Equivalent Inductance: Redrawing Circuits and Solving for Leq

Click For Summary

Discussion Overview

The discussion revolves around calculating the equivalent inductance (Leq) of a circuit involving inductors in series and parallel configurations. Participants are addressing a homework problem that includes redrawing the circuit based on the position of a switch and determining the impact on the equivalent inductance.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant expresses uncertainty about redrawing the circuit correctly when the switch is open and questions the implications for the equivalent inductance.
  • Another participant confirms the correctness of the redrawn circuit for part A and suggests simplifying the circuit by combining parallel inductors early in the process.
  • A third participant proposes a method for redrawing the circuit for part B, suggesting to visualize it with fewer inductors and to simplify in manageable steps.
  • One participant seeks clarification on the assumption that moving an empty branch into the middle of the circuit is valid due to the parallel configuration of the inductors.
  • A later reply agrees with the previous participant's approach, indicating a shared understanding of the circuit simplification.

Areas of Agreement / Disagreement

Participants generally agree on the approach to redrawing the circuit and simplifying the inductors, but there are still questions about specific configurations and assumptions regarding the circuit layout.

Contextual Notes

Some assumptions about the circuit configuration and the effects of the switch position remain unverified, and there may be unresolved steps in the mathematical simplification process.

RoKr93
Messages
21
Reaction score
0

Homework Statement


2013-06-30154521_zps69c24c59.jpg


Homework Equations


Leq = L1 + L2 + ... + Ln (series)

Leq = 1/((1/L1) + (1/L2) + ... + (1/Ln)) (parallel)

The Attempt at a Solution



2013-06-30154553_zps12231bfe.jpg


For part A, I redrew the circuit (above), taking into account that the switch was open, but I'm not entirely certain I did it right. If that is correct, then it would appear that for part B, a redrawn circuit would be the same except for one more parallel branch with nothing on it (representing the closed switch), which would cause a short circuit. This doesn't seem right. Did I redraw the circuit incorrectly?
 
Physics news on Phys.org
You did fine for part A.

Note that the two sets of 12H and 24H inductors remain parallel pairs regardless of the switch position, so you might as well replace them with their equivalents right away.

Now, for part B, when the switch closes it ties the top rail to the bottom rail, making them all one node. So you can "fold" the circuit about the horizontal mid-line bringing the bottom rail up to the top and merging the connections along the resulting single rail. That should put some more inductors in parallel...
 
I make it 7L for part (a), also. For (b), why don't you redraw it showing 4 inductors, two of them being 8L, and go from there? Simplify it in small, easily-manageable, absolutely-irrefutable steps and you can't go wrong! :smile:
 
2013-06-30180901_zpsf01e5142.jpg


It's...the same thing? Combining the 12L and 24L inductors definitely helped clear up my mental picture. But am I correct in assuming that since those 3 branches are parallel, I can swing the empty one into the middle like I did?
 
Right.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 35 ·
2
Replies
35
Views
4K
  • · Replies 108 ·
4
Replies
108
Views
37K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 29 ·
Replies
29
Views
4K