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ECE 221Electric Circuit Analysis I

Chapter 12Superposition

Herbert G. Mayer, PSUStatus 2/26/2015

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Syllabus

Source Goal Remove CCS Remove CVS Currents Superimposed Conclusion Exercise

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Source

First sample taken from [1], pages 122-124

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Goal

Linear electrical systems allow superposition: compute separate electrical units for each source, then add them after individual computations

This does not apply when dependent sources are included!

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 that 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 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

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

Removal requires the CVS connectors to be short-circuited; 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’’

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To Remove CCS from C1

A Constant Current Source provides the same, constant current to a circuit

Regardless of the circuit loads With varying loads, a CCS causes a different voltage

drop at that same, 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 Then no current flow, yet a voltage drops is 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 provides the same, constant 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 open Then no voltage drops, 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 = 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 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 = 2 A

i”2 = v3/3 = -12/3 =-4 A

i”3 = (v3-v4)/2 = 6 A

i”4 = v4/4 = -24/4 =-6 A

i1 = i’1+i”1 =15+2 = 17 A

i2 = i’2+i”2 =10+-4 = 6 A

i3 = i’3+i”3 =5+6 = 11 A

i4 = i’4+i”4 =5-6 = -1 A

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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’’ = 4 A - 3 A

i1 = 1 A

i2 = i2’ + i2’’ = 2 A + 2 A

i2 = 4 A

i4 = i4’ - i4’’ = 6 A - 1 A

i4 = 5 A