basic electrical 1
TRANSCRIPT
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
=Δ
××==
− 619 10106.1..
q = Charge that flows through section in time Δ t. Δ
Uniform flow of charges ⇒ direct current ( dc )
-1-
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
-2-
Power :- Time rate for doing work .
IVtq
qw
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
-3-
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
-
-4-
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
-5-
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
-6-
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
-7-
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
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
-8-
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
-9-
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 ?
-10-
-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
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
-12-
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-
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
-14-
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 .
-15-
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 ∴
-16-
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-
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-
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
+−
=∴
-19-
Example: For the following circuit diagram, Find I using:-
a) Ohm's law.
b) K.V.L.
7Ω
10Ω
40V
10V
20V10V
17Ω
6Ω
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
==⇒
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 . ⇒ ⇒
-21-
Example :- For the following circuit diagram , find the current ?
Solution :
3Ω15V
8V
2Ω2Ω
2Ω
2Ω
2Ω2Ω2Ω
2Ω
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 .
-22-
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-
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
-24-
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 :-
-25-
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.
-26-
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-
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
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-
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-
R1
R2
R1
R2
20 V
E = 20 V
⇒
R1
R2
-12 V
E = -12 V
R1
R2
⇒
⇒
-31-
⇒
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-
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-
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-
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-
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-
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-
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-
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-
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-
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-
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-
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-
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-
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-
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-
Example :- for the following cct. Network , Given that (V= 24 v), Find E ?
8Ω
E
24Ω
8Ω
4Ω
4Ω
16Ω12Ω
6Ω
6Ω
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
8Ω
E
24Ω
8Ω
4Ω
4Ω
16Ω12Ω
6Ω
6Ω
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-
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
2Ω
3Ω
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-
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-
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-
⇒
⇒
AREI 4
832
11 ===
Is = I1 + I2 = 4 + 6 = 10 A
Rs = R1 // R2 ARR
RIILs
ssL 3
1466*106
24824*8
=+
=+
=⇒∴Ω=+
=
-51-
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-
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-
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-
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-
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 4Ω
⇒
B
A
C12Ω
12Ω12Ω
C
CACBBAAB R
RRRRRRR ++=
A
CACBBABC R
RRRRRRR ++=
B
CACBBACA R
RRRRRRR ++=
-55-
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-
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-
Example :- Find I for the following cct. network :-
12 V
5Ω
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-
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
===
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-
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-
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-
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-
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
5Ω
60Ω20A
10A
5A
Solution:- The above diagram can be reduced to the following diagram;
أكمل الحل
-64-
3Ω 15Ω
40Ω
5Ω
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
6Ω
20V
I1 I2
6AVo
أكمل الحل
-65-
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-
Example ( 7 ) : Solve the following circuit , using loop current method :-
6Ω
8Ω
12V
7Ω
9Ω
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-
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-
Example, (Sheet 4 – Q. 25): Solve the following circuit diagram using loop
current:
30Ω
15Ω
13V
7Ω
9V
3Ω
12Ω
1.2A
0.8A
20Ω
6V
Solution:-
15 Ω // 30 Ω = Ω=+
10301530*15 ; 12 + 3 = 15 Ω
13V
7Ω
15Ω
1.2A
10Ω
6V
9V
16V
20Ω
Ib
Ic
Ia
Loop a:-
-( 15+7 )Ia + 6 + 9 + 7Ib - 13 = 0
------------------- ( 1 )
-69-
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-
Solution:
120Ω
80Ω
5Ω
2Ω 10V
1Ω
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-
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-
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-
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-
Example 3 :- Solve the following circuit using nodal voltage method:
9Ω 50Ω
5Ω
30Ω
15Ω
20Ω
6Ω
15V
3Ω
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-
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-
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-
):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:-
3Ω
6Ω
7Ω
25V
3A4Ω
I5
I2
I1I4
I3
-79-
Solution :-
1.) Effect of 25 V source :-
AI 5.237
251 =
+=′
AI 5.264
252 =
+=′
AIII 5213 =′+′=′
2.) Effect of 3 A source :-
3Ω
6Ω
7Ω
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-
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-
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:-
4Ω
15V
8Ω
5Ω
2A
12V
I4
I2
I3
I1
I5
-82-
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:-
8Ω
5Ω
12V
I"2
I"1
I"3
4Ω
AI 57.1
484*85
121 =
++
=′′
AI 52.048
4*57.12 =+
=′′
AI 04.148
8*57.13 =+
=″
-83-
3.) Effect of 2 A source:-
4Ω8Ω
5Ω
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-
):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-
Example 1:- For the following circuit diagram, find the current in ( 6Ω )
resistor?
5Ω4Ω
2Ω
25V
6Ω3Ω
7A
Solution:-
5Ω4Ω
2Ω
25V
3Ω
7A
A B
1.) Find Rth :
5Ω4Ω
2Ω
3Ω A B
Rth
( )
Ω=+=+⎭⎬⎫
⎩⎨⎧
+=
++=
22.759205
454*5
54//32thR
-86-
2.) Find Eth :
A925
( )
VE
VV
V
VVV
th
oc
oc
oc
89.23
89.23359
100
05*7925*4
021
=∴
−=−=
=+⎟⎠⎞
⎜⎝⎛−
=+−
6Ω
Rth = 7.22 Ω
Eth = 23.89 V
A
B
I
A
RREI
Lth
th
8.1622.7
89.23=
+=
+=
-87-
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-
AI
RREI
Lth
th
4567.0
252067.0
=+
=∴
+=
Example 3:- Find I in the ( 9Ω ) resistor for the following cct. diagram?
9Ω
8Ω
7Ω
6A
4A
10Ω
25V
Solution :-
Ω=++= 258107thR
-89-
8Ω
7Ω
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-
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-
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-
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-
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-
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-
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-
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-
(
For Max. power RL = Rth
Ω=∴ 14LR
12Ω
6Ω
3Ω
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:-
8Ω
7Ω
4Ω
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-
( ) ( )
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-
Example 3 (sheet 5, fig. 20):- Find the maximum power in ( R ), for the
following cct. diagram?
1Ω
1Ω 2Ω
6Ω 6Ω
2A4Ω
10V
6V
2Ω
2Ω 4Ω
R
6V
Solution:-
1-) Find Rth.:-
-100-
⇒
RRR theq =Ω=+++==∴ 102314.
2.) Find Eth :-
-101-
1Ω
1Ω 2Ω
6Ω 6Ω
2A4Ω
10V
6V
2Ω
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-
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
5Ω
12Ω
R
15Ω
0.86V
0.3V
8Ω
20mA
Solution:-
First find Req.:- // )515( +)1812(. +=eqR //8 = 4.8Ω
-103-
Second find IN:-
18Ω
5Ω
12Ω
15Ω
0.86V
0.3V
0.36V
1.2V
8Ω
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-
A2
3.58
mWP 1.48.4*10*2
3.58 23
.max =⎟⎠⎞
⎜⎝⎛=∴ −
-105-