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Network Theorems
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Circuit analysis
Mesh analysis
Nodal analysis
Superposition
Thevenins Theorem
Nortons Theorem Delta-star transformation
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An active network having two terminals A and Bcan be replaced by a constant-voltage sourcehaving an e.m.f Vth and an internal resistance
Rth. The value of Vth is equal to the open-circuited
p.d between A and B.
The value of Rth is the resistance of the network
measured between A and B with the loaddisconnected and the sources of e.m.f replacedby their internal resistances.
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Networks to illustrate Thevenin theorem
V
R2
R
R1
R3
A
B
R2
Rth
R1
R3
A
B
V
R2
Vth
R1
R3
A
B
Vth
R
Rth
A
B
(a)(b)
(c)(d)
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31
3RR
VIR
31
3
RR
VRV
th
31
3
3
RR
VRV
R
Since no current in R2, thus
Refer to network (b), in R2 there is not complete circuit, thus no
current, thus current in R3
And p.d across R3 is
31
31
2 RR
RRRR
th
RR
VI
th
th
Thus current in R(refer network (d))
Refer to network (c) the resistance at AB
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R3=10
R1=2 R
2=3
E1=6V E
2=4V
C
D
A B
R1=2 R
2=3
E1=6V E
2=4V
C
D
A B
V
I1
ARR
I 4.032
246
31
1
VV 2.524.06
Calculate the current through R3
Solution
With R3 disconnected as in figure below
p.d across CD is E1-I1R1
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continue
R1=2 R
2=3
C
D
A B r
r=1.2
R3=10
C
D
V=5.2V
I
2.1
32
32r
AI 46.0102.1
2.5
To determine the internal resistance we
remove the e.m.f s
Replace the network with V=5.2V
and r=1.2, then the at terminal CD,
R3, thus the current
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Determine the value and direction of the current in BD, using(a) Kirchoffs law (b) Thevenin theorem
A
B
C
DE=2V
10
40
20 15
30I1 I
1
-I3
I3
I2
I2+I
3
311 30102III
3130402 II
Solution
(a) Kirchoffs law
Using K.V.L in mesh ABC + the voltage E
31323 3015_400 IIIII
3214020100 III
Similarly to mesh ABDA
For mesh BDCB
321 8515300III
..(a)
(b)
..(c)
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31 460900 II 31 111.5 II
Continue
Multiplying (b) by 3 and (c) by 4and adding the two expressions,
thus
321 12060300 III
mAAI 5.110115.03
Since the I3 is positive then the direction in the figure is
correct.
321340601200 III
Substitute I1 in (a)
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continue
A
B
C
D
E=2V
10
2015
30
By Thevenin Theorem
VVAD
143.11520
202
VVBD
643.05.0143.1
VVAB
5.03010
102
P.D between A and B (voltage divider)
P.D between A and D (voltage divider)
P.D between B and D
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continue
A
B
C
D
10
2015
30
r
16.07
0.643V10
57.8
1520
1520
07.1657.85.7r
For effective resistance,
5.7
3010
3010
AI 0115.01007.16
643.03
Substitute the voltage, resistance r and 10W as in figure below
DtoBfrom5.11 mA
10 parallel to 30
20 parallel to 15
Total
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E
RS
RL
IL
s
S
R
EI
S
Ls
s
R
RR
R
E
Ls
LI
RR
R
RR
EI
s
Ls
s
Another of expressing the current IL
Where IS=E/RS is the current would flow in a short circuit
across the source terminal( i.e when RL is replaced by short
circuit)
Then we can represent the voltage source as equivalent
current source
E
RS
IS
RS
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1A
5
15
5
Rs
Vo
VVo
15151
20155 s
R
Calculate the equivalent constant-voltage generator for thefollowing constant current source
Vo
Current flowing in 15 is 1 A, therefore
Current source is opened thus the 5 W and 15 W are in series, therefore
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Node 1
5
4V
reference
node
V2
Node 2
6V15
V1I1
I2
I4
I5
I3
4V
5
50.8A
6V
0.5A
Analysis of circuit using constant current source
From circuit above we change all the voltage sources to current sources
A
R
VI 5.0
12
6A
R
VI 8.0
5
4
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continueNode 1
reference
node
V2
Node 2
15
V1
I2
I4
I3
0.8A 0.5A12 5
I1 I
5
101558.0 2111
VVVV
12
1
10
1
8
1
10
5.02
1V
V
1012151260 21 VV
1010
1
15
1
5
1
8.02
1
V
V
21 332624 VV
101285.0 2122
VVVV
At node 1 At node 2
21371260 VV 21 31124 VV ..(a) (b)
X 30 X 120
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continue
65.3155.2324111 V
2727.338.86 V
AV
I 32.08
55.2
8
24
21
273.3128.26 VV 11
12)( a
VV 55.22
( c )
(c) + (b)
Hence the current in the 8 is
So the answers are same as before
VV 88.211
65.311
From (a)
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Calculate the potential difference across the 2.0resistor in the following circuit
10V 20V
2.0
8.0
8.04.0
10V 20V
8.04.0 AI 5.2
0.4
101
67.2
0.80.4
0.80.40.8//0.4
sR
AIIIs
55.25.221
20.820 I
10.410 I
( c )
I2
First short-circuiting the branch containing 2.0 resistor
AI 5.20.8
202
I1
Is
i
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continue
AI 06.151067.2
67.2
VV 1.20.206.1
Redraw for equivalent current constant circuit
Hence the voltage different in 8 isUsing current division method
5A
8.0
2.0
Is
I
V
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Calculate the current in the 5.0 resistor in thefollowing circuit
10A 8.0
2.0
4.0
6.0
10A 8.0
2.0
4.0
6.0
Is
AIs
0.810
0.20.8
0.8
Short-circuiting the branch that containing the 5.0 resistor
Since the circuit is short-circuited
across the 6.0 and 4.0 so they have
not introduced any impedance. Thus
using current divider method
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continue
8.0
2.0
4.0
6.0
5.05.08.0A
0.5
0.40.60.80.2
0.40.60.80.2s
R
AI 0.40.80.50.5
0.5
The equivalent resistance is a parallel (2.0+8.0)//(6.0+4.0)
Hence the current in the 5 is
Redraw the equivalent constant current circuit with the load 5.0
I
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A
C BR1
Ra
Rb
Rc
R3
R2
BC
A
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321
2131
RRR
RRRRRR
ba
321
21
RRR
RRR
c
baAB RRR
321
213
RRR
RRRR
AB
321
13
RRR
RRR
b
From delta cct , impedance sees from AB
Thus equating
Delta to star transformation
321
32
RRR
RRR
a
Similarly from BC
321
3221
RRR
RRRRRR
ca
321
3121
RRR
RRRR
RR cb
321
2132
RRR
RRRR
RR ca
From star cct , impedance sees from AB
and from AC
(a)
(b)
(c)
(b) (c) (d)
By adding (a) and (d) ; (b) and (d) ;and (c) and (d) and then divided
by two yield
(e) (f)(g)
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1
3
R
R
R
R
c
a
b
a
R
RR
R1
2
1
2
R
R
R
R
b
a Dividing (e) by (f)
Similarly
Delta to star transformation
c
ba
ba
R
RRRRR
3
Similarly, dividing (e) by (g)
a
cb
cb
R
RRRRR
1
c
a
R
RRR 13
b
ac
ac
R
RRRRR 2
therefore
We have
(i)
(j)
(j)
Substitude R2 and R3 into (e)
(k)
(l)
(m)
(n)Similarly
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A
B
C D
R1
16
R3
6
R2
8
R4
12
R5
20
C
B
D
B'
R2
8
R4
12
R5
20
1
2
3
4
Rc
Ra Rb
Find the effective resistance at
terminal between A and B of thenetwork on the right side
Solution
R = R2 + R4 + R5 = 40
Ra = R2 x R5/R = 4
Rb = R4 x R5/R = 6
Rc = R2 x R4/R = 2.4
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Substitute R2, R5 and R4 with Ra, Rb dan Rc:
R1+Ra 20 R3+Rb12
A
B
R3 6R116
Rc 2.4
Ra
4
Rb
6
A
B
Rc 2.4
RAB = [(20x12)/(20+12)] + 2.4 = 9.9