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1 ECE 221 Electric Circuit Analysis I Chapter 12 Superposition Herbert G. Mayer, PSU Status 2/26/2015

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2 Syllabus Source Source Goal Goal Remove CCS Remove CCS Remove CVS Remove CVS Currents Superimposed Currents Superimposed Conclusion Conclusion Exercise Exercise

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

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4 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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5 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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6 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’’ In the end, superimpose all i’ and i’’

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7 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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8 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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9 Original Circuit C1

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10 Removed CCS from C1

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11 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 R eq 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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12 Removed CCS, v1 Via Voltage Division R eq =2Ω v1=120 * 2 / ( 2 + 6 ) v1=120 * 1/4=30 V

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13 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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14 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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15 Removed CVS from C1

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16 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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17 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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18 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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19 Partial Currents Superimposed i”1=-v3/6=12/6= 2A i”2=v3/3=-12/3=-4A i”3=(v3-v4)/2= 6A i”4=v4/4= -24/4=-6A i1=i’1+i”1=15+2=17A i2=i’2+i”2=10+-4= 6A i3=i’3+i”3=5+6=11A i4=i’4+i”4=5-6=-1A

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20 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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21 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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22 Exercise: Original Circuit

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23 Exercise: Remove 7 V CVS

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24 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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25 Exercise: Remove 28 V CVS

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26 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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27 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

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