ECE 3144 Lecture 4
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Transcript ECE 3144 Lecture 4
ECE 3144 Lecture 8
Dr. Rose Q. Hu
Electrical and Computer Engineering Department
Mississippi State University
1
Reminder for Lecture 7
• Wye-Delta or Delta-Wye transformation
a
a
Ra
R1
R2
R3
c
Rc
b
c
Rb
b
R R Rb Rc Ra Rc
R1 a b
Rb
Ra
R1 R2
R1 R2 R3
R R Rb Rc Ra Rc
R2 a b
Rc
Rb
R2 R3
R1 R2 R3
R R Rb Rc Ra Rc
R3 a b
Ra
Rc
R1 R3
R1 R2 R3
• Circuits with dependent sources
– treat the dependent source as though it were an independent
source.
– write the equations that specifies the relationship of the dependent2
source to the controlling
Problem solving strategy:
Delta-Wye transform (HW 2.73)
2
Io
3
9
9
2
6
12
36 V
4
Io
36 V
12
18
5
12
18
I
9
2
Io
3
2
I
36 V
9 3 2 2 6
4A
9 3
Io
I 3A
9 3 2 2
36 V
6
3
Problem solving strategy: Delta-Wye transform
• if possible, simplify the circuits first by using seriesparallel combination techniques
• Repeat step 1 until no series or parallel combination
existing in the network.
• Re-draw the shape of the circuit to the one you are
more familiar with if necessary
• Identify the Delta shape components or Wye shape
components in the simplified circuit.
• Determine where and how to do Wye-Delta or Deltato-Wye transforms.
• Perform Wye-to-Delta (or Delta-to-Wye)
transformations so that the circuit seen is represented
as a series-parallel interconnection of resistors.
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It’s your turn: how to solve the problem?
6 k
6 k
I
HW 2.72: Given the network, find I
21 V
12 k
2 k
18 k
•First if there are resistors in series or in parallel in the circuit network
given?
•If there are, how to do the simplification;
•If there are not, do you feel comfortable or familiar with the shape of
the circuit given?
•If not, redraw the circuit network.
•Identify the Delta-shape or the Wye-shape which you can do the
transformations.
5
Problem solving strategy:
Circuit with dependent sources (HW
Problem 2.85 in the textbook: for the network shown, choose the values of and such that
is maximized. What is the resulting ratio , Vo /Vs ?
RS
Ro
+
VS
Rin
Vin
RS Rin
VS
Rin
Vin
_
+
Vin
RL
Vo
_
(voltage divider for resistors in series)
R
R R
Vo L Vin in L VS
Ro RL
RS Rin Ro RL
(voltage divider for
resistors in series)
Rin RL
Vo
VS
R
R
R
R
in o
L
S
6
Problem solving strategy: HW 2.69
1 k
Io
4 k
IS
10 k
2 k
12 V
+
+
I2
I3
V2
-
1 k +
I1
Io
4 k
IS
V3
_
10 k
2 k
V1
_
12 V
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Summary for Chapter 2
•
•
•
•
•
•
•
•
•
•
Resistor, Resistance, Resistivity
Ohms’law V = I*R: only linear resistors satisfy Ohm’s law.
Power P = V*I= V2/R = I2R
Kirchoff’s current law (KCL): The algebraic sum of the
currents leaving (entering) a node is zero.
Kirchoff’s voltage law: The algebraic sum of the voltages
around any loop path is zero.
Single loop circuit: resistors in series and voltage divider
Single node circuit: resistors in parallel and current divider
Circuits containing a single source and a series-parallel
interconnection of resistors
Wye-delta or Delta-to-Wye transformations
Circuits containing dependent sources
8
Exam 1: Feb 6
• Exam 1 covers Chapter 1 and Chapter 2.
• Chapter 1: Basic concepts
– Charge (q), current (i), and their relationship
– Voltage (v), work/energy (w), power (p), and their
relationships.
– Passive sign convention
• The positive reference of voltage v(t) is at the same terminal
the the current variable i(t) is entering.
• How to determine absorbing or supplying energy?
– Active elements and passive elements
– Active elements
• Independent sources
• Dependent sources.
• Chapter 2: See slide 7
9
Homework for Lecture 8
• Problems 2.79,2.80, 2.81, 2.83, 2.87
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