Equivalent resistance
R1
iS
R3
R2
VS
R4
R5
iS
+
–
VS
equivalent ?
Interested only in iS. Not
interested in details of
individual resistor currents
and voltages.
Same applied VS must
give same resulting iS.
(Same power supplied.)
iS
VS
EE 201
+
–
Req
=
equivalent resistance – 1
Test generator (or test source) method
R1
R3
R2
Req = ?
R4
R5
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.
R1
itest
Vtest
EE 201
+
–
R3
R2
R4
R5
Apply Vtest.
Determine itest.
=
equivalent resistance – 2
Series combination
R1
itest
Apply test source.
Define voltages
Vtest +
–
and currents.
R2
Req = ?
+ vR1 –
R3
iR1
iR2
iR3
+
vR2
–
– vR3 +
By KCL:
=
=
Series connection.
=
By KVL:
=
Series combination:
use Ohm’s law:
=
(
+
+
)=
=
=
=
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=
+
+
equivalent resistance – 3
Parallel combination
itest
Req = ?
R1
R2
By KVL:
=
By KCL:
=
use Ohm’s law:
Apply Vtest.
+
Vtest –
Define
voltages and
currents.
R3
=
+
iR1
+ iR2
vR1
+ iR3
vR3
+
vR3
–
–
–
(Parallel connection)
=
+
=
+
=
+
+
Parallel combination:
+
=
=
=
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=
+
+
equivalent resistance – 4
Series combination: Easy to calculate.
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.
Special cases for parallel combinations:
Two resistors only:
=
+
=
=
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+
+
=
+
(product over sum)
equivalent resistance – 5
More special cases for parallel combinations:
Two resistors, R1 = R2 = R:
=
Two resistors, R2 = 2R1:
=
=
=
One small resistor: R1 << R2, R3, R4,...
=
+
+
+
+ ...
(Equivalent is approximately
equal to smallest.
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equivalent resistance – 6
Combination circuits
R1
itest
Vtest
R3
R2
+
–
R4
R5
Test generator method always works.
Sometime necessary (with dependent sources in circuit).
For purely resistive circuits, there is a faster method – inspection.
R1
R3
R2
Ohm’s
eye
Inspect structure of network.
R4
R5
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.
EE 201
equivalent resistance – 7
R1
R3
R2
1. Recognize and replace the series
branch with the three resistors.
R4
R5
=
+
+
R1
R2
2. Recognize and replace the
parallel combination.
R345
=
R1
=
+
3. We are left with a
simple series pair.
R2345
=
=
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||
Req
+
+
( +
+ )
+( +
+ )
equivalent resistance – 8
Example
R3 50 !
R1
Req = ?
R2
25 !
25 !
R4
100 !
R3
R1
R2
R4
=
+
=
+
=
=
R1
R2
R34
=
=
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+
.
+
Req
11.5 !
equivalent resistance – 9
Example
R3 3 k!
=
(
=
)(
+
)
=
6 k!
R1
3 k!
Req = ?
R2
R4
R5
4 k!
R7
R6
5 k!
=
5 k!
=
(
)(
+
)
= .
0.5 k!
R34
R1
R2
R56
=
R7
R1
R2
R37
+
+ .
=
=
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=
= .
+
+ .
=
Req
+
+
1.28 k!
equivalent resistance – 10
Example
a
R1
R3
680 ! R2
Req = ?
b
1 k!
470 !
R5
R4
470 ! R6
c
c
680 !
330 !
d
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.
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Req
1022 !
equivalent resistance – 11
Example
a
R1
R3
680 ! R2
1 k!
470 !
R5
R4
470 ! R6
c
680 !
330 !
b
Req = ?
d
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.
c 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.
Req
354 !
d
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4. R25 and R6 are in parallel.
Req = R25||R6 = 739.5 Ω || 680 Ω = 354 Ω.
equivalent resistance – 12
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.
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equivalent resistance – 13