Transcript I 2 - McGraw Hill Higher Education

6-1
Electricity
Principles & Applications
Eighth Edition
Richard J. Fowler
Chapter 6
Complex-Circuit Analysis
(student version)
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6-2
INTRODUCTION
• Superpositon Theorem (Page 156)
• Simultaneous Equations (Page 136)
• Voltage Sources (Page 160)
• Thevenin’s Theorem (Page 161)
• Norton’s Theorem (Page 169)
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6-3
Dear Student:
This presentation is arranged in segments. Each segment
is preceded by a Concept Preview slide and is followed by a
Concept Review slide. When you reach a Concept Review
slide, you can return to the beginning of that segment by
clicking on the Repeat Segment button. This will allow you
to view that segment again, if you want to.
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6-4
Concept Preview
• Two or more sources are required to
use the superposition theorem. (Page 156)
• Superposition theorem requires only
series-parallel rules and procedures
to determine values in a complex
circuits. (Page 157)
• Superposition theorem will solve for
all currents and voltages in the circuit.
(Page 157)
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6-5
Superposition Technique
20 W
B1
2A
110 V
5A
10 W
30 W
3A
B2
Rty
10 V
V
u 1110
Replace B2 with a short and calculate the currents.
20 W
Rty
110 V
u 110
30 W
3A
10 W
4A
110 V
1A
Replace B1 with a short and calculate the currents.
20 W
110 V
1A
8A
10 W
30 W
7A
110 V
Algebraically add the two currents for each resistor.
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6-6
Verification of Superposition Results
Using Circuit Measurements.
The dark bars on the ammeters are the negative terminals.
Notice that both the directions and magnitudes of the currents
agree with those obtained by the superposition method.
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6-7
Simultaneous equations
A procedure for determining the values of I1 and I2 using two independent
equations is shown below.
The independent equations are:
#1
#2
8.5V = 5W(I1) - 1W(I2)
2.5V = -1W(I1) + 3W(I2)
Multiply equation # 2 by 5 and add the results to equation #1.
#2
[2.5V = -1W(I1) + 3W(I2)] x 5 yields 12.5V = -5W(I1) + 15W(I2)
#1
8.5V = 5W(I1) - 1W(I2)
21V = 14W(I2)
Now solve for I2 which yields
I2 = 21V / 14 W = 1.5A
Finally, using this value of I2
in either equation #1 or #2,
solve for I1 which yields
I1 = (8.5V + 1.5V) / 5W = 2A
In summary the values are:
I1 = 2A
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and
I2 = 1.5A
© 2013 The McGraw-Hill Companies Inc. All rights reserved.
6-8
Concept Review
• Two or more sources are required to
use the superposition theorem.
• Superposition theorem requires only
series-parallel rules and procedures
to determine values in a complex circuits.
• Superposition theorem will solve for
all currents and voltages in the circuit.
• Simultaneous equations techniques can
be used to solve loop equations.
Repeat Segment
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6-9
Concept Preview
• Thevenin’s theorem reduces a complex
circuit to an equivalent-circuit voltage
source with a load. (Page 161)
• Norton’s theorem reduces a complex
circuit to an equivalent-circuit current
source with a load. (Page 169
• Norton’s theorem and Thevenin’s
theorem may not solve for all values
of current and voltage. (Page 171)
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6 - 10
Voltage Sources
(Page 160)
Ideal Source
15 V
V
Voltage does
not change
when loaded.
V
15 V
No load
Loaded
Equivalent-Circuit Source
V
15 V
No load
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Voltage changes
when loaded.
V
13 V
Loaded
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6 - 11
Thevenizing A Complex Circuit
20 W
B1
110 V
R
R2TH
2 = 36.7 V
V
30
RTHW= 6.7 W
10 W
(Page 161)
B2
110 V
Replace
Remove
B1 and
R2 and
Be2 calculate
with shorts
VTHand
. calculate RTH.
RTH
6.7 W
-
VTH
36.7 V +
R2
30 W
30 V
1A
Draw the Thevenin equivalent circuit.
Load the circuit with R2.
Calculate the load V and I.
20 W
1A
110 V
8A
10 W
R2
7A
30 W
30 V
110 V
Return to the original circuit and calculate
the other currents.
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6 - 12
Applying Norton’s Theorem to a Complex Circuit
(Page 169)
•Select a resistor to be the load.
•Replace the load with a short.
•Calculate the current through the short. This is IN.
•Remove the short from the load terminals.
•Replace all sources with shorts.
•Calculate the resistance between the load terminals. This is RN.
•Use IN and RN for the Norton circuit.
•Connect the load to the Norton circuit.
•Calculate Vload and Iload.
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6 - 13
Equivalency Of Norton and Thevenin Circuits Page
(172)
+
+
+
RTH
8W V
16 V
VTH
16 V
The open-circuit
voltages are equal.
RTH
8W A
2A
VTH
16 V
The short-circuit
currents are equal.
RTH
8W
VTH
16 V
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IN
2A
RN
8W
IN
2A
RN
8W
RL When equally loaded,
the voltages
24 W
IN
and currents
12 V
2A
0.5 A
are equal.
RN
8W
V
16 V
A
2A
RL
24 W
12 V
0.5 A
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6 - 14
Complex-Circuits Quiz
Which theorem(s) require(s) algebraically
adding the currents for each resistor? ____
Superposition
Which theorem(s) require(s) shorting all
voltage sources at the same time? ____
Thevenin
Norton
Which theorem(s) use(s) an equivalentcircuit current source? ____
Norton
Which theorem(s) use(s) an equivalentcircuit voltage source? ____
Thevenin
Which theorem(s) determine(s) the current
for all resistors in the circuit? ____
The voltage of a(n) ____ voltage source is
independent of the load current.
Solutions by simultaneous equations require
____ equations.
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Superposition
ideal
independent
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6 - 15
Concept Review
• Thevenin’s theorem reduces a complex
circuit to an equivalent-circuit voltage
source with a load.
• Norton’s theorem reduces a complex
circuit to an equivalent-circuit current
source with a load.
• Norton’s theorem and Thevenin’s
theorem may not solve for all values
of current and voltage.
Repeat Segment
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6 - 16
REVIEW
• Superpositon Theorem
• Simultaneous Equations
• Voltage Sources
• Thevenin’s Theorem
• Norton’s Theorem
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