ECE 221
Electric Circuit Analysis I
Chapter 12
Superposition
Herbert G. Mayer, PSU
Status 11/26/2014
For use at Changchun University of Technology CCUT
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Syllabus
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Source
Goal
Remove CCS
Remove CVS
Currents Added Up
Conclusion
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Source
Sample taken from [1], pages 122-124
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Goal
 Linear systems allow superposition, the
separate computation of electrical units for
each separate constant source, all to be
added up after individual computations
 This does not apply when dependent
sources are included! Those can not be
removed to simplify computation
 Goal of the stepwise removal and stepwise
computation is the simplification of overall
problem
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Goal
 Linear systems means, that all currents are
direct function of voltages
 Or vice versa: Voltages are direct functions
of current
 But not to a power different than 1
 Finally, when all separate sources have been
considered, the final result can be computed
by adding all partial results
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Goal
 Goal is to compute currents i1, i2, i3, and i4
 Step1: When first the CCS is removed, we compute 4
sub-currents, i’1, i’2, i’3, and i’4, solely created by
the remaining CVS
 Removal of a CCS requires the connectors to be left
open for CCS; since the current through the CCS is
created by solely that CCS, not by other sources
 Step2: The CCS is added again, and the CVS is
removed
 Removal requires the CVS connector to be shortcircuited; since voltage at its terminals is solely
created by that CVS
 We then compute 4 sub-currents, i’’1, i’’2, i’’3, and
i’’4, solely created by the remaining CCS
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Original Circuit
7
Removed CCS
8
Removed CCS
 Once we know the Node-Voltage across the 3 Ω
resistor, we can easily compute all partial currents i’
 We name this voltage v1
 We compute v1 via 2 methods: first the Voltage
Division, then using the Node-Voltage method
 v1 drops across the 3 Ω resistor, but also across the
series of the 2 Ω + 4 Ω resistors
 The equivalent resistance Req of 3 Ω parallel to the
series of 2 Ω + 4 Ω is: 2 Ω
 Try it: 3 // ( 2+4 ) = 3 // 6 = 3*6 / ( 3+6) = 18 / 9 = 2 Ω
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Removed CCS, v1 Via Voltage Division
Req =
2Ω
v1
= 120 * 2 / ( 2 + 6 )
v1
= 120 * 1/4 = 30 V
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Removed CCS, v1 Via Node-Voltage
v1/3 + (v1-120)/6 + v1/(2+4) = 0 // *6
2*v1 + v1 + v1 = 120
4*v1
= 120
v1
= 120/4
v1
= 30 V
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Removed CCS, Compute Currents i’
i’1
= (120 - v1) / 6
i’1
= 90 / 6
i’1
= 15 A
i’2
= v1 / 3
i’2
= 10 A
i’3
= v1 / 6
i’3
= 5 A
i’4
= i’3
i’4
= 5 A
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Removed CVS
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Removed CVS
 Now we compute the node-voltages in the
two nodes 1 and 2 via two methods:
 First via Ohm’s Law
 Then using the Node-Voltage Method
 Both nodes have 3 currents, which later we
compute using the node-voltage method
 We name the voltage drop at the 3 Ω
resistor v3
 And the voltage drop at the 4 Ω resistor v4
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Removed CVS, Using Ohm’s Law
At node 1:
6 || 3 = 2 Ω
At node 2:
2 + 2= 4 Ω
4 || 4 = 2 Ω
yields:
v4
= -2 * 12 = -24 V
v3
= v4 / 2 = -12 V
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Removed CVS, Using Node-Voltage
At node 1:
v3/3 + (v3-v4)/2 + v3/6 = 0
At node 2:
v4/4 + (v4-v3)/2 + 12
= 0
v4 * ( 1/4 + 1/2 )
= -12 + v3/2
v4
= 2*v3/3 - 16
yields:
v3
= -12 V
v4
= -24 V
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Currents Added Up
i”1
= -v3/6= 12/6
= 2 A
i”2
= v3/3
i”3
= (v3-v4)/2
= 6 A
i”4
= v4/4
= -6 A
i1
= i’1+i”1 = 15+2
i2
= i’2+i”2 = 10+-4= 6 A
i3
= i’3+i”3 = 5+6
= 11 A
i1
= i’4+i”4 = 5-6
= -1 A
= -12/3= -4 A
= -24/4
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= 17 A
Conclusion
 Superposition allows us to break a complex
problem into multiple smaller problems that
are simpler each
 Applicable only in linear systems
 Resistive circuits are linear
 The same principle also applies to circuits
with capacitances and inductances
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