Download 4. Network thoerem 2

Network Theorems
Circuit analysis
Mesh analysis
 Nodal analysis
 Superposition
 Thevenin’s Theorem
 Norton’s Theorem
 Delta-star transformation
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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
(a)
(b)
A
R2
V
A
R2
V
R
R3
R1
Vth
R3
R1
B
(d)
(c)
A
A
R2
R3
B
Vth
Rth
R
Rth
R1
B
B
Refer to network (b), in R2 there is not complete circuit, thus no
current, thus current in R3
V
I R3 
R1  R3
And p.d across R3 is
VR3
VR 3 
R1  R3
Since no current in R2, thus
VR3
Vth 
R1  R3
Refer to network (c) the resistance at AB
Thus current in R (refer network (d))
R1 R3
Rth  R2 
R1  R3
Vth
I
Rth  R
C
Calculate the current through R3
Solution
E1=6V
A
With R3 disconnected as in figure below
B
R1=2
R3=10
R2=3
D
64
2
I1 

 0.4 A
R1  R3 2  3
I1
p.d across CD is E1-I1R1
C
E1=6V
A
V  6  0.4  2  5.2V
E2=4V
E2=4V
B
R1=2
V
R2=3
D
continue
To determine the internal resistance we
remove the e.m.f s
23
r
 1 .2 
23
C
A
B
r
R1=2
R2=3
D
Replace the network with V=5.2V
and r=1.2, then the at terminal CD,
R3, thus the current
5.2
I
 0.46 A
1.2  10
I
C
V=5.2V
R3=10
r=1.2
D
Determine the value and direction of the current in BD, using
(a) Kirchoff’s law (b) Thevenin theorem
Solution
(a) Kirchoff’s law
Using K.V.L in mesh ABC + the voltage E
2  10I1  30I1  I 3 
2  40 I1  30 I 3
…..(a)
B
10
A
For mesh BDCB
0  40I 3 _  15I 2  I 3   30I1  I 3 
0  30 I1  15I 2  85I 3 …..(c)
I1
I3
C
40
I2
Similarly to mesh ABDA
0  10 I1  20I 2  40I 3 ……(b)
30
I1-I3
I2+I3
15
20
D
E=2V
Continue……
Multiplying (b) by 3 and (c) by 4and adding the two expressions,
thus
0  30 I1  60 I 2  120 I 3
0  120I1  60I 2  340I 3
0  90I1  460I 3
I1  5.111I 3
Substitute I1 in (a)
I 3  0.0115 A  11.5mA
Since the I3 is positive then the direction in the figure is
correct.
continue
B
By Thevenin Theorem
P.D between A and B (voltage divider)
VAB
10
 2
 0.5V
10  30
P.D between A and D (voltage divider)
VAD
20
 2
 1.143V
20  15
P.D between B and D
VBD  1.143  0.5  0.643V
30
10
A
C
15
20
D
E=2V
For effective resistance,
10 parallel to 30 
continue
B
10  30
 7.5
10  30
30
10
A
C
r
20 parallel to 15 
20  15
 8.57
20  15
15
20
D
r  7.5  8.57  16.07
Total
Substitute the voltage, resistance r and 10W as in figure below
16.07
0.643
I3 
16.07  10
 0.0115 A
 11.5mA from B to D
0.643V
10
RS
Another of expressing the current IL
E
IL 

Rs  RL
E
Rs
Rs  RL
Rs
Rs

 IS
Rs  RL
E
IL
RL
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
RS
E
IS
RS
E
IS 
Rs
Calculate the equivalent constant-voltage generator for the
following constant current source
5
5
Rs
1A
15
Vo
Vo
Current flowing in 15 is 1 A, therefore
Vo  115  15V
Current source is opened thus the 5 W and 15 W are in series, therefore
Rs  5  15  20
Analysis of circuit using constant current source
I1
5
Node 1
V1
I3
Node 2
V2

4V

I4
I2
I5
6V

15
reference
node
From circuit above we change all the voltage sources to current sources
5

0.8A
0.5A
5
4V
6V
I
V 4
  0.8 A
R 5
I
V
6

 0.5 A
R 12

continue
Node 1
V1
Node 2
V2
I3
I1
0.8A
I2
5
15

I4

reference
node
At node 1
V V V V
0.8  1  1  1 2
5 15
10
1  V
1 1
0.8  V1  
  2
 5 15 10  10
I5
12
0.5A
At node 2
0.5 
0.5  
V2 V2 V1  V2


8 12
10
V1
1 1 1 
 V2    
10
 8 10 12 
X 30
X 120
24  V1 6  2  3  3V2
60  12V1  V2 15  12  10
24  11V1  3V2
60  12V1  37V2
…..(a)
……(b)
continue
(a) 
12
11
(c) + (b)
26.8  12V1  3.273V2
………( c )
86.8  33.727V2
V2  2.55V
From (a)
11V1  24  3 2.55  31.65
V1 
31.65
 2.88V
11
Hence the current in the 8  is
I4 
V2 2.55

 0.32 A
8
8
So the answers are same as before
Calculate the potential difference across the 2.0
resistor in the following circuit
4.0
8.0
8.0
20V
10V
………( c )
2.0
First short-circuiting the branch containing 2.0 resistor
4.0
10V
I1
Is
8.0
10  4.0I1
I2
20  8.0I 2
20V
10
 2.5 A
4.0
20
I2 
 2.5 A
8.0
I1 
I s  I1  I 2  2.5  2.5  5 A
4.0  8.0
Rs  4.0 // 8.0 
 2.67
4.0  8.0
continue
Redraw for equivalent current constant circuit
I
Is
8.0
5A

V
Using current division method
I
2.67
 5  1.06 A
2.67  10
2.0
Hence the voltage different in 8  is
V  1.06  2.0  2.1 V
Calculate the current in the 5.0 resistor in the
following circuit
2.0
10A
6.0

8.0
4.0
Short-circuiting the branch that containing the 5.0  resistor
2.0
10A
8.0
6.0
Is
4.0
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
Is 
8.0
10  8.0 A
8.0  2.0
continue
2.0
6.0
8.0
The equivalent resistance is a parallel (2.0+8.0)//(6.0+4.0)
Rs 
4.0
2.0  8.06.0  4.0  5.0
2.0  8.0  6.0  4.0
Redraw the equivalent constant current circuit with the load 5.0
I
8.0A
5.0
Hence the current in the 5  is
5.0
I
5.0
 8.0  4.0 A
5.0  5.0
A
R2
A
Ra
R3
Rc
C
R1
B
C
Rb
B
Delta to star transformation
From delta cct , impedance sees from AB RAB 
From star cct , impedance sees from AB
R3 R1  R2 
R1  R2  R3
RAB  Ra  Rb
Thus equating
R1 R3  R1 R2
Ra  Rb 
R1  R2  R3
Similarly from BC
Rb  Rc 
and from AC
(b) – (c)
Ra  Rc 
Ra  Rc 
R1 R2  R1 R3
R1  R2  R3
R1 R2  R2 R3
R1  R2  R3
R2 R3  R1 R2
R1  R2  R3
(a)
(b)
(c)
(d)
By adding (a) and (d) ; (b) and (d) ;and (c) and (d) and then divided
by two yield
R1 R2
R3 R1
(g)
R2 R3
(f) Rc 
(e) Rb 
Ra 
R1  R2  R3
R1  R2  R3
R1  R2  R3
Delta to star transformation
Ra R2

Rb
R1
Dividing (e) by (f)
R1 Ra
R2 
therefore
Rb
Ra R3

Similarly, dividing (e) by (g)
Rc
R1
We have
Substitude R2 and R3 into (e)
(i)
(j)
(j)
R1 Ra
R3 
Rc
R1  Rb  Rc 
(k)
Rb Rc
Ra
(l)
Similarly
Rc Ra
R2  Rc  Ra 
Rb
(m)
Similarly
Ra Rb
R3  Ra  Rb 
Rc
(n)
A
Find the effective resistance at
terminal between A and B of the
network on the right side
C
Solution
B
Ra
C
R2
8
B
20
D
20
Rb
1
R4
12
6
R5
R2
8
R5
R3
R1
16
D
R4
12
3
Rc
B'
R = R2 + R4 + R5 = 40 
Ra = R2 x R5/R = 4 
Rb = R4 x R5/R = 6 
Rc = R2 x R4/R = 2.4 
2
4
Substitute R2, R5 and R4 with Ra, Rb dan Rc:
A
A
R1 16
R3
Rb
Ra
4
Rc
B
6
R1+Ra
6
2.4
20
12
Rc 2.4
B
RAB = [(20x12)/(20+12)] + 2.4 = 9.9 
R3+Rb