Download presentation

Presentation is loading. Please wait.

Published byCarli Adgate Modified about 1 year ago

1
1 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

2
2 Syllabus Source Source Goal Goal Remove CCS Remove CCS Remove CVS Remove CVS Currents Added Up Currents Added Up Conclusion Conclusion

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

4
4 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

5
5 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

6
6 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 short- circuited; 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

7
7 Original Circuit

8
8 Removed CCS

9
9 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 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

10
10 Removed CCS, v1 Via Voltage Division R eq =2Ω v1=120 * 2 / ( 2 + 6 ) v1=120 * 1/4=30 V

11
11 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

12
12 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

13
13 Removed CVS

14
14 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

15
15 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

16
16 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

17
17 Currents Added Up 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 i1=i’4+i”4=5-6=-1A

18
18 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

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google