- #61
builder_user
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gneill said:
l=0.012
What is the instantaneous value of UL after commutation?
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Homework Help Overview
The discussion revolves around finding the instantaneous value of UL after commutation in a circuit involving inductors and resistors. The participants are exploring the implications of commutation on circuit behavior, particularly focusing on the inductor's response during the transition between states.
Discussion Character
- Exploratory, Assumption checking, Problem interpretation
Approaches and Questions Raised
- Participants discuss the conditions before and after commutation, questioning the behavior of current and voltage across components like resistors and inductors. There are inquiries about the definitions and differences between classical and operator methods for solving the problem.
Discussion Status
The discussion is ongoing with various interpretations being explored. Some participants have offered insights into the behavior of inductors during commutation and the concept of Norton equivalents, while others are seeking clarification on the methods and the specific calculations needed for the problem.
Contextual Notes
There are mentions of potential confusion regarding the definitions of classical and operator methods, as well as the need to derive equivalent circuits for different states of the circuit. Some participants express uncertainty about the initial conditions and the role of the inductor before and after the switch is closed.
l=0.012
- #62
gneill
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builder_user said:
If you say so. In your first post you had:
K1=1.2
L = 10*K1
You did not indicate units (for anything!) so I assumed Ohms for resistances and Henries for inductance.
- #63
- #64
builder_user
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gneill said:
I forgot.In every task I have millihenries and so I've not written it there.
- #65
gneill
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- #66
gneill
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builder_user said:
Okay. Perhaps in future you should just state the component values. All the K multipliers and assumed units is confusing.
- #67
builder_user
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gneill said:
Thevenin equivalent after commutation?Or at the moment of commutation?One of the equivalents that was found before?Or not found in this topic?
- #68
gneill
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builder_user said:
After the switch closes, so yes, after commutation.
The circuit at the moment the switch closes remains the same thereafter.
It's only the initial conditions (the existing current in the inductor) that makes the moment of commutation special.
- #69
builder_user
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How to do inverse Laplace transform in MathCAD?
- #70
gneill
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builder_user said:
Select the 's' variable and then Symbolics --> Transform --> Inverse Laplace.
- #71
builder_user
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Thanks.
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