 0
...
3mA
20 30
-104-

58
...


⎛ 58
...
8 = 4

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