Methods of solution of DC circuits

Electronics - Devices

Oct 7, 2013 (4 years and 7 months ago)

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1
Circuits 1
Methods of solution of DC circuits
Prof. Dr.
Mohamed Fathy
Copyright 2009 Prof. Dr. Mohamed FathyAbu El-Yazeed

2
Methods of solution of DC
circuits
•Definition :
The problem of solution of DC circuits is
finding the response(voltage or current) in
any network element due to inputexcitation
(voltage and/or current sources).
3
Methods of solution of DC circuits
•Method 1
KCL & KVL
•Circuit Topology
•N Nodes L Loops B=N+L-1 Branches
–Write N-1equations using KCL
–Write Lequations using KVL
–Solve the N+L-1equations to find all branch
currents.
4
Example
•Nodes:
N=3
N=3
non simple nodes
•Loops:
L=3
independent loops
12501, 23052, 34603
•Branches
B=5
012, 250, 23, 30, 3460
•B=N+L-1
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R5
E1
R1
R2
R3
RL
R4
R6
1
3
E2
2
4
5
0
I1
I2
I
3
I
5
I4
5
Example
•Nodes:
N=3
N=3
non simple nodes
Write N-1 equations
using KCL (usually
node 0 is excluded)
6
R5
E1
R1
R2
R3
RL
R4
R6
1
3
E2
2
4
5
0
I1
I2
I
3
I5
I4
Node 2
Node 3
*****
6
Example
•Loops:
L=3
independent loops
12501, 23052, 34603
Write L equations
using KVL
6
R5
E1
R1
R2
R3
RL
R4
R6
1
3
E2
2
4
5
0
I1
I2
I
3
I
5
I4
Loop 12501
VVVV0
12255001
+++=
*****
7
Example
•Loops:
L=3
independent loops
12501, 23052, 34603
Write L equations
using KVL
6
R5
E1
R1
R2
R3
RL
R4
R6
1
3
E2
2
4
5
0
I1
I2
I
3
I
5
I4
Loop 23052
*****
8
Example
•Loops:
L=3
independent loops
12501, 23052, 34603
Write L equations
using KVL
6
R5
E1
R1
R2
R3
RL
R4
R6
1
3
E2
2
4
5
0
I1
I2
I
3
I
5
I4
Loop 34603
*****
9
Example
1122231
IR+IR+IR-E=0(3)
5556244
IR+IR+E-IR=0(5)
123
0(1)III−−=
345
0(2)III−−=
3L442322
IR+IR-IR-IR=0(4)
10
Circuits 1
Methods of solution of DC
circuits
Simplification Method
Prof. Dr.
Mohamed Fathy
Copyright 2009 Prof. Dr. Mohamed FathyAbu El-Yazeed

11
Methods of solution of DC circuits
•Method 2
Step by Step Simplification
1.Transforming current sources into equivalent
voltage sources or visa versa
2.Combination of active elements
3.Replacing series and parallel resistors by its
equivalent (All resistors are assumed linear)
4.Star-Delta or Delta-Star transformation
Copyright 2009 Prof. Dr. Mohamed FathyAbu El-Yazeed
12
Method 2
Step by Step Simplification
–Source Equivalence
–Combination of active elements
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Source Equivalence

0
0
V
I=I-
R
V= V0
–I R
0
V= I
0 R0
–I R
0
121332
V=V=V+V
00
=- IR+V
000
V=IR
14
000
V=IR
00
VR
00
IRVV−=
000
IRRIV−=
0
I
15
Example
•Convert the shown voltage source
into its
equivalent current source.
Solution
()
Ω==100&3
00
RvV
0
0
0
3
30
0.1
V
ImA
R
===
16
Combination of active elements
i)Series Connection

3210
VVVV−+=
3210
RRRR++=
17
Example
•Convert the shown current sources
into one equivalent current source.
()
Ω==Ω==50,200,50,100
2211
RandmAIRmAI
18
Solution
()
1122
100,50,200,50ImARImAandR==Ω==Ω
*****
19
Combination of active elements
ii) Parallel Connection
3210
IIII+−=
3210
1111
RRRR
++=
20
Example
•For the shown current, use simplification
methodto find the load current I
L. (All
voltages are in Volts & all resistors are in
k Ω).
1
0
2
21
Solution
22
*****
12
0
23
Circuits 1
Simplification Method (Cont.)
Star-Delta Transformation
Prof. Dr.
Mohamed Fathy
Copyright 2009 Prof. Dr. Mohamed FathyAbu El-Yazeed

24
Star -Delta Transformation
Delta (Pi or π) Star (Y or T)
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∆to Y Transformation
122331
V+V+V=0
121223312331
0IRIIRR++=
()()
12122212331121
IR+R+R-IIII+=0
3123
1212
122331122331
RR
I=I-I
R+R+RR+R+R
12311223
1212
122331122331
RRRR
V=I-I
R+R+RR+R+R
121212
VIR=
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∆to Y Transformation
1214241122
V=V-V=IR-IR
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∆to Y Transformation
•For the ∆
For the Y
Similarly
12311223
1212
122331122331
RRRR
V=I-I
R+R+RR+R+R
1231
1
122331
RR
R=
R+R+R
2312
2
122331
RR
R=
R+R+R
3123
3
122331
RR
R=
R+R+R
121122
V=IR-IR
28
Y to ∆Transformation
123
I+I+I=0
()()
()
114224334
GV-V+GV-V+GV-V=0
()
4123112233
VG+G+G=GV+GV+GV
112233
4
123
GV+GV+GV
V=
G+G+G
3
ii
i=1
3
i
i=1
GV
=
G

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Y to ∆Transformation
112233
4
123
GV+GV+GV
V=
G+G+G
()



112233
111411
123
GV+GV+GV
I=GV-V=GV-
G+G+G
211111
1
12
2213
3
33
GVGGV++-GV--
=G
G+G+G
GVVGV



()()
23311
1
2
11
23
GV--
I=G
G+G+G
GV-VV



30
Yto ∆Transformation
()
()
I=I-I=GV-V-GV-V
1123112123131
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∆to Y Transformation
•For the Y
For the ∆
Similarly
()
()
212331
11
123
GV-V-GV-V
I=G
G+G+G



()
()
I=GV-V-GV-V
112123131
31
12
1231
123123
GG
GG
G=&G=
G+G+GG+G+G
23
23
123
GG
G=
G+G+G
32
∆to Y Transformation
31
12
1231
123123
GG
GG
G=&G=
G+G+GG+G+G
23
23
123
GG
G=
G+G+G
123
12
1212
123
G+G+G
RR
R==R+R+
GGR
33
∆-Y Transformation
1231
1
122331
RR
R=
R+R+R
2312
2
122331
RR
R=
R+R+R
3123
3
122331
RR
R=
R+R+R
112233
4
123
GV+GV+GV
V=
G+G+G
34
∆-Y Transformation
Special case
122331
R=R=R=R
1231
1
122331
RR
R=
R+R+R
2312
2
122331
RR
R=
R+R+R
3123
3
122331
RR
R=
R+R+R
123
3
R
RRR===
123
4
V+V+V
V=
3
35
Y -∆Transformation
12
1212
3
RR
R=R+R+
R
31
3131
2
RR
R=R+R+
R
2
2323
1
RR3
R=R+R+
R
Special case
123
R=R=R=R
122331
R=R=R=3R
36
Example
For the shown circuit , Find:
a) The input resistance between the two terminals
1 and 3
b) V23
37
Solution 1
By converting Y 123-4 ⇒∆123
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*****
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*****
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Solution 2
By converting ∆123 ⇒Y 123-4
*****
Can node 4 always be selected as the 4th
point for the Y ?
41
Solution 2
By converting ∆123 ⇒Y 123-4
'
42
hasnoeffectNotethatR
'
42
R
*****