network theorems. circuit analysis mesh analysis nodal analysis superposition thevenin’s theorem...
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Network Theorems
Circuit analysis
Mesh analysis Nodal analysis Superposition Thevenin’s Theorem Norton’s Theorem Delta-star transformation
An active network having two terminals A and B can be replaced by a constant-voltage source having 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 load disconnected and the sources of e.m.f replaced by their internal resistances.
Networks to illustrate Thevenin theorem
VR 2
R
R 1
R 3
A
B
R 2
R th
R 1
R 3
A
B
VR 2
V th
R 1
R 3
A
B
V th
R
R th
A
B
(a)(b)
(c)(d)
313 RR
VIR
31
3
RR
VRVth
31
33 RR
VRVR
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
312 RR
RRRRth
RR
VI
th
th
Thus current in R (refer network (d))
Refer to network (c) the resistance at AB
R 3=10
R 1=2 R 2=3
E 1=6V E 2=4V
C
D
A B
R 1=2 R 2=3
E 1=6V E 2=4V
C
D
A BV
I1
ARR
I 4.032
246
311
VV 2.524.06
Calculate the current through R3
Solution
With R3 disconnected as in figure below
p.d across CD is E1-I1R1
continue
R 1=2 R 2=3
C
D
A B r
r=1.2 R 3=10
C
D
V=5.2V
I
2.132
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
Determine the value and direction of the current in BD, using (a) Kirchoff’s law (b) Thevenin theorem
A
B
C
DE=2V
10
40
20 15
30 I1 I1-I 3
I3
I2I2+I 3
311 30102 III
31 30402 II
Solution
(a) Kirchoff’s lawUsing K.V.L in mesh ABC + the voltage E
31323 3015_400 IIIII
321 4020100 III Similarly to mesh ABDA
For mesh BDCB
321 8515300 III
…..(a)
……(b)
…..(c)
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.
321 340601200 III
Substitute I1 in (a)
continue
A
B
C
DE=2V
10
20 15
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
continue
A
B
C
D
10
20 15
30
r
16.07
0.643V 10
57.81520
1520
07.1657.85.7r
For effective resistance,
5.73010
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
E
R S
R L
IL
sS R
EI
SLs
s
RRR
RE
LsL I
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
R S
ISR S
1A
5
15
5
R s
V o
VVo 15151
20155 sR
Calculate the equivalent constant-voltage generator for the following 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
Node 1
5
4V
referencenode
V 2
Node 2
6V15
V 1I1
I2I4
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
AR
VI 5.0
12
6A
R
VI 8.0
5
4
continueNode 1
referencenode
V2
Node 2
15
V1
I2I4
I3
0.8A 0.5A12 5
I1 I5
101558.0 2111 VVVV
12
1
10
1
8
1
105.0 2
1 VV
1012151260 21 VV
1010
1
15
1
5
18.0 2
1
VV
21 332624 VV
101285.0 2122 VVVV
At node 1 At node 2
21 371260 VV 21 31124 VV …..(a) ……(b)
X 30 X 120
continue
65.3155.232411 1 V
2727.338.86 V
AV
I 32.08
55.2
82
4
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)
Calculate the potential difference across the 2.0 resistor in the following circuit
10V 20V
2.0
8.0
8.0 4.0
10V 20V
8.0 4.0 AI 5.20.4
101
67.20.80.4
0.80.40.8//0.4sR
AIII s 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
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
Calculate the current in the 5.0 resistor in the following circuit
10A 8.0
2.0
4.0
6.0
10A 8.0
2.0
4.0
6.0
Is
AI s 0.8100.20.8
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
continue
8.0
2.0
4.0
6.0
5.0 5.0 8.0A
0.50.40.60.80.2
0.40.60.80.2sR
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
A
C BR 1
R a
R bR c
R 3R 2
BC
A
321
2131
RRR
RRRRRR ba
321
21
RRR
RRRc
baAB RRR
321
213
RRR
RRRRAB
321
13
RRR
RRRb
From delta cct , impedance sees from AB
Thus equating
Delta to star transformation
321
32
RRR
RRRa
Similarly from BC
321
3221
RRR
RRRRRR ca
321
3121
RRR
RRRRRR cb
321
2132
RRR
RRRRRR 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)
1
3
R
R
R
R
c
a b
a
R
RRR 1
2 1
2
R
R
R
R
b
a Dividing (e) by (f)
Similarly
Delta to star transformation
c
baba R
RRRRR 3
Similarly, dividing (e) by (g)
a
cbcb R
RRRRR 1
c
a
R
RRR 1
3
b
acac R
RRRRR 2
therefore
We have
(i)
(j)
(j)
Substitude R2 and R3 into (e)
(k)
(l)
(m)
(n)Similarly
A
B
C D
R116
R3
6
R28
R412
R5
20
C
B
D
B '
R28
R412
R5
201
2
3
4
Rc
Ra Rb
Find the effective resistance at terminal between A and B of the network 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
Substitute R2, R5 and R4 with Ra, Rb dan Rc:
R1+Ra20 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