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 -
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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
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
-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 .