Equivalent Resistance: Series or Parallel Triangles?

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SUMMARY

The discussion revolves around determining the equivalent resistance of a circuit with resistors ab and ae connected to a voltage source. The user is confused about whether these resistors are in series or parallel due to the current division at node a. The solution involves recognizing that neither configuration applies directly, and suggests using Kirchhoff's Voltage Law (KVL) to find the current or applying Delta-Y transformations to simplify the circuit. These methods provide a pathway to calculate the equivalent resistance effectively.

PREREQUISITES
  • Understanding of basic circuit theory, including series and parallel resistor configurations.
  • Familiarity with Kirchhoff's Voltage Law (KVL) for circuit analysis.
  • Knowledge of Delta-Y and Y-Delta transformations for resistor networks.
  • Ability to manipulate circuit equations and solve for equivalent resistance.
NEXT STEPS
  • Study Kirchhoff's Voltage Law (KVL) and its application in circuit analysis.
  • Learn about Delta-Y and Y-Delta transformations in electrical circuits.
  • Practice problems involving complex resistor networks to enhance problem-solving skills.
  • Explore advanced circuit analysis techniques, such as mesh and nodal analysis.
USEFUL FOR

Students studying electrical engineering, circuit designers, and anyone looking to deepen their understanding of resistor networks and equivalent resistance calculations.

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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