Finding Req - Understanding Professor's Solution

  • Thread starter zr95
  • Start date
In summary, the conversation discusses the process of solving a circuit problem involving series and parallel resistors. The professor's solution involves immediately adding two resistors together, while the student redraws the circuit to find the parallel connections. There is also a question about whether the resistors would still be in parallel if the circuit was not redrawn, and the importance of recognizing short-circuited nodes and using symmetry to simplify the circuit. The principle used to justify immediately adding resistors is that if there is no path between two points that does not pass through the resistors in question, then those resistors must be in series.
  • #1
zr95
25
1

Homework Statement


upload_2016-9-25_14-53-41.png


Homework Equations


Series: Req = r1 + r2...
Parallel: 1/req = 1/r1 + 1/r2...

The Attempt at a Solution


upload_2016-9-25_14-53-26.png

This is my professors solution for the problem and of course without any explanation it doesn't make sense to me. Under what principle can he justify immediately adding 10 + 8?

upload_2016-9-25_14-56-38.png

For my solution I redrew the circuit and did:
5||20 = 4
10+4 = 14

3 || 6 = 2
2+8 = 10

10+14 = 24

Another question: Had I not redrawn the circuit I still would've gotten 14 and 10 but I would've thought they would be in parallel with each other with the way its drawn out?

I understand redrawing it is the easiest way to make sure I have it correct but there won't be enough time unless I immediately recognize what it looks like.
 

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  • #2
zr95 said:
Under what principle can he justify immediately adding 10 + 8?

Is there any path that you can draw between a and b (following circuit pathways of course) that does not pass through both the 10 and 8 Ohm resistors? If not, they must be in series between a and b.

zr95 said:
Another question: Had I not redrawn the circuit I still would've gotten 14 and 10 but I would've thought they would be in parallel with each other with the way its drawn out?

I understand redrawing it is the easiest way to make sure I have it correct but there won't be enough time unless I immediately recognize what it looks like.

The trick is to look at the nodes that are bridged by a short circuit. In this circuit nodes c and f are connected together via a wire. That means you are free to move all the connections at one of those nodes to the other node. Symmetry also helps here, as you can put nodes c and f together by "folding" the circuit along a line passing through d-a-b-e. Once you make that fold you've got your parallel resistor connections that you and your professor found, while the two series resistors are left at the fold.
 

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