Circuit with capacitors and inductors

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Homework Help Overview

The discussion revolves around a circuit containing capacitors and inductors, focusing on the behavior of these components in a steady-state DC scenario. Participants explore the implications of long-term connections in the circuit and the resulting current and voltage characteristics.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the removal of inductors and capacitors in steady-state analysis, questioning the validity of current division with the remaining resistors and current sources. There is also inquiry into the effect of a specific current source on the circuit.

Discussion Status

The conversation is active, with participants providing insights into circuit simplifications and questioning the roles of various current sources. Some guidance has been offered regarding current division and the behavior of ideal sources, but no consensus has been reached on all points raised.

Contextual Notes

Participants are operating under the assumption that the circuit has been connected for a long time, leading to steady-state conditions. There are discussions about the implications of this assumption on the analysis of the circuit components.

eurekameh
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The question states that the circuit has been connected for a very long time. I'm assuming that then, the current will remain constant and thus the voltage for the inductor is vL = Ldi/dt = 0. Thus, using nodal analysis, I was able to solve (a) by eliminating all of the inductors from the circuit. For (b), however, there is a capacitor for which I do not know how to handle, although I would guess that a mesh analysis would be involved from the looks of the current sources.
 
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When capacitors reach steady state in a DC circuit they reach a constant voltage and pass no current. You can remove capacitors and short out inductors...
 
I have removed the capacitors and shorted out the inductors. All that remains are the resistors. Would it be valid to sum the current sources and put the 12 A source to the left of everything and use current division?
 
eurekameh said:
I have removed the capacitors and shorted out the inductors. All that remains are the resistors. Would it be valid to sum the current sources and put the 12 A source to the left of everything and use current division?

No, you can't sum the current sources as they are not in parallel. But you don't need to: the 10A supply is the only supply to the 2k and 5k resistors which are in parallel (now that the inductor is shorted). Apply current division.
 
Why doesn't the 2 A source have no added effect? I would assume from intuition that it would contribute to the current i,x going through the 5 k,ohm resistor.
 
Last edited:
eurekameh said:
Why doesn't the 2 A source have no added effect? I would assume from intuition that it would contribute to the current i,x going through the 5 k,ohm resistor.

The 2A source can't have any effect on those components because the 10A current source will pass 10 amps, and ONLY 10 amps no matter what. An ideal source cannot do anything but exactly what it's specified to do. If an ideal source is putting 10A into a node, you better believe that all 10A are going somewhere! :smile:

If you were to draw a current arrow on the bottom rail that is below the 10A source, what current value would you label it with? Where does that current come from ?
 
10 A going to the left?
 
eurekameh said:
10 A going to the left?

Yes. Where does that current come from (through which components must it flow)?
 

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