Electrical Circuits - Introduction
description
Transcript of Electrical Circuits - Introduction
-
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. . & .
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 =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 =
=
The direction of motion of positive charges is opposite to direction of motion of
negative charges.
Amper ( ) Itq
tSecCoul =
===
619 10106.1..
q = Charge that flows through section in time t. 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===
=
Work & Energy:
dFw = .Where:
w d: 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 wExternal force work done on charges charges gain energy.
( Like Sources )
Work done by charges charges lose energy
(Like loads )
= Joules = N.m w
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 =
=qwv
w q= work done for transporting a total charge .
Potential difference
Voltage
Potential
-2-
-
Power :- Time rate for doing work .
IVtq
qw
qq
twP
twP
.=
=
=
=
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 ( )
The resistance of any material is depends on four factors:
1. The length of conductor ( l ) lR 2. Cross section area of conductor ( A ) A
R 1 3. The nature of material conductor.
4. The temperature of conductor.
AR l
AR l= = constant is called resistivity or specific resistance unit of resistivity.
-
-
V I
I
-
l
A
-5-
-
mmmRA
AR .==== l or
.l
2
cm. 2
2
2==A dr
r = radians of section.
d = diameter of section.
Con ctance ( G ):-du
ll A
AR
=== 11 (Siemens (S)) G
Prefix :-
pico 10 P -12
nano 10-9 n
micro 10-6
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 cm .
Solution:
( ) ( )( ) === 9106 6AR 10310108.1 326l
-6-
-
Example: What is the resistance of 100 m length of copper wire with a
diameter of (1 mm) and resistivity 0.0159 m .
Solution:
( ) ( ) =
02.210010 23
6
22 101
2
dA===
0159.0100100159.0 6R l
Effect of Temperature on a resistance :-
slop = TR = constant =
1
1
212 TTTT212
TTRRRRRR
=
=
Example: The resistance of material is 300 at 10Co, and 400 at 60Co. Find
its resistance at 50 Co ?
Solution:
oCTTRR /2
1060300400
12
12 ==
= slop
=+==
==
=3803008080300
40300
10503002
1
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 ==
l 101
1TTTT
A
l
ple: Aluminum conductor has resistance 0.25 at 10 Co . Find its
resistance at 65 Co ?
Solution:
Exam
( )( )
.31.024630125.0
236102366525.02
01
0212
01
02
1
2
=
=
=
=
=
R
TTTTRR
TTTTR
The following table illustrate the value of resistivity (
R
) for different materials t 20 Co temperature.
a
Material at 20 Co , .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:
( ) ( )
=== m15.14105.141105083.2 44
==
AR
105.110751083.2
6
28l
. : 20Co
( )( )
.236202369015.
209015
1
2
++
TTR
r method
18= m14=236 236 .142 =R
01
02
= TTR
Anothe
++=
=2362023690
209001
02
1
2
TTTT
( )
=
=
++
==
0 8
m
AR
15.14
183.2
105.1
10752362023690
90
20
6
220
90
l
-9-
-
Driving :
( )
( )
+=
+=
+=
+=
0112 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 = temperature coefficient of resistance at a temperature T1
( )[ ]12112 1 TTRR +=
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
++=
11
1TT +=
Example:
a) Find the value of 1 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 at 40 Co ?
-10-
-
-11-
Solution:
a) ( ) 00364.05.2741
5.2344011
011 ==== TT 1/K
Or 00364.05.274
1405.234
11
11 ==+=+= TT 1/K
b) ( )[ ]12112 1 TTRR += ( )[ ] =+= 8.33407500364.0130
.
Example: If the resistance of a copper wire at freezing ( 0 Co ) is 30 , Find its
sistance at -40 Co ?
Solution:
re
( )( )
== 88.245.19430
=
=
5.234
5.23405.2344030
01
0212 TT
TTR
Or
R
+ 20 TT +
=101
2
TTRR
== 88.245.23430
5.194
40 += 05.23430 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=
V = constant = R I
IVR = ; 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 ( ) .
O.C.01 ==
=G
R
-12-
-
Electrical Energy ( W ) :-
tPWtWP .==Q KWh
( )( )
tRV
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 ===
mR
G 2.010511
3 ===
( ) mWVIP 18030106. 3 === or ( ) ( ) mWRIP 180105106. 332 === or ( ) ( ) mWGVP 180102.030. 322 === or ( )( ) mWRVP 180102.0 30 3
22
===
-13-
-
Efficiency ( ) :-
Wi/p = Wo/p + Wloss
ttt+= WWW losspopi //
Pi/p = Po/p + Ploss
%100powerInput powerOutput Efficiency ( ) =
%100=i
o
PP
%100=i
o
WW
nT = ......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
PP
-14-
-
WPP ii
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-
-
VE
E
V
I
V = E = constant voltage element
V
IIo
IIo
I = Io = constant current element
.
.
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
REIT
====
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
AREIIT
TT 45
20 ====
Kirchoff's voltage law ( K.V.L. ):-
The algebraic sum of all voltages around any closed path is zero.
=
=m
mmV
10
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.
7
10
40V
10V
20V10V
17
6
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 * D
A 14 .
Take the loop FEDAF to find the potential difference between Node C & F .
5 + 12 V V2 = 0 5 + 12 14 V2 = 0 7V2 = 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 :
315V
8V
22
2
2
222
2
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 , P4 , P6
, PE , verify by K.V.L. ?
R1 R2
V1 V2
I
E=20V
4 6
Solution :-
RT = R1 + R2 = 4 + 6 = 10
AREIT
21020 ===
-23-
-
VIRV 84211 === VIRV 126222 ===
( ) WRIP 1642 2124 === ; or ( ) wRVP 16
48 2
1
21
4 ===
( ) WRIP 2462 2226 === ; or ( ) wRVP 24
612 2
2
22
6 ===
WIEPE 40202 === ; or WPPPE 40241664 =+=+= 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
b) Ro = 8
c) Ro = 16
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 and 4A , and load two with 10
& 6A . Find the current for load three which have 30 ?
Solution :-
1) 20 & 4A
2) 10 & 6A
3) 30 & 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 ; 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
=== 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 ==
=== 75.02015
IVR LL
-
Voltage divider Rule :-
RT = R1 + R2
TREI =
TT RRER
RERIV 1111
... =
==
TRRER
RERIV 22
222
... =
==
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 = 2K , R2 = 5K ,
R3 = 8K , E = 45 V ?
Solution :-
R1 R2
V1 V2
I
E
R3
V3
V4
VRERVT
1510*15
45*10*53
32
2 ===
T
nn R
ERV =
VRERVT
610*15
45*10*23
31
1 ===
-29-
-
VRERVT
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 :-
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
AREIT
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 GIV 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 IIOr , At any point , the algebraic sum of entering and leaving current is zero .
= 0I
I1
I5
I2I3
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 .L.
= V1G1 + V2G2 = V1 ( G1 + G2 )
or I = V ( G1 + G2 )
I = VGT
-36-
-
Where GT = G1 + G2
Hence 21
21
21 .111
RRRR
RRRT
+=+=
or 21
21.RRRRRT +=
In the same minner , if we have three resistors in parallel , then:
321
1111RRRR T
++=
321
213132
.....1
RRRRRRRRR
RT
++=
213132
321
.....
RRRRRRRRRRT ++=
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?
816 V
IT
8 8 8
Solution :-
=== 248
NRRT
AREIT
T 8216 ===
AREIbranch 28
16
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 ?
Solution :-
a)
===
=
++=
++=
++=
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
AREIT
T
22040
104
40
22
22 ====
===
d) ( ) ( ) WRIP 8020.2 22222 ===
or 2
22
2 RVP = , or P2 = I2V2
-39-
-
Current division Rule :-
21
21.RRRRIV +=
1
21
21.
RRRRRI
RVIT
+==
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 :-
==+=+= 0999.01.10010
1.01001.0*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
===
210*2110*42
3
3
1
11 I
VR
or
=
+=+=
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 , also find the voltage regulation ?
Rint. = 2
RL = 13VLE = 30V
Solution :-
ARR
EIL
L 213230
int
=+=+=
VRIEV LL 262*230.int === ( ) ( ) WRIP Lloss 82.2 2.int2 === %385.15%100
262630%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 = 6
I1
I = 42 mA
R2 =24 R3 =24
Solution :- All resistance in parallel , so if we define that R = R2 // R3 then :-
=+=+= 12242424*24
32
32
RRRRR
Hence
( ) mARR
RII 28612
1210*42 31
1 =+=+=
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
REIT
6.310*5
183
1
====
VERIV 18. 1 ===
Example :- For the following cct. Network , find RT , IA , IB , IC , VA , VB , I1 ,
I2 ?
R1 = 9
R2 = 6
R5 = 3R4 = 6
R3 = 4
R6 = 316.8 V
IA
I1
I2 IB IC
Solution :-
RC = 3
IC
IB
IA
RB = 6
RA = 3.6=+=+= 6.3696*9
21
21
RRRRRA
=++= 6363*94RB = R3 + R4 // R5
RC = 3
RT = RA + RB // RC
=++= 6.5363*66.3
ARE
TA 36.5
8.16 === I
-44-
-
Apply C.D.R.
ARR
RIICB
CAB 163
3*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
ARRRII A 2.1
963*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 ?
Solution :-
From K.C.L.
Ish = 1.2 0.12 = 1.08 A
+=+=
sh
shA
Ash
R
RRRII
7.27.22.108.1
( )
=
=
+=+=
+=
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 ?
8
E
24
8
4
4
1612
6
6
Solution :-
AI 22.1
241 == V
The same voltage ( V = 24 V ) on the
resistor Ra = 4 +16 + 4 = 24
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
8
E
24
8
4
4
1612
6
6
V
V1
V2
I5B
V3C
I6
I4
D
I2AI3I1
V1 = 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)Vs = IRL = 10 * 2 = 20 V if RL = 2 VsVs = IRL = 10 * 5 = 50 V if RL = 5
Example :- Calculate V1 , V2 , Vs for the following cct.:
Solution :-
R1
R2
VsI = 5A
V1
V2
2
3
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:
ILSolution :-
Rs = 2
RL = 4A
RREI
LsL 142
6 =+=+= E = 6V
For the current source cct. A
RRRII
Ls
sL 142
23 =
+=+=
RL = 4
IL3A
Rs = 2
A
REIs
326 ==
=
Example :- Convert the current source in the cct. Shown below to a voltage
source and determine IL for each cct.:
Solution :-
IL
.
For the current cct. ( )
mAIRR
RII
L
Ls
sL
310*610*3
10*310*9 333
3
=
+=+=
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 // 6 = 2
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 ARRRII
Ls
ssL 3146
6*10624824*8 =+=+==+=
-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
( )( ) ADDI 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 ( ) and Delta Star ( ) transformation :-
1. ) Delta Star ( ) transformation :- 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 ==== RRRRR CBA 3
or 3
= RR
BCCAAB
CAABA RRR
RRR ++=
BCCAAB
BCABB RRR
RRR ++=
BCCAAB
BCCAC RRR
RRR ++=
-54-
-
BA
C30
3030
B
A
C
10
10 10
2. ) Star Delta ( ) transformation :-
If the value of RA , RB , RC are known ,
and we need to get the values of RAB ,
RCA , RBC ; as follows :-
If === RRRR CBA , in this case RAB = RCA = RBC = R = 3 RY
B
A
C
4
4 4
B
A
C12
1212
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 :-
=++= 5.21510515*5
aR
=++= 51510515*10
bR
=++= 67.11510510*5
cR
25 V
Rc Rb Ra
5 8 10
{ ( ) ( ) } ( ){ ( ) ( ) } ( ){ }
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
====+=
+
+=
+=++++=++++=
-56-
-
Example :- For the following network , find I ?
Solution :- The resistances ( 6 , 3 , 3 ) are delta , can convert to star connection
as follows:
=++= 5.13363*6
aR , =++= 5.13363*6
bR , =++= 75.03363*3
cR
( ) ( )[ ]( ) ( )[ ]
75.05.35.55.3*5.5
5.3//5.52//4
+
+=
+=+++=
c
cbaT
RRRRR
AREIT
077.2889.26 ===
-57-
-
Example :- Find I for the following cct. network :-
12 V
5
2015
12.5 10
30
I
Solution :- The resistors ( 5 , 10 , 20 ) are star convert to delta
I
15
12.5
30
b
a
c
Rbc
Rab
Rac
12 V
=++=
=++=
=++=
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 // Rac ) and (15 // Rbc ) and (30 // Rab ) , hence the
circuit can be reduce to the following network :-
-58-
-
12 VR3
I
R1
R2
( )
-59-
=+== 3.75.175.125.17*5.12//5.121 acRR
( ) =+== 5.10351535*15//152 bcRR
( ) =+== 21703070*30//303 abRR
( )( )
=+==+=
+=
634.9218.1721*8.1721//8.17
21//5.103.7
// 321 RRRRT
AREIT
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 resistor of the network
shown:
Solution :-
- * Loop + * Loop m
) Loop = ( .
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
==
=
==
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
4 3
6 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 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 11 =
DDI 22 =
DDI 33 =
V15 = I15 * R15
= ( I3 I4 ) * 15
= ( I3 2 ) * 15
Example( 5 ):- Solve the following circuit diagram .
3 15
40
5
6020A
10A
5A
Solution:- The above diagram can be reduced to the following diagram;
-64-
-
3 15
40
5
60
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 11 =
DDI 22 =
Example( 6 ):- Solve the following circuit diagram:
10
42
6
20V
I1 I2
6AVo
-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-
-
-: dohtem tnerruc pool gnisu , tiucric gniwollof eht evloS : ) 7 ( elpmaxE
6
8
V21
7
9
bI
cI
oV A3
aI 5
:noituloS
-: a pooL 0 = oV - cI8 + aI) 8+5 (- ) 1 ( -------------------
-: b pooL 0 = 21 oV + cI7 + bI) 7+6 (- ) 2 ( -------------------
-: c pooL 0 = bI7 + aI8 + cI) 9+7+8 (- ) 3 ( -------------------
3 = aI bI ) 4 ( -------------------
oV -: . oV
) oV
oV ( oV
.
-76-
-
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 862.11402
2610
3112478
1513133102470
151312
1 ==
==
DDIb 2=
DDIc 3=
------------------- (1 )
) ------------------- ( 2------------------- ( ) 3
------------------- (1 ) ------------------- ( ) 2
) ------------------- ( 3
-68-
-
Example, (Sheet 4 Q. 25): Solve the following circuit diagram using loop
current:
30
15
13V
7
9V
3
12
1.2A
0.8A
20
6V
Solution:-
15 // 30 = =+ 10301530*15 ; 12 + 3 = 15
13V
7
15
1.2A
10
6V
9V
16V
20
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 )
Example, (Sheet 4 Q. 7): Solve the following circuit diagram:
-70-
-
Solution:
120
80
5
2 10V
1
10 2
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 )
-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
-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 , VC
-74-
-
Example 3 :- Solve the following circuit using nodal voltage method:
9 50
5
30
15
20
6
15V
3
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
25
15
35
17V
15
25
12V
0.8A
0.3A
20 A
C
D
B
DD
A I1
IB
Solution:
First; let D reference:-
A : 03.02515
4012
4025201
251
401 =+++++
++ CBA VVV
B : 02515
25351
151
251 =+
++ IVV AB
C : 08.04040
12401 =+++
IVV AC
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 =++++++
++ BBA VVV
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 =+++++
++ DA VV
D : 08.03.02020
1351
151 =+
++ AD VV
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:( . .
/
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:-
3
6
7
25V
3A4
I5I2
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 :-
3
6
7
3A4
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:-
30
6040
150V 270V
I1
I2
I3
Solution:-
1.) Effect of 150 V source:-
30
6040
150V
I'1
I'2
I'3
=++= 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:-
=++= 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:-
4
15V
8
5
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:-
8
5
12V
I"2
I"1
I"3
4
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:-
48
5
2A
I'"3
I'"2
I'"5
I'"4
I'"1
Vo
=++= 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:( . .
/
2-) Thevenin's Theorems:-
.
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 ( 6 )
resistor?
54
2
25V
63
7A
Solution:-
54
2
25V
3
7A
A B
1.) Find Rth :
54
2
3 A B
Rth
( ){ }
=+=+
+=
++=
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
===
=+
=+
6
Rth = 7.22
Eth = 23.89 V
A
B
I
A
RREI
Lth
th
8.1622.7
89.23 =+=
+=
-87-
-
Example 2:- Find the current in the 25 resistor for the following circuit
network?
10 40
25
20 20
2V
Solution:-
1.) Find Rth :
10 40
20 20
A B
( ) ( )
=+++=
+=
20204020*40
201020*10
20//4020//10
th
th
R
R
2.) Find Eth :
10 40
20 20
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 ( 9 ) resistor for the following cct. diagram?
9
8
7
6A
4A
10
25V
Solution :-
=++= 258107thR
-89-
-
8
7
6A
4A
10
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.
ththRE
RN = Rth as before .
IN = Isc = short circuit current between the two terminals of the active network.
Example 1:- Find the current in 25 resistor for the following circuit network
using Norton's Theorem?
10 40
25
20 20
2V
-91-
-
Solution:-
First find RN :-
10 40
20 20
A B
Second find IN :-
10 40
20 20
A B2VIN
I1 I2
I4I3
I
)( ) (
=+++=
+=
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 0147.0252020*033.0 =+=
-92-
-
Example 2:- Find I in 50v voltage source, for the following circuit using
eorem? Norton's Th
25
12
20
17
30
65V 50V45V
Solution:-
1.) Find RN :-
12
20
17
30
25
A
B
R1 R2 R3
30 20A
B
-93-
-
== 8.754
25*171R , == 78.354
17*122R , == 56.554
12*253R
2.) Find IN :-
( ) ( )[ ][ ] =+=
+++=
1978.356.25//8.37
20//30 231 RRRRN
12
20
17
30
65V
25
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 IRI
-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 RRR
ERIP *2
2.max
+==
thth
th RRE *
4 22
=
RNIN RL
LLN
NNLLL RRR
RIRIP *2
2.max
+==
NN
NN RREI *
4 22
2=
RL = Rth RL = RN
Under Max. Power transfer conditions, the efficiency is:-
%100*i
o
PP % =
%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
*2
% ==L
LL
VVV
h efficiency will always be 50% under max. power transfer conditions .
Practical example:-
T e *
RLEth
RthI
et Rth = 3 & RL = 1 & Eth = 15 V L
WPL 141*1315 2
+=
For RL = 2 WPL 182*515 2 =
=
For RL = 3 WPL 75.183*615 2 =
=
LL RIP
2=
LLth
RRR
E *
+=
2
-96-
-
For RL = 4 WP 36.184*15 2 == L 7
WPL 57.175*815 2 =
=For RL = 5
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
= 14LR 12
6
3
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:-
8
7
4
6Rth
a
b
)
=+ +=
+
141263
12
th
R = 6//3th
6*3R
VVE abth 12366*18 =+==
( ) WREP
th
th 57.214*4
124
22
.max ===
-98-
-
( ) ( )
V
EVV thaboc
7215*25
120
78*7846
120
=
++++===
=
( ) WRIP LL 2166*6672
22 =
+==
or
RL= 6
Eth = 72 V
Rth = 6
a
b
( ) ( )
L
theq
R
RR
==+=
+==
6101510*15
46//7//8.
WREP
th
thL 2166*4
74
22
===
-99-
-
Example 3 (sheet 5, fig. 20):- Find the maximum power in ( R ), for the
following cct. diagram?
1
1 2
6 6
2A4
10V
6V
2
2 4
R
6V
Solution:-
1-) Find Rth.:-
-100-
-
RRR theq ==+++== 102314. 2.) Find Eth :-
-101-
-
1
1 2
6 6
2A4
10V
6V
2
2 4
6V
A
B
IY
IX
IZ
2A
Voc
( )( )
( ) AIIAII
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 = 10 RL= 10
Eth = 15 V
B
AI
Example 4 (sheet 5, fig. 21):- Find the maximum power in ( R ), for the
following cct. diagram?
( ) WRIP
625.510*75.0 2
2.max
==
=
80mA18
5
12
R
15
0.86V
0.3V
8
20mA
Solution:-
First find Req.:- // )515( +)1812(. +=eqR //8 = 4.8
-103-
-
Second find IN:-
18
5
12
15
0.86V
0.3V
0.36V
1.2V
8
Isc A
B
30
20
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-