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ECE 3144 Lecture 7
Dr. Rose Q. Hu
Electrical and Computer Engineering Department
Mississippi State University
Reminder for Lecture 6
• Resistors in parallel
N
  1
Rn
R
n

1
parallel
1
or
• Current divider
in 
Gn  i
N
 Gn
n1
• Current sources in parallel
n
i(t )   ik (t )
k 1
N
G
 Gn
parallel n
1
Problem solving strategy
Find RAB in the circuit
given
12 k
A
2 k
RAB
6 k
4 k
12 k
B
A
B
2 k
RAB
4 k
12 k
A
12 k
B
6 k
Wye  Delta Transform
I1
R1
V1
+
R2
R3
-
R4
R5
R6
•
•
•
•
•
We have learned how to simplify the network with series parallel resistors
combination.
In the circuit network given, nowhere is the resistor in series or parallel
with another.
Techniques learned so far do not apply here.
We can replace one portion of the network with the equivalent network.
The conversion will reduce the network to series parallel combination of
resistors, which we are already familiar with.
One conversion technique we are learning here is wye-to-delta or delta-towye transformation.
Wye  Delta Transform
a
a
Ra
R1
R2
Rc
R3
c
c
Figure (1)
•
•
•
•
Rb
b
b
Figure (2)
Notice that resistors figure (1) forms a  (delta) and resistors in
figure (2) forms a Y (wye).
Both of these two configurations are connected to the same
terminals a, b,c
It is possible to relate these two networks to each other such that
terminal characteristics are the same. The relationship is called
wye-delta (Y- ) transformation.
The transformation must keep terminal characteristics the same.
At each corresponding pair of terminals, the resistance at the
corresponding terminals must be the same.
Wye  Delta Transform
Rab  Ra  Rb 
R2 ( R1  R3 )
R2  R1  R3
Rbc  Rb  Rc 
R3 ( R1  R2 )
R3  R1  R2
Rca  Rc  Ra 
R1 ( R2  R3 )
R1  R2  R3
Wye  Delta Transform
Ra 
R1 R2
R1  R2  R3
R2 R3
Rb 
R1  R2  R3
Rc 
R1 R3
R1  R2  R3
R1 
Ra Rb  Rb Rc  Ra Rc
Rb
R2 
Ra Rb  Rb Rc  Ra Rc
Rc
R3 
Ra Rb  Rb Rc  Ra Rc
Ra
Special case: suppose the delta-connected load is balanced, that is,
R1=R2=R3=R. The equivalent wye-connected load is also balanced, so
Ra=Rb=Rc=RY. Then we have
RY
R

3
Wye  Delta Transform
I1
R1
V1
+
Ra
R2
Rc
R3
-
V1
R4
R
b
+
R5
R6
R4
R6
•Rc and R4 are in series => Rc4
•Rb and R5 are in series => Rb5
•Rc4 and Rb5 are in parallel => Rc4b5
•Rc4b5 and Ra are in series =>Rc4b5a
R5
Example
Is
Is
Ra=6k
R1=12 k
12 V
+
-
R3
6 k
R2=18 k
Rc =2 k
12 V
+
4 k
9 k
R1 R2
12 *18

 6k
R1  R2  R3 12  18  6
R2 R3
6 *18
Rb 

 3k
R1  R2  R3 12  18  6
R1 R3
12 * 6
Rc 

 2k
R1  R2  R3 12  18  6
Ra 
Is 
Rb=3 k
12
12

 1.2mA
6k  R ps 6k  4k
4 k
9 k
Circuits with dependent sources
• When writing the KVL and/or KCL equations for
the network, treat the dependent source as though
it were an independent source.
• Write the equations that specifies the relationship
of the dependent source to the controlling
parameters.
• Solve the equations for the unknowns. Be sure that
number of linearly independent equations matches
the number of the unknowns.
Example
I1
+ 12 V
-
3 k
VA = 2000 I1
- +
+
Vo
5 k
-
Determine the voltage Vo in
the circuit
Example
+
2 k
3 k
10 mA
+
Vs
4 k
V0
-
4 I0
I0
-
Given the network,
find voltage V0
Homework for Lecture 7
• Problems 2.57, 2.64,2.65,2.69, 2.71, 2.72,
2.73, 2.75, 2.77