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

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The discussion focuses on analyzing an RLC circuit's behavior immediately after a switch is opened. 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 voltage across the capacitor of 10V. Confusion arises regarding the inductor's current, with a simulation yielding 3A instead of the expected 1A. It is clarified that the current from other sources must be considered, and the total current through the inductor should include contributions from all sources. The importance of using superposition to analyze circuit contributions is emphasized for accurate results.
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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.

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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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