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Engineering 43. 2 nd Order Transient Ckt Parallel, Single Node-Pair. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]. The Problem. Find for the Circuit Below i o (t), and Plot The Single-Node-Pair. Use Parallel- Ckt , KCL Analysis - PowerPoint PPT Presentation
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ENGR-25_Prob_9_3_Solution.ppt1
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Engineering 43
22ndnd Order OrderTransient Transient
CktCktParallel, Parallel, SingleSingle
Node-PairNode-Pair
ENGR-25_Prob_9_3_Solution.ppt2
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
The ProblemThe Problem
Find for the Circuit Below io(t), and Plot
The Single-Node-Pair
t = 0 1A 1 1H 2/5 F 5
iL ( t)i0 ( t)
2Li
• Use Parallel-Ckt, KCL Analysis
• Assume that the SWITCH is a BETTER Short-Circuit than the INDUCTOR for t = 0− Steady-State Analysis
ENGR-25_Prob_9_3_Solution.ppt3
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt4
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt5
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt6
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt7
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt8
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt9
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt10
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ENGR-25_Prob_9_3_Solution.ppt11
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
The Excel PlotThe Excel Plot
See File: ENGR44_Lec_06-1_Last_example_Fall03.xls
0.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
-1 0 1 2 3 4 5 6 7 8 9 10
Cu
rre
nt (
A)
Time (s)
Parallel RLC Circuit Forced Response (P 6.69)
i(t) (A)
file = Engr44_Lec_06-1_Last_example_Fall03..xls
