Equivalent resistance - Gary Tuttle's ISU web...
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EE 201 equivalent resistance – 1 Equivalent resistance V S R 1 i S R 3 R 4 R 5 R 2 Interested only in i S . Not interested in details of individual resistor currents and voltages. + – V S i S equivalent ? Same applied V S must give same resulting i S . (Same power supplied.) + – V S R eq i S 5 HT = 9 6 L 6
Transcript of Equivalent resistance - Gary Tuttle's ISU web...
EE 201 equivalent resistance – 1
Equivalent resistance
VS
R1iSR3
R4R5
R2Interested only in iS. Not interested in details of individual resistor currents and voltages.
+–VS
iS
equivalent ?
Same applied VS must give same resulting iS. (Same power supplied.)
+–VS Req
iS
5HT =96L6
EE 201 equivalent resistance – 2
Test generator (or test source) method
R1 R3
R4R5
R2Req = ?
Equivalent resistance must be defined between 2 nodes of the network. A different pair of nodes gives different Req.
Apply a test generator between the two nodes of interest.
Vtest
R1itestR3
R4R5
R2+–
Apply Vtest. Determine itest.
5HT =9WHVWLWHVW
EE 201 equivalent resistance – 3
Series combination
R1
R2R3
Req = ? Vtest
itest
+–
iR1 iR2iR3
+
+
+
–
–
–
vR1
vR2
vR3
Apply test source. Define voltages and currents.
By KCL: LWHVW = L5� = L5� = L5� Series connection.
By KVL: 9WHVW � Y5� � Y5� � Y5� = �
use Ohm’s law: 9WHVW � L5�5� � L5�5� � L5�5� = �
9WHVW � LWHVW (5� + 5� + 5�) = �
5HT =9WHVWLWHVW
= 5� + 5� + 5�
Series combination:
5HT =1�
P=�5P
EE 201 equivalent resistance – 4
Parallel combination
R1 R2 R3Req = ?
Apply Vtest. Define voltages and currents.
–+Vtest
itest
–
vR1+iR1
–
vR3+iR2
–
vR3+iR3
By KVL: 9WHVW = Y5� = Y5� = Y5� (Parallel connection)
By KCL: LWHVW = L5� + L5� + L5�
use Ohm’s law: LWHVW =Y5�5�
+Y5�5�
+Y5�5�
LWHVW =YWHVW5�
+YWHVW5�
+YWHVW5�
�5HT
=LWHVW9WHVW
=�5�
+�5�
+�5�
Parallel combination:
�5HT
=1�
P=�
�5P
EE 201 equivalent resistance – 5
Series combination: Easy to calculate.
Special cases for parallel combinations:
Series: equivalent is always bigger than any resistor in the string.
Req > Rm.
Parallel: equivalent is always smaller than any single resistor the parallel branches.
Req < Rm.
Two resistors only: �5HT
=�5�
+�5�
5HT =5�5�
5� + 5�(product over sum)
=5�5�5�
+5�5�5�
=5� + 5�5�5�
EE 201 equivalent resistance – 6
Two resistors, R1 = R2 = R: 5HT =5��5 =
5�
Two resistors, R2 = 2R1: 5HT =�5���5 =
��5�
One small resistor: R1 << R2, R3, R4,...
5HT � 5�
More special cases for parallel combinations:
�5HT
=�5�
+�5�
+�5�
+�5�
+ ...
� �5�
(Equivalent is approximately equal to smallest.
EE 201 equivalent resistance – 7
Vtest
R1itestR3
R4R5
R2+–
Combination circuits
Test generator method always works.
Sometime necessary (with dependent sources in circuit).
For purely resistive circuits, there is a faster method – inspection.
R1 R3
R4R5
R2
Inspect structure of network.
Use parallel & series combinations to sequentially reduce pieces of the network to single resistances.
With practice, you will be able to find Req in one step.
Ohm’s eye
EE 201 equivalent resistance – 8
R1 R3
R4R5
R21. Recognize and replace the series branch with the three resistors.
5��� = 5� + 5� + 5�
R1
R345R2
2. Recognize and replace the parallel combination.
5���� = 5�||5��� =5�5���5� + 5���
R1
R2345
3. We are left with a simple series pair.
5HT = 5� + 5����
= 5� +5� (5� + 5� + 5�)5� + (5� + 5� + 5�)
Req
EE 201 equivalent resistance – 9
Example
R1 R2 R4
R3
Req = ? 25 ! 25 !
50 !
100 !
R1 R2 R4
R3
5�� = 5� + 5�
R1 R2 R345HT = 5��5��5��
= ��ż + ���ż = ���ż
�5HT
=���ż +
���ż +
����ż
5HT = ��.�ż
Req11.5 !
EE 201 equivalent resistance – 10
Example
Req = ?
=(�Nż) (�Nż)
�Nż + �Nż = �Nż5�� = 5��5�
R1
R3
R4 R5 R6
R7
R23 k! 4 k!
3 k!
6 k!
5 k! 5 k!
0.5 k!
5�� = 5��5�
=(�Nż) (�Nż)
�Nż + �Nż = �.�Nż
R1
R34
R56
R7
R2 5�� = 5�� + 5�� + 5�= �Nż + �.�Nż + �.�Nż = �Nż
R1 R37R25HT = 5��5��5���5HT
=�
�Nż +�
�Nż +�
�Nż5HT = �.��Nż
Req1.28 k!
EE 201 equivalent resistance – 11
Example
Req = ?
R1
R2
R3
R4
R5
R6a
b
c c
d
680 !
470 !
1 k!
330 !
470 !680 !
1. R5 and R6 are in series. R56 = R5 + R6 = 1150 Ω.
2. R4 is in parallel with R56.R46 = R4||R6 = 256 Ω.
3. R3 is in series with R46.R36 = R2 + R46 = 1000 Ω + 256 Ω = 1256 Ω.
4. R2 is in parallel with R36.R26 = R2||R36 = 470 Ω || 1256 Ω = 342 Ω.
5. R1 is in parallel with R26.Req = R1 + R26 = 680 Ω + 342 Ω = 1022 Ω.
Req1022 !
EE 201 equivalent resistance – 12
Example
Req = ?
R1
R2
R3
R4
R5
R6a
b
c
d
680 !
470 !
1 k!
330 !
470 !680 !
Find the Req referenced between the nodes c and d. Note that in this case R1 is dangling (unconnected). No current will flow there – it has no effect on the rest of the circuit, and we can ignore it.
1. R2 and R3 are in series. R23 = R2 + R3 = 470 Ω + 1000 Ω = 1470 Ω.
2. R23 and R4 are in parallel. R24 = R23||R4 = 1470 Ω || 330 Ω = 269.5 Ω.
3. R24 and R5 are in series. R25 = R24 + R5 = 269.5 Ω + 470 Ω = 739.5 Ω.
4. R25 and R6 are in parallel. Req = R25||R6 = 739.5 Ω || 680 Ω = 354 Ω.
Reqc
d354 !
EE 201 equivalent resistance – 13
To study:
1. Work at least a dozen of the equivalent resistance practice problems on the web site, making sure you can get the correct answer each time.
2. Sketch out your own crazy resistor network and see if you can calculate the equivalent resistance.
3. “The equivalent resistance of a parallel combination is always less than the value of any of the individual resistors.” Make sure that you understand this statement and why it is true.
4. Use a test generator alone with KCL and KVL to work any of the examples shown in this lecture. Show that you obtain the same result.
5. As noted, the test generator could be a current source. Then the goal would be to find the corresponding voltage. Re-work the series and parallel cases using a test current generator. Show that you obtain the same result.
6. Work through the first circuit (bottom of slide 2) using the test generator method. Show that you obtain the same equivalent resistance as the “inspection” method.
