Equivalent Resistance: Series or Parallel Triangles?

AI Thread Summary
The discussion centers on reducing a circuit with resistors ab and ae, where confusion arises over whether they are in series or parallel. One participant believes they are in series, while another argues they are in parallel due to current division at node a. It is clarified that neither method applies directly, as the configuration does not fit standard series or parallel definitions. Suggested approaches include using Kirchhoff's Voltage Law (KVL) to determine the equivalent resistance or applying Delta-Y transformations to simplify the circuit. The conversation emphasizes the need for alternative methods when traditional approaches are ineffective.
sparkleshine
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Homework Statement


1jlfr.jpg


Okay. I thought I knew how to do these type of questions, but here goes. The node at a is attached to the positive terminal of a voltage source and c is attached to the negative. I'm completely bemused as to how to reduce this circuit into a single-resistor equivalent.

I thought that resistors ab and ae could be reduced into a series resistor, but my study partner says they are in parallel because the current does not go through them one after another but divides at a. However, they can't be reduced to parallel resistors because of the set-up of the junctions (or can they?) I'm actually stumped by how to approach this problem.

Homework Equations



1/RP=1/R1+1/R2...

RS=R1+R2...

R|| = R1R2/R1+R2

The Attempt at a Solution



The circuit has already been reduced from a more complex one and I've tried searching through the questions posted in homework for inspiration. Any help would be much appreciated!
 
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sparkleshine said:

Homework Statement


1jlfr.jpg


Okay. I thought I knew how to do these type of questions, but here goes. The node at a is attached to the positive terminal of a voltage source and c is attached to the negative. I'm completely bemused as to how to reduce this circuit into a single-resistor equivalent.

I thought that resistors ab and ae could be reduced into a series resistor, but my study partner says they are in parallel because the current does not go through them one after another but divides at a. However, they can't be reduced to parallel resistors because of the set-up of the junctions (or can they?) I'm actually stumped by how to approach this problem.

Homework Equations



1/RP=1/R1+1/R2...

RS=R1+R2...

R|| = R1R2/R1+R2

The Attempt at a Solution



The circuit has already been reduced from a more complex one and I've tried searching through the questions posted in homework for inspiration. Any help would be much appreciated!

Hi sparkleshine, Welcome to Physics Forums.

This is one of those cases where there are no resistors that are in parallel or in series, so progress is blocked if you stick to the those two methods.

Fortunately there are several possible ways to proceed. One way would be to assume some voltage V between terminals a and c as you suggest and then write and solve the KVL loop equations to find the current I that the source drives through the network. Then Req = V/I.

A second way forward is to employ what is called a Delta-Y transformation on one of the groups of resistors that's laid out in the form of a Delta (Δ). So, look up Delta-Y and Y-Delta transformations.
 
Thank you very much! :)
 

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