# DC NETWORK THEOREM Charge I =

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DC NETWORK THEOREM Charge I =
```DC NETWORK THEOREM
Current: “Rate of flow of electric charge.”
Charge
I=
Coulombs/Sec or Ampere
time
Note:1.
Direction of current is same as the direction of motion of +Ve charge or opposite to the direction of
motion of –Ve charge.
Voltage: “Energy required in transferring a charge of one coulomb from one point to
another point.”
Energy(W)
V=
Joule/Coulomb or Volts
Charge(Q)
EMF (Electromotive force): “The EMF of a voltage source is the energy imparted by
the source to each coulomb of the charge passing through it.”
Energy(W)
Joule/Coulomb or Volts
E=
Charge(Q)
Potential Difference: “The pd between two points is the energy required in transferring a
charge of coulomb from one point to another point.”
Energy(W)
pd =
Joule/Coulomb or Volts
Charge(Q)
Voltage drop: “The voltage drop between two points is the decrease in energy required
in transferring a charge of coulomb from one point to another point.”
Energy(W)
Voltage drop =
Joule/Coulomb or Volts
Charge(Q)
Resistance: “Electric resistance is the property of material which offers opposition to the
flow of current and dissipates energy.”
l
Ohm or Ω
(Law of resistance)
R=ρ
a
l = Length of the wire
Where
a = cross − sectional area of the wire
ρ = Resistivit y or Specific Resistance of the material
Note:1.
Resistance also depends on temperature.
Ohm’s Law: “The current passing through a conductor is directly proportional to
potential drop across its ends provided physical conditions are same.”
I ∞V
1
I= V
Or
R
Or
V = IR
1
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Conductance (G): It is reciprocal of resistance
G=
a
a
1
=
=σ
l
Resistance (R) ρl
Power & Energy in an electric circuit:
Power P = VI
= I 2R
V2
=
R
Energy E = P × time
mho, Ω −1 , Siemens
Watt, KW, MW
Joule, watt-sec, watt-hours, KWH
Series Circuit:
I
R1
R2
R3
V1
V2
V
(a)
V3
We know
Total voltage
V1 = IR1
Req
⇒
I
V
(b)
Fig.1
V2 = IR2
V3 = IR3
V = V1 + V 2 + V 3
⇒ IReq = IR1 + IR2 + IR3
⇒ Req = R1 + R2 + R3
Note:1.
Same current means resistances are in series.
Example 1: Three resistors are connected in series across a 12V battery. The one
resistance has a value of 1 ohm, second has a voltage drop of 4 Volts & third has power
dissipation of 12 W. Calculate value of each resistance & circuit current.
Solution: Hint
V = V1 + V2 + V3
12 = I + 4 + 12 / I
I = 2 Or 6 Amperes
When I=2 Amp
R1 = 1
R2 = 2
R3 = 3
When I=6 Amp
R2 = 2/3
R3 = 1/3
R1 = 1
2
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Parallel Circuit:
I1 R1
I2
R2
I3
R2
Req
⇒
I
V
(b)
I
V
(a)
We know
I1 =
V
R1
Total current
Fig.2
V
V
I3 =
R2
R3
I = I1 + I 2 + I 3
V
V V V
⇒
= +
+
Req R1 R2 R3
I2 =
⇒
1
1
1
1
= +
+
Req R1 R2 R3
Note:1.
2.
Same voltage means resistances are in parallel.
If two resistances are in parallel and R1=R2=R then
Req =
R
2
Series Parallel circuit:
I1 R1
A
I3 R3
I2 R2
B
C
I4
R4
I5
R5
D
I6 R6
A RAB B RBC C RCD D
VAB VBC VCD
I
V
(b)
I
V
(a)
I
V
(C)
(c)
Fig.3
RAB =
So
R1R2
R1 + R2
RAD = RAB + RBC + RCD And I =
VAB = IRAB
VBC = IRBC
VCD = IRCD
3
1
1
1
1
=
+
+
RCD R4 R5 R6
RBC = R3
V
From fig.3b
From fig.3b
From fig.3b
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I1 =
VAB
R1
I2 =
VAB
R2
I3 =
VBC
=I
R3
From fig.3a
VCD
V
V
I 5 = CD
I 6 = CD
R4
R5
R6
Power consumed by whole circuit
I4 =
From fig.3a
From fig.3c
Or
= I R AB + I R BC + I R CD
From fig.3b
Or
= I12 R1 + I 22 R2 + I 32 R3 + I 42 R4 + I 52 R5 + I 62 R6
From fig.3a
2
2
2
Example 2: Two resistances of 20 & 30 ohms respectively are connected in parallel.
These two parallel resistances are further connected in series with a third resistance of 15
ohm. If current through 15 ohm resistance is 3 Amperes. Find
Current through 20 & 30 ohm resistances
(i)
(ii)
Voltage across whole circuit
I1 20
(iii) Total power consumed.
3 A 15
30
Solution: Hint
I2
B
C
Req = (20||30) + 15
A
= 12 + 15 =37
I
V AB = 12x3 = 36
I 1 = 36 / 20 = 1.8 A
I 2 = 36 / 30 = 1.2 A
(i)
V
(ii)
V AC = V AB + V BC = 36 + 15 × 3 = 81Volts
Fig. 4
(iii)
Power = VI = 243 Watts
Circuit or Network Elements: R, L, C
Network or Circuit: “Path followed by an electric current.”
Or
“Any interconnection of circuit elements with or without energy sources.”
Note:1.
2.
A circuit must have a closed energized path.
A network may not have a closed path i.e. T-Network
3.
So every network may not be circuit (i.e. T-Network) but every circuit is a network.
Fig.5-T-Network
Loop: “Any closed path in the network.”
Mesh: “A closed path which does not have any closed path inside it.”
Node: “It is a junction where 2 or more branches are connected together.”
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Note:1.
2.
A junction where 3 or more branches connected together is known as principal node or
essential node.
For example
B
A
C
Loop but not Mesh
H
G
F
D
E
Loop or Mesh
Loop or Mesh
Fig.6
Total no of nodes
No of principal nodes
No of Loops
No of Meshes
= 6+2 = 8
=2
=3
=2
(A, B, C, D, E, F, G, H)
(B & E)
(ABEFA, BCDEB, ABCDEFA)
(ABEFA, BCDEB)
Energy Sources:
Basically Two types
(A)
Independent Energy Sources
a. Voltage Source
i. AC/DC
ii. Ideal/Practical
b. Current Source
i. AC/DC
ii. Ideal/Practical
(B)
Dependent Energy Sources
a. Voltage Source
b. Current Source
(A) Independent Energy Sources:
a. Voltage Source
r
V
V
(a) Ideal DC Voltage Source
(b) Practical DC Voltage Source
(Internal Resistance, r=0)
(With Internal Resistance, r)
r
+
V
_
(c)Ideal AC Voltage Source
(Internal Resistance, r=0)
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+
V_
Fig.7
(d) Ideal AC Voltage Source
(Internal Resistance, r)
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b. Current Source
I
R
I
(a) Ideal Current Source
(b) Practical Current Source
(Internal Resistance, R=infinity)
(Internal Resistance, R)
Fig.8
(B) Dependent Energy Sources
Consider some branch of a circuit
I1 R1
V1
aV1
+
_
bI1
+
_
Volts
Volts
(b) Current Dependent Voltage
Source (CDVS)
(a) Voltage Dependent Voltage
Source (VDVS)
dI1
cV1
Amperes
Amperes
(d) Current Dependent Current
Source (CDCS)
(c) Voltage Dependent Current
Source (VDCS)
Fig.9
Kirchhoff’s Current Law (KCL):
“At any instant of time, the algebraic sum of currents at any node is zero.”
∑I = 0
Or
“Total incoming currents = Total outgoing currents.”
Kirchhoff’s Voltage Law (KVL):
“At any instant of time, the algebraic sum of voltages in a closed path is zero.”
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Consider following circuit
A
I1
R1
B I2
R6
R3
I3
V1
F
E
R2
C
V2
R4
D
Fig.10
Apply KCL at node B
I1 - I2 - I3 = 0 Or
I1 = I2 + I3
Apply KVL in loop ABEFA
I 1 R 1 + I 3 R3 − V1 + I 1 R6 = 0
Apply KVL in loop BCDEB
I 2 R 2 − V 2 + I 2 R 4 − I 3 R3 = 0
Apply KVL in loop ABCDEFA
I 1 R1 + I 2 R2 − V2 + I 2 R4 − V1 + I 2 R6 = 0
Example 3: In the following circuit calculate current in each branch by KCL & KVL
(Fig.11a).
Solution:
I1+I2
B
A
C
I1
I2
2Ω
2Ω
4Ω
4Ω
6Ω
6Ω
40V
40V
44V
44V
F
D
E
(a)
(b)
Fig.11
Apply KVL in loop ABEFA
− 40 + 2 I 1 − 4 I 2 + 44 = 0 − − − − − − − −(1)
− 44 + 4 I 2 + 6( I 1 + I 2 ) = 0 − − − − − − − (2)
Apply KVL in loop BCDEB
34
28
62
Solving (1) & (2)
I1 =
A
I2 =
A
I1 + I 2 =
A
11
11
11
Mesh Analysis:
Steps are
1. Identify the total meshes
2. Assume some mesh current in each mesh (clockwise or anticlockwise)
3. Apply KVL in each mesh.
4. Solve the above equations.
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Example 4: Solve Example 3 by Mesh Analysis.
Solution:
Let the two meshes are having clockwise currents as shown in following figure
Apply KVL in loop 1
− 40 + 2 I 1 − 4( I 1 − I 2 ) + 44 = 0 − − − −(1)
2Ω
Apply KVL in loop 2
− 44 + 4( I 2 − I 1 ) + 6 I 2 = 0 − − − −(2)
40V
Solving (1) & (2)
34
28
62
I1 =
A
I 2 − I1 =
A
I2 =
A
11
11
11
4Ω
I2
I1
6Ω
44V
Fig.12
Example 5: Find the current through R3=4 ohm resistance by mesh analysis (Fig.13a)
Solution:
2Ω
2
2Ω
R3
10V
5A 10V
(a)
2
I1
R3
I2
I3
5A
(b)
Fig.13
Let the three meshes are having clockwise currents as shown in following figure
Apply KVL in mesh 1
− 10 + 2 I1 + 2( I1 − I 2 ) = 0 − − − − − −(1)
Apply KVL in mesh 2
+ 2( I 2 − I1 ) + 4( I 2 − I 3 ) = 0 − − − − − −(2)
Clearly From mesh 3
I 3 = −5 A − − − − − −(3)
Solving (1), (2) & (3)
Current in R3 = I2-I3 = (-3)-(-5) = 2 A
I 1 = −1 A
I 2 = −3
Node Analysis:
Steps are
1. Identify the total Principal Nodes
2. Assume one node as reference node (Voltage of this node = 0 Volts)
3. Assume some node voltages for other remaining nodes w.r.to
reference node. (V1, V2, V3 etc).
4. Assume some branch currents in different branches.
5. Apply KCL at different nodes and make the equations in terms of node
voltages and circuit elements.
6. Solve the above equations.
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Example 6: Solve Example 3 by Node Analysis.
Solution:
Apply KCL at node V1
 V1 − 40 − 0   V1 − 44 − 0   V1 − 0 

+
+
=0
2
4

 
  6 
372
V1 =
Volts
11
2
34
I1 = −
A
11
40V
28
I2 = −
A
11
62
I3 =
A
11
So
I1
V1
I2
4
6
44V
Fig.14
Reference or
Datum Node, V=0
I4
I1 V1
2Ω
I2
Example 7: Solve Example 5 by Node Analysis.
Solution:
Apply KCL at node V1
10V
I3
I3
R3
2Ω
5A
 V1 − 10 − 0   V1 − 0   V1 − 0 

+
+
=5
2

  2   4 
V1 = 8 Volts
V=0
Fig.15
V − 0 
I3 =  1
 = 2 Amps
 4 
Superposition theorem:
“In a linear circuit, containing more then one independent
energy sources, the overall response (Voltage or current) in any branch of the circuit is
equal to sum of the response due to each independent source acting one at a time while
making other source in-operative.”
So
I
R2
R1
V1
I’
R2
R1
I1 ⇒
V1
(a)
(b)
+ R1
I”
R2
(c)
Fig.16
So according to superposition theorem
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I=I’+I”
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I1
Example 8: Solve example 5 by superposition theorem.
Solution:
I’
R3
2
2
I”
R3
2
2
10V
5A
(a)
(b)
Fig.17
Consider 10 V Voltage source only (Fig.17a)

10
 2 
 = 1 Amps
I'= 

 2 + 4  2 + (2 || 4) 
Consider 5A Current Source only (Fig.17b)
 1 
I" = 
5 = 1 Amps
1 + 4 
So I=I’+I”=1+1=2 Amps
Example 9: Find I in the following circuit by superposition theorem (Fig.18a).
Solution:
5
5 I1
Ω
I
I’
20A
20A
12
12
4
8
4
8
40 V
10V
(a)
Consider 20A current source only (Fig.18b)
By current division rule


8
20 = 10
I 1 = 
 8 + 5 + (4 || 12) 
Again bye current division rule
5
 4 
I'= 
10 = Amps
2
 4 + 12 
(b)
5
10V
I2
8
Ω
4
(c)
5
V1
Consider 10V Voltage source only (Fig.18c)


10
5
 = −
I 2 = −
8
 8 + 5 + (4 || 12) 
By current division rule
5
 4  5 
I" = 
Amps
 −  = −
32
 4 + 12  8 
10
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I”
12
4
8
I”
’12
40V
(d)
Fig.18
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Consider 40V Voltage source only (Fig.18d)
Apply KCL at node V1
 V1 − 0   V1 − 40 − 0   V1 − 0 

+
=

4
 8+5  
  12 
6240
V1 =
V
256
 V − 0  65
I '''=  1
 = Amps
 12  32
 5   5   65  35
I = I'+ I' '+ I' ' ' =   +  −  +   =
Amps
 2   8   32  8
So
Thevenin’s theorem:
“Any linear two terminal circuits can be replaced by an
equivalent network consisting of a voltage source (VTh) in series with a resistance (RTh).”
Any Linear
Two
terminals
Network
A
A
IL
RL
⇒
IL
RL
RTh
VTh
B
(a)
Fig.19
B
(b) Thevenin’s Equivalent
Circuit
Where
VTh = Open circuit voltage at load terminals
RTh = Equivalent resistance of the network at load terminals when
VTh
IL =
RTh + RL
And
Example 10: Solve example 5 by Thevenin’s Theorem.
Solution:
VTh
2
2
2
2
Ω
RTh
10V
Ω
5
A
V=0
(a)
Fig.20
11
VTh
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(b)
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Find RTh (Fig.20a)
RTh = (2 || 2) = 1Ω
Find VTh (Fig.20b)
Apply KCL at node VTh
A
So I L =
IL
RL
RTh
 VTh − 10 − 0   VTh − 0 

+
=5
2

  2 
VTh = 10 Volts
VTh
B
(c) Thevenin’s Equivalent
Circuit
VTh
10
=
= 2 Amps
RTh + RL 1 + 4
Example 11: Find Thevenin’s equivalent circuit across AB of following circuit
(Fig.21a).
Solution:
4A
A
A
3Ω
5Ω
20 Ω
3Ω
5Ω
20 Ω
30V
2Ω
2Ω
B
(a)
4A
A
Find VTh(Fig.21c)
 30 
VTh = −3 × 4 + 5

 20 + 5 
VTh = −6 Volts
B
(b)
Find RTh (Fig.21b)
RTh = (20 || 5) + 3 + 2 = 9 Ω
VTh = −3 × 4 + 5 I
RTh
3Ω
5Ω
20 Ω
30V
I
2Ω
B
(c)
B
RTh
So Its Thevenin’s equivalent circuit is
shown in fig.21d, It is to be noted here
that terminal B is +Ve and A is –Ve.
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VTh
A
(d) Thevenin’s Equivalent
Circuit
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Norton’s theorem:
Fig.21
“Any linear two terminal circuits can be replaced by an
equivalent network consisting of a current source (IN) in parallel with a resistance (RN).”
A
A
Any Linear
Two
terminals
Network
IL
RL
⇒
IN
IL
RL
RN
B
(b) Norton’s Equivalent
Circuit
B
(a)
Fig.22
Where
IN = Isc = Short circuit current at load terminals
RN = Equivalent resistance of the network at load terminals when
RN
IL =
IN
R N + RL
And
Example 12: Solve example 5 by Norton’s Theorem.
Solution:
Find RN (Fig.22a)
Same as RTh RTh = (2 || 2) = 1Ω
Find IN (Fig.22b)
10V
10
I N = 5A + A
2
I N = 10 Amps
RN
IL =
IN
R N + RL
1
IL =
10 = 2 Amps
1+ 4
the
2
2
IN = Isc
5A
(a)
A
IN
RN
IL
RL
B
(b) Norton’s Equivalent
Circuit
Fig.23
Example 13: Find current and voltage across 5 ohm resistance by Norton’s theorem
(Fig.24a).
Solution:
4Ω
3Ω
2Ω
15V
6A
(a)
13
3Ω
4Ω
5Ω
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2Ω
RN
(b)
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Find RN (Fig.24b)
V1
13
R N = (4 || 2) + 3 = Ω
3
Find IN (Fig.24c)
15V
Apply KCL at node V1
 V1 − 15 − 0   V1 − 0 
V − 0 

+
−6+ 1
=0
4

  2 
 3 
117
V1 =
Volts
13
V − 0 
IN =  1
 = 3Amps
 3 
(13 / 3)
13
3 = Amps
IL =
(13 / 3) + 5
9
13
VL = I L RL = Volts
3
4Ω
3Ω
6A
2Ω
(c)
IN
A
IN
IL
RL
RN
B
(d) Norton’s Equivalent
Circuit
Fig.24
Example 14: Find Norton’s equivalent circuit at A-B of example 11.
12 V
Solution:
V1
Find RN (Fig.21b)
3Ω
20 Ω
RN = RTh = (20 || 5) + 3 + 2 = 9 Ω
5Ω
IN
Find IN (Fig.25a)
30V
Apply Node analysis at node V1 after applying
2Ω
source transformation
V=0
 V1 − 30 − 0   V1 − 0   V 1 − 12 − 0 
(a)
+
+
=
0

 
 

A
20
3

  5  

76
V1 = V
IN
RN
3
12
0
2
V
−
−


IN =  1
= A
20
 3

B
(b) Norton’s Equivalent
Fig.25
Circuit
Example 15: Use Norton’s theorem to find out current in 6 ohm resistance and verify it
with Thevenin’s theorem (Fig.26a).
Solution:
4
4
4
I3
A
12 V
2
4
2A
4
6
B
(a)
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12V
4
2A
2
(b)
2
I1
I2
(c)
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By Norton’s Theorem:
Find RN (fig.26b)
RN=2 Ohms
Find IN (fig 26c)
Clearly from mesh 1
I1 = 2 A
Apply KVL in mesh 2
2(I2+I1)-12=0
So
I2 = IN = 4 Amp
2
IL =
4 = 1 Amps
2+6
By Thevenin’s Theorem:
Find RTh (fig.26b)
RTh = RN=2 Ohms
Find VTh (fig 26d)
Clearly from mesh 1
I1 = 2 A
So
VTh = 12-2I1= 8 Volts
8
IL =
= 1 Amps
2+6
4
12V
4
2A
A
2
I1
B
(d)
Fig.26
Hence verified
Maximum Power Transfer Theorem:
“Maximum power is transferred by a circuit to a
load resistance (RL), when RL is equal to Thevenin’s equivalent resistance (RTh) of that
network.”
A
So for maximum power
IL
RL = RTh
RTh
RL
And maximum power will be
VTh
2
V
Pmax = Th
4 RL
B
Fig.27 Thevenin’s Equivalent
Circuit
Proof:
IL =
VTh
− − − − − (1)
RTh + R L
Power
P = I L2 RL
2
 VTh 
 RL
= 
 RTh + RL 
RL
= VTh2
− − − − − ( 2)
( RTh + RL ) 2
Differentiating equation equa (2) w. r. t. RL and put dP/dRL = 0
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 ( R + RL ) 2 × 1 − RL × ( RTh + RL ) 
dP
= VTh2  Th
 − − − − − (3)
dRL
( RTh + RL ) 4


RL-RTh = 0
RL = RTh
Put this in equation (2)
2
V
Pmax = Th
4 RL
Or
Example 16: Find out the value of R for maximum power transfer to this load and find
out the value of maximum power (fig.28a).
Solution:
6V
6
3
15
2A
15
R
6
3
RTh
8V
(a)
(b)
6V
Find the RTh (Fig.28b)
RTh = (15 + 6) || 3 =
21
Ω
8
2A
I1
15
6
I2
+
VTh
3
Find VTh (Fig.28c)
8V
-
(c)
Clearly
I1 = 2 A
By KVL in mesh 2
+ 6 I 2 + 6 + 3I 2 + 15( I 2 − I1 ) = 0
Fig.28
So
I2 = 1 A
VTh = 3I 2 + 8 = 11Volts
Hence for maximum power transfer to R
21
V2
RL = RTh = Ω
Pmax = Th = 11.524Watt
and
8
4 RL
Star-Delta and Delta-Star transformation:
1
1
R1
R3
R31
⇔
R2
R12
3
2
R23
(b) Delta ( ∆ ) Connection
2
3
(a) Star or WYE (Y)
Connection
Fig.29
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Star to Delta
R R + R2 R3 + R3 R1
R12 = 1 2
R3
Delta to Star
R12 R31
R1 =
R12 + R23 + R31
R23 =
R1 R2 + R2 R3 + R3 R1
R1
R2 =
R12 R23
R12 + R23 + R31
R31 =
R1 R2 + R2 R3 + R3 R1
R2
R3 =
R23 R31
R12 + R23 + R31
Proof:
(Equivalent resistance at 1 - 2)Y = (Equivalent resistance at 1 - 2 )∆
R1 + R2 = (R23 + R31 ) || R12
R1 + R2 =
R12 (R23 + R31 )
− − − − − (1)
R12 + R23 + R31
Similarly
(Equivalent resistance at 2 - 3)Y = (Equivalent resistance at 2 - 3)∆
R2 + R3 = (R12 + R31 ) || R23
R (R + R31 )
R2 + R3 = 23 12
− − − − − ( 2)
R12 + R23 + R31
(Equivalent resistance at 3 - 1)Y = (Equivalent resistance at 3 - 1)∆
R3 + R1 = (R12 + R23 ) || R31
R (R + R23 )
R3 + R1 = 31 12
− − − − − (3)
R12 + R23 + R31
Delta to star
Equation (1) + (2) + (3)
2(R12 R23 + R23 R31 + R31 R12 )
R12 + R23 + R31
R R + R23 R31 + R31 R12
R1 + R2 + R3 = 12 23
− − − − − ( 4)
R12 + R23 + R31
R12 R31
R1 =
− − − − − (5)
(4) - (2)
R12 + R23 + R31
R12 R23
R2 =
(4) - (3)
− − − − − ( 6)
R12 + R23 + R31
R23 R31
R3 =
(4) – (1)
− − − − − (7 )
R12 + R23 + R31
2(R1 + R2 + R3 ) =
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Star to Delta
From Equation (5), (6) & (7)
R12 R23 R31 (R12 + R23 + R31 )
(R12 + R23 + R31 )2
R12 R23 R31
R1 R2 + R2 R3 + R3 R1 =
− − − − − (8)
(R12 + R23 + R31 )
R R + R2 R3 + R3 R1
R12 = 1 2
(8) / (7)
R3
R R + R2 R3 + R3 R1
R23 = 1 2
(8) / (5)
R1
R R + R2 R3 + R3 R1
R31 = 1 2
(8) / (6)
R2
Example 17: Find the equivalent resistance between the terminals a-b of the bridge
circuit of the fig.30a
Solution:
a
R1=2
a
6
4
2
R3=2/3
R =1
R1 R2 + R2 R3 + R3 R1 =
⇒
b
10
14
(a)
2
b
10
14
(b)
Fig. 30
Apply delta to star transformation
R12 R31
6× 4
R1 =
=
= 2Ω
R12 + R23 + R31
12
R12 R23
6× 2
R2 =
=
= 1Ω
R12 + R23 + R31
12
R23 R31
4× 2 2
R3 =
=
= Ω
R12 + R23 + R31
12
3


2
Rab = 2 + (1 + 14 ) ||  + 10 

3

Rab = 8.234Ω
18
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