questions question 1 animation: applying … question 1 animation: applying th´evenin’s theorem...
TRANSCRIPT
![Page 1: Questions Question 1 Animation: Applying … Question 1 Animation: Applying Th´evenin’s theorem This question consists of a series of images (one per page) ... We may use Thevenin’s](https://reader036.vdocuments.site/reader036/viewer/2022062504/5ae653e67f8b9a3d3b8d0e9a/html5/thumbnails/1.jpg)
Questions
Question 1
Animation: Applying Thevenin’s theorem
This question consists of a series of images (one per page) that form an animation. Flip the pages withyour fingers to view this animation (or click on the ”next” button on your viewer) frame-by-frame.
The following animation shows the steps involved in ”Thevenizing” a circuit.
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RloadR1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
2
![Page 3: Questions Question 1 Animation: Applying … Question 1 Animation: Applying Th´evenin’s theorem This question consists of a series of images (one per page) ... We may use Thevenin’s](https://reader036.vdocuments.site/reader036/viewer/2022062504/5ae653e67f8b9a3d3b8d0e9a/html5/thumbnails/3.jpg)
RloadR1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
This is our original circuit:
3
![Page 4: Questions Question 1 Animation: Applying … Question 1 Animation: Applying Th´evenin’s theorem This question consists of a series of images (one per page) ... We may use Thevenin’s](https://reader036.vdocuments.site/reader036/viewer/2022062504/5ae653e67f8b9a3d3b8d0e9a/html5/thumbnails/4.jpg)
RloadR1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
We may use Thevenin’s theoremto simplify this portion of the circuit . . .
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Rload
We may use Thevenin’s theoremto simplify this portion of the circuit . . .
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
5
![Page 6: Questions Question 1 Animation: Applying … Question 1 Animation: Applying Th´evenin’s theorem This question consists of a series of images (one per page) ... We may use Thevenin’s](https://reader036.vdocuments.site/reader036/viewer/2022062504/5ae653e67f8b9a3d3b8d0e9a/html5/thumbnails/6.jpg)
Rload
We may use Thevenin’s theoremto simplify this portion of the circuit . . .
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
6
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Rload
We may use Thevenin’s theoremto simplify this portion of the circuit . . .
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
7
![Page 8: Questions Question 1 Animation: Applying … Question 1 Animation: Applying Th´evenin’s theorem This question consists of a series of images (one per page) ... We may use Thevenin’s](https://reader036.vdocuments.site/reader036/viewer/2022062504/5ae653e67f8b9a3d3b8d0e9a/html5/thumbnails/8.jpg)
RloadR1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
8
![Page 9: Questions Question 1 Animation: Applying … Question 1 Animation: Applying Th´evenin’s theorem This question consists of a series of images (one per page) ... We may use Thevenin’s](https://reader036.vdocuments.site/reader036/viewer/2022062504/5ae653e67f8b9a3d3b8d0e9a/html5/thumbnails/9.jpg)
RloadR1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
To this Thevenin equivalent circuit . . .
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RloadR1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
To this Thevenin equivalent circuit . . .
VTH
RTH
10
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RloadR1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
VTH
RTH
. . . to which we may attachthe same load and analyze.
11
![Page 12: Questions Question 1 Animation: Applying … Question 1 Animation: Applying Th´evenin’s theorem This question consists of a series of images (one per page) ... We may use Thevenin’s](https://reader036.vdocuments.site/reader036/viewer/2022062504/5ae653e67f8b9a3d3b8d0e9a/html5/thumbnails/12.jpg)
RloadR1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
VTH
RTH
. . . to which we may attachthe same load and analyze.
Rload
12
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RloadR1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
VTH
RTH
. . . to which we may attachthe same load and analyze.
Rload
13
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RloadR1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
VTH
RTH
. . . to which we may attachthe same load and analyze.
Rload
14
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RloadR1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
VTH
RTH
. . . to which we may attachthe same load and analyze.
Rload
15
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RloadR1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
VTH
RTH
Rload
16
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RloadR1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
VTH
RTH
Rload
the load resistor.First we disconnect
17
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RloadR1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
VTH
RTH
the load resistor.First we disconnect
18
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Rload
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
VTH
RTH
the load resistor.First we disconnect
19
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Rload
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
VTH
RTH
the load resistor.First we disconnect
20
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Rload
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
VTH
RTH
21
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Rload
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
VTH
RTH
Then we calculate howmuch voltage appearsacross the open load
terminals.
22
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Rload
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
VTH
RTH
Then we calculate howmuch voltage appearsacross the open load
terminals.+
V-
23
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Rload
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
VTH
RTH
Then we calculate howmuch voltage appearsacross the open load
terminals.+
V-
24
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Rload
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
VTH
RTH
Then we calculate howmuch voltage appearsacross the open load
terminals.+
V- (18 volts)
10 kΩ
(14 kΩ + 12 kΩ + 10 kΩ)
25
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Rload
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
VTH
RTH
Then we calculate howmuch voltage appearsacross the open load
terminals.+
V- (18 volts)
10 kΩ
(14 kΩ + 12 kΩ + 10 kΩ)
= 5 volts
26
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Rload
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
VTH
RTH
+V
-5 V
27
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Rload
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
VTH
RTH
+V
-5 V
This voltage becomesour Thevenin sourcevoltage . . .
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Rload
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
RTH
+V
-5 V
This voltage becomesour Thevenin sourcevoltage . . .
5 V
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Rload
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
RTH
+V
-5 V
This voltage becomesour Thevenin sourcevoltage . . .
5 V. . . in the Thevenin equivalent circuit.
30
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Rload
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
RTH
5 V
31
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Rload
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
RTH
5 V
Now we replace each sourcein the original circuit with itsown internal resistance.
32
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Rload
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
RTH
5 V
Now we replace each sourcein the original circuit with itsown internal resistance.
For voltage sources, thismeans a short-circuit.
33
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Rload
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
RTH
5 V
Now we replace each sourcein the original circuit with itsown internal resistance.
For voltage sources, thismeans a short-circuit.
34
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Rload
R1
R2
R3
18 V
10 kΩ
12 kΩ
14 kΩ
RTH
5 V
Now we replace each sourcein the original circuit with itsown internal resistance.
For voltage sources, thismeans a short-circuit.
35
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
RTH
5 V
Now we replace each sourcein the original circuit with itsown internal resistance.
For voltage sources, thismeans a short-circuit.
36
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
RTH
5 V
37
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
RTH
5 V
resistance across the openload terminals.
. . . and we calculate
38
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
RTH
5 V
resistance across the openload terminals.Ω
. . . and we calculate
39
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
RTH
5 V
resistance across the openload terminals.Ω
. . . and we calculate
40
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
RTH
5 V
resistance across the openload terminals.Ω
. . . and we calculate
(14 kΩ + 12 kΩ) // 10 kΩ
41
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
RTH
5 V
resistance across the openload terminals.Ω
. . . and we calculate
(14 kΩ + 12 kΩ) // 10 kΩ= 7.22 kΩ
42
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
RTH
5 V
Ω 7.22 kΩ
43
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
RTH
5 V
Ω 7.22 kΩ
This resistance becomesour Thevenin source
resistance . . .
44
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
Ω 7.22 kΩ
This resistance becomesour Thevenin source
resistance . . .
7.22 kΩ
45
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
Ω 7.22 kΩ
This resistance becomesour Thevenin source
resistance . . .
7.22 kΩ
. . . in the Thevenin equivalent circuit.
46
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ
18 V
47
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ
Now that we have an equivalent circuit to
work with, we may insertthe load there to see what
happens!
18 V
48
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ
Now that we have an equivalent circuit to
work with, we may insertthe load there to see what
happens!
18 V
49
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ
Now that we have an equivalent circuit to
work with, we may insertthe load there to see what
happens!
18 V
50
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ
Now that we have an equivalent circuit to
work with, we may insertthe load there to see what
happens!
18 V
51
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ
Now that we have an equivalent circuit to
work with, we may insertthe load there to see what
happens!
18 V
52
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ
Now that we have an equivalent circuit to
work with, we may insertthe load there to see what
happens!
Calculate:
18 V
53
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ
Now that we have an equivalent circuit to
work with, we may insertthe load there to see what
happens!
Calculate:
Vload
18 V
54
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ
Now that we have an equivalent circuit to
work with, we may insertthe load there to see what
happens!
Calculate:
Vload
Iload
18 V
55
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ
Now that we have an equivalent circuit to
work with, we may insertthe load there to see what
happens!
Calculate:
Vload
Iload
Pload
18 V
56
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ
Calculate:
Vload
Iload
Pload
18 V
57
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ
Calculate:
Vload
Iload
Pload
18 VThese load calculations willreflect what happens in theoriginal circuit!
58
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ
Calculate:
Vload
Iload
Pload
18 VThese load calculations willreflect what happens in theoriginal circuit!
Rload
59
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ
Calculate:
Vload
Iload
Pload
18 VThese load calculations willreflect what happens in theoriginal circuit!
RloadVload
(same)
60
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ
Calculate:
Vload
Iload
Pload
18 VThese load calculations willreflect what happens in theoriginal circuit!
RloadVload
(same)
Iload
61
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ
Calculate:
Vload
Iload
Pload
18 VThese load calculations willreflect what happens in theoriginal circuit!
RloadVload
(same)
Iload
Pload
62
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ
Calculate:
Vload
Iload
Pload
18 VThese load calculations willreflect what happens in theoriginal circuit!
RloadVload
Iload
Pload
63
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ Vload
Iload
Pload
18 V
Rload
Vload
Iload
Pload
64
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ Vload
Iload
Pload
18 V
Rload
Vload
Iload
Pload
65
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Rload
R1
R2
R3 10 kΩ
12 kΩ
14 kΩ
5 V
7.22 kΩ Vload
Iload
Pload
18 V
Rload
Vload
Iload
Pload
The load cannot ‘‘tell’’any difference between the
Thevenin equivalent circuit.original circuit and the
file 03261
66
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Answers
Answer 1
Nothing to note here.
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Notes
Notes 1
The purpose of this animation is to let students see how Thevenin’s theorem may be applied to the
simplification of a resistor network.
68