Transcript Superposition Method

ECE 221
Electric Circuit Analysis I
Chapter 12
Superposition
Herbert G. Mayer, PSU
Status 10/25/2015
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Syllabus
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Source
Goal
Remove CCS
Remove CVS
Currents Superimposed
Conclusion
Exercise
2
Source
First sample taken from [1], pages 122-124
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Goal
 Linear electrical systems allow superposition
of multiple sources: that is computing
separate electrical units for each source, then
adding them after individual computations
 This does not apply with dependent sources!
 Goal of the stepwise removal is the reduction
of complex to simpler problems, followed by
the addition of partial results
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Goal
 Linear system means, that all currents in
such a system are linear functions of
voltages
 Or vice versa: Voltages are linear
functions of currents
 But never functions of a different power
than 1, meaning a different exponent!
 Finally, when all separate sources have
been considered and computed, the final
result can simply be added
(superimposed) from all partial results
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Goal With Circuit C1
 Goal in Circuit C1 below 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
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Removal of a CCS requires the connectors to be left
open at former CCS; since the current through the CCS
is created by solely that CCS
 Step2: The CCS is added again, and the CVS is
removed
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Removal requires the CVS connectors to be shortcircuited; since voltage at its terminals is solely created
by that CVS
Then compute 4 sub-currents, i’’1, i’’2, i’’3, and i’’4, solely
created by the remaining CCS
 In the end, superimpose all i’ and i’’, that means,
add them all up
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To Remove CCS from C1
 A Constant Current Source (CCS) provides invariant
current to a circuit
 Regardless of the circuit loads
 With varying loads, a CCS causes a different voltage
drop with its constant current
 Regardless of other constant power sources, each
causing voltage drops or currents or both
 Removing a CCS means, leaving the 2 CCS
terminals open
 With open terminals no current flows, yet voltage
drop is certainly possible at the now open terminals,
regardless of other sources and circuit elements
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To Remove CVS from C1
 A Constant Voltage Source (CVS) provides invariant
voltage to a circuit
 Regardless of the circuit loads
 With varying loads, a CVS causes a different current
at that same, constant voltage
 Regardless of other constant power sources, each
causing voltage drops or currents or both
 Removing a CVS means, short-circuiting the 2 CVS
terminals
 Then there is no voltage drop, yet a current flow is
possible at the now connected terminals, regardless
of other sources and circuit elements
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Original Circuit C1
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Removed CCS from C1
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Removed CCS from C1
 Once we know the Node Voltage across the 3 Ω
resistor, we can easily compute all partial currents i’
 We name this voltage along the 3 Ω v1
 We compute v1 via 2 methods: first using Ohms’
Law and Voltage Division; secondly 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
v1
= 120 * 2 / ( 2 + 6 )
= 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 from C1
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Removed CVS from C1
 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 from C1, Use Ohm’s
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 from C1, Use 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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Partial Currents Superimposed
i”1
= -v3/6= 12/6
=
2A
i”2
= v3/3
i”3
= (v3-v4)/2
=
i”4
= v4/4
= -6 A
i1
= i’1+i”1 = 15+2
i2
= i’2+i”2 = 10+-4=
i3
= i’3+i”3 = 5+6
= 11 A
i4
= i’4+i”4 = 5-6
= -1 A
= -12/3= -4 A
= -24/4
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6A
= 17 A
6A
Conclusion
 Superposition allows breaking 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
 But not to circuits that contain dependent
power sources, Op Amps, etc.
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Exercise
 Use superposition in the circuit below to
compute the currents in the 3 resistors
 First short-circuit only the 7 V CVS, and
compute currents i1’, i2’, and i4’
 There is no i3 
 Then short-circuit only the 28 V CVS, and
compute currents i1’’, i2’’, and i4’’
 Finally superimpose the two reduced circuits
to add up i1, i2, and i4
 But consider the opposite current directions
for: i1’ vs. i1’’, and for i4’ vs. i4’’
 Not used for i2 !! Those 2 currents add up!
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Exercise: Original Circuit
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Exercise: Remove 7 V CVS
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Exercise: Remove 7 V CVS
 With the 7 V CVS removed, the equivalent
resistance of the 3 resistors is:
 1 Ω || 2 Ω + 4 Ω = 2/3 + 4 = 14 / 3 Ω
 Hence i4’ = 28 / ( 14 / 3 ) = 6 A
 Current-division of 6 A in the 1 Ω and 2 Ω
results in i2’ = 2 A, and i1’ = 4 A
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Exercise: Remove 28 V CVS
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Exercise: Remove 28 V CVS
 With the 28 V CVS removed, the equivalent
resistance of the 3 resistors is:
 1 Ω + 2 Ω || 4 Ω = 4/3 + 1 = 7 / 3 Ω
 Hence i1’’ = 7 / ( 3 / 3 ) = 3 A
 Current-division of 3 A in the 2 Ω and 4 Ω
results in i4’’ = 1 A, and i2’’ = 2 A
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Exercise: Final Result
i1 = i1’ - i1’’
i1 = 1 A
= 4A-3A
i2 = i2’ + i2’’
i2 = 4 A
= 2A+2A
i4 = i4’ - i4’’
i4 = 5 A
= 6A-1A
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