Analyzing DC Steady State in an RLC Circuit with a Closed and Opened Switch

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Discussion Overview

The discussion revolves around analyzing the behavior of an RLC circuit in a DC steady state, specifically focusing on the values of the capacitor voltage vC(t) and the inductor current iL(t) immediately after a switch is opened. The context includes homework-related queries and circuit analysis techniques.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant states that in DC steady state, the capacitor acts as an open circuit and the inductor as a short circuit, leading to a calculated current of 1A and a capacitor voltage of 10V.
  • The same participant expresses confusion regarding the inductor current, questioning whether the 2A current source affects it and how to reconcile their simulation results with theoretical calculations.
  • Another participant suggests using Thevenin's equivalent to clarify the circuit behavior.
  • A later reply indicates an understanding that the 2A source should be added to the existing 1A from the 20V source, implying that all sources must be considered in the analysis.
  • Further advice is given to consider superposition to analyze the contributions of each source individually.

Areas of Agreement / Disagreement

Participants generally agree on the importance of considering all sources in the circuit analysis, but there remains uncertainty regarding the specific values of the inductor current and how to approach the problem effectively.

Contextual Notes

There are unresolved aspects regarding the assumptions made about the circuit components and the impact of the current sources on the inductor's behavior. The discussion reflects varying interpretations of circuit analysis techniques.

Who May Find This Useful

Students and practitioners interested in circuit analysis, particularly those dealing with RLC circuits and DC steady state conditions.

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Homework Statement



In the Figure Q1 (c), the switch S was closed for very long time before it is opened at time t = 0. Find the value of vC(t) and iL(t) at t = 0+ c L i.e. immediately after the switch was opened.

2ewzh1i.jpg

Homework Equations



V = IR

The Attempt at a Solution



From DC steady state, we know that C acts as an open circuit while L acts as a short wire, hence, we will have:

Current in circuit = 20 / (10+10) = 1A
vC(t) is thus 20 - (10*1) = 10V.

However, for iL(t), I am a bit confused, because since this is a short wire, will the 2A current affect the current in this wire? This is because my answer is 1A, however, when I simulate this circuit in LTSpice, it gave me an answer of 3A, which is a bit puzzling. The direction of current would be different, and thus, even if the 2A current source affects the current in the inductor (which is a short), it should be 1-2 = -1A right?

Thanks for any assistance rendered! :D
 
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Try replacing the 2A source and 50Omh resistor with their Thevinin equivalent - see if that clears it up for you.
 
Simon Bridge said:
Try replacing the 2A source and 50Omh resistor with their Thevinin equivalent - see if that clears it up for you.

otngo7.jpg


I think I know why now. I must add 2A to the 1A already present from the 20V source. Thanks! So that means, unless the right of the inductor are purely resistors, I cannot just ignore the right branches even though the inductor is shorting, right?
 
In general you can't ignore any sources that might contribute to the voltage or current you're interested in. When in doubt, try looking at the circuit via superposition (consider the contributions of each source individually).
 
gneill said:
In general you can't ignore any sources that might contribute to the voltage or current you're interested in. When in doubt, try looking at the circuit via superposition (consider the contributions of each source individually).

I see. I will keep that in mind. Thank you! :D
 

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