Download 3. Network thoerem

Network Theorems
Mesh analysis
 Nodal analysis
 Superposition
 Thevenin’s Theorem
 Norton’s Theorem
 Delta-star transformation

Direct application in conjunction with
Ohm’s law
 Indirect application in conjunction with
resistance
 Simultaneous equations

I
Determine current and source
e.m.f
E
V2
V3
Since R3 and R4 are in parallel
V1
R3
16
R1
8
R2
6 I =3A
4
R4
8
V3  I 4 R4  3  8  24V  I 3 R3  I 3 16
Therefore
24
I3 
 1.5 A
16
By Kirchoff’s first law
I  I 3  I 4  1.5  3  4.5 A
By Kirchoff’s second law
Also
V1  IR1  4.5  8  36 V
V2  IR2  4.5  6  27 V
E  V1  V2  V3  36  27  24  87 V
Determine I1, E, I3 and I
V1 27
I1 

 3A
R1 9
V2  I1R2  315  45V
By Kirchoff’s second law
E  V  V1  V2  27  45  72V
V 72
Also
I3 

 9A
R3 8
By Kirchoff’s first law
I  I1  I 3  3  9  12 A
I
I1
V1
V
I3
R1

V3
E
V2
R2

R3

Power dissipated in R3 is
20W. Calculate I3, R1,I1, I2
and E
a
5A
V10  110  10V
P  20W  10  I 3

E
20
I3 
 2A
10
I1
1
V1
b

1A
I3
2
R1
3
R3


20  I 3 R3  2  R3
2
I2
2
d
c
By Kirchoff’s first law in node a
R3  5 
I1  5  I 2  5  3  2 A
By Kirchoff’s first law in node b
V1 22
R1  
 11
I1
2
I 2  I3 1  2  1  3A
By Kirchoff’s second law in loop 2
V1  6  10  6  22V
P.D across 1  is 5 X 1=5V
E  V  5  22  27V
I
Determine current I and I4
E=87V
R1
8
R2
6
V1
V2
First find the total effective resistance
R3 R4
16  8
Re 

 5.33 
R3  R4 16  8
R3
16
V3
R4
8
Rt  R1  R2  Re  8  6  5.33  19.33 
Then
V
87
I

 4.5 A
Rt 19.33
Using current division
R3
16
I4 
I
 4.5  3 A
R3  R4
16  8
I
V
Rt
Determine VAB
Effective resistance for parallel
resistor 10 // 15  and
16//16
10 
15 
VAB
A
12V
10 15
Ra 
6
10  15
VAC
VBC
6

12  6V
66
8

12  8V
8 4
B
6
16 
16 
C
16 16
Rb 
8
16  16
Using voltage division
4
6
4
VAB
A
12V
VAC
B
6  VBC
8
C
Then
VAB  VAC  VBC  6  8  2V
I=5A
A
Calculate the current in each resistor
Applying Kirchoff’s 2nd law for loop 1
3
1

D
B
40V
40  3I1  14( I1  I 2 )
40  17I1 14I 2

3

---(a)
Applying Kirchoff’s 2nd law for loop 2
C
0  28I  I1   8I 2  3I1
 28I  31I1  8I 2
But
Thus
I  5A
140  31I1  8I 2

2
I
A
3I1
---(b)
1
I2
D
I1
2


I-I1
B
40V
I1-I2
---(b)
I-I1+I2
3


C
continue
Solving the 2 simultaneous equations
(a) X 4
160  68I1  56I 2
---(c)
(b) X 7
980  217I1  56I 2
---(d)
Then (c) + (d)
1140  285I1
I1  4 A
In 3 resistor
Substitute in (b)
140  124  8I 2
I2  2A
In 8 resistor
In 28 resistor
I  I1  5  4  1A
In 14  resistor
I1  I 2  4  2  2 A
In 4  resistor
I  I1  I 2  5  4  2  3A
I1
Calculate the current in the network
I3
Applying Kirchoff’s 2nd law for loop 1
10  19I1  18I 2
1
---(a)
10V
Applying Kirchoff’s 2nd law for loop 2
20  2I 2  18I1  I 2 
20  18I1  20I 2
---(b)
100  190I1  180I 2 ---(c)
(b)X 9
180  162I1  180I 2
I1  2.85 A

2
I 3  I1  I 2
20V
Substitute I1 in(a)
(a)x10
(d)-(c) we get
18
1
10  1I1  18( I1  I 2 )
I2
---(d)
10  54.34  18I 2
I 2  3.57 A
Current in 18  resistor
I 3  3.57  2.85  0.72 A
Calculate the current in the network
I1
I3
Applying Kirchoff’s 2nd law for loop 2
Current in 18  resistor
18
1
20  18I 3
I2
1
2
10V
20
I3 
 1. 1 A
18
Applying Kirchoff’s 2nd law for outside loop
20 10  I1 1
I1  10 A
Current in 1  resistor
I 2  (10)  1.1  11.1A
20V
A
The network shown is a 3 cells having
an internal resistance of 30 . Calculate
the current in the network
R=30
E=1.5V
E=1.5V
R=30
Applying Kirchoff’s 2nd law
C
1.5  I  30 1.5  I  30 1.5  I  30  0
B
R=30
E=1.5V
4.5  90I  0
4.5
I
 0.05 A
90
The voltage drop due to internal resistor is 0.05 x30=1.5V
Thus there is no potential different between two terminals
Create loop’s current rather than branch
current
 Use Kirchoff’s second (voltage ) law
 Ohm’s law to calculate p.d
 Branch is calculated by taking the
algebraic sum of the loop currents

Calculate the current in each branch
First create loop current ,i.e I1 , I2, I3 as shown
20
30
20
40
3
30
60
50
10
60
20V
50V
40
50
1
100V
I3
2
I1
100V
20V
10
I2
50V
continue
In loop 1
100  20  I1 60  30  50  I 2 50  I 3 30
80  140I1  50I 2  30I 3
---(a)
In loop 2
50  20  I 2 50  40  10  I1 50  I 3 40
20
70  50I1  100I 2  40I 3
3
---(b)
In loop 3
30
0  I 3 30  20  40  I1 30  I 2 4060
0  30I1  40I 2  90I 3
---(c)
I3
40
50
1
2
I1
100V
20V
10
I2
50V
continue
Solving these equations
I 3  1.50 A
I1  1.65 A
I 2  2.16 A
Current in 60
 I1  1.65 A
In direction of I1
Current in 30
 I1  I 3  0.15 A
In direction of I1
Current in 50
 I 2  I1  0.51A
In direction of I2
Current in 40
 I 2  I 3  0.66 A
In direction of I2
Current in 10
 I 2  2.16 A
In direction of I2
Current in 20
 I 3  1.50 A
In direction of I3
Choose reference node where all nodes
can refer
 Assign currents going to/out the nodes
 Assign voltage at nodes as V1 , V2,
V3….which refer to reference node
 Apply Kirchoff’s current law at each node
 Relate the voltage , resistance andcurrent
using ohm’s law
 Solve the equations obtained

node 1
Calculate V1 and V2
At node 1
1A
I3  I 4
V2

5
reference
I1 node 1
V1 I2
Simplified
At node 2
node 2
V1
I1  I 2  I 3
V1 V1  V2
1 
5
3
 1 1  V2
V1      1
5 3 3

….(a) 1A

node 2
I3
I4

5
reference
V1  V2 V2

3
7
V2
Simplified
V1
1 1
V2     0
3
3 7
continue
…..(b)
8V1  5V2  15
Substitute V2 we have
35V1
8V1 
 15
10
node 2
I3
I4
reference
7
V2  V1
10
10
V1  V
3
7 10 7
V2    V
10 3 3
V2

1A
7V1 V2 7  3  0
7V1  10V2
(a) X 15

5
Solve for equations (a) and (b)
(b) X 21
I1 node 1
V1 I2
Node 1
V1
Calculate V1 and V2 and
current in 8
At node 1
I1  I 2  I 3
4  V1 V1 V1  V2
 
5
15
10
5
24  6V1  2V2  3V1  3V2
….(a)
6V

Node 1
V1
I3
I2
4V

15
I1
Simplified
11V1  3V2  24

5
4V
Node 2
V2
reference
node
Node 2
V2


I4

15
reference
node
I5
6V
continue
I1
At node 2
I3  I 4  I5
5
Node 1
V1
I3
I2
V1  V2 V2 V2  6
 
10
8
12
4V
Node 2
V2

I5

I4

15
reference
node
Simplified
12V1 12V2  15V2  10V2  60
12V1  37V2  60
….(b)
Solving the simultaneous equations (a) and (b)
V1  2.88V
V2  2.55V
I 8
V2

 0.32 A
8
6V
The superposition states that in any network
containing more than one source , the current
in , or the p.d. across in any branch can be
found by considering each source separately
and adding their effects: omitted sources of
e.m.f are replaced by resistance equal to
their internal resistances.
Separating the network into several circuit contenting
only one source
I1
Original network
1
I1+I2
Separating into 2 networks
I1b
10V

18
10V
1
I2
I2b
18 
I1b+I2b
20V
I1c
1
I2c
18
I1c+I2c

20V
I1b
Network 1
Total resistance
2 18
1
 2.8 
2  18
1
10V
Thus
V
10
I1b 

 3.57 A
Rtotal 2.8
18

 3.57  3.21A
2  18
and
I 2b
Also
I1b  I 2b  3.57  3.21  0.36 A
I2b
18 
I1b+I2b
I1c
Network 2
Total resistance
118
2
 2.95 
1  18
Thus
V
20
I 2c 

 6.78 A
Rtotal 2.95
and
18
I1c  
 6.78  6.42 A
1  18
Also
1
I 2c  I1c  6.78  6.42  0.36 A
I2c
18
I1c+I2c

20V
I1
combination
1
10V
I2
18
I1+I2

20V
Thus
I1  I1b  I1c  3.57  6.42  2.85 A
and
I 2  I 2b  I 2c  3.21  6.78  3.57 A
Also
I1  I 2  2.85  3.57  0.72 A