Find equivalent resistance R in the circuit

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SUMMARY

The discussion focuses on calculating the equivalent resistance (R) in a complex circuit involving resistors and the application of Kirchhoff's laws. Participants clarify that resistors must be directly connected to be considered parallel and discuss various circuit analysis methods, including Kirchhoff's Current Law (KCL), Kirchhoff's Voltage Law (KVL), mesh analysis, and delta-wye transformations. The delta-wye transformation method is highlighted as a successful approach for solving the circuit problem, particularly when dealing with multiple unknowns. The importance of careful algebraic manipulation is emphasized to avoid errors in calculations.

PREREQUISITES
  • Understanding of Kirchhoff's Current Law (KCL)
  • Familiarity with Kirchhoff's Voltage Law (KVL)
  • Knowledge of delta-wye transformations
  • Basic circuit analysis techniques, including mesh and nodal analysis
NEXT STEPS
  • Study the application of delta-wye transformations in circuit analysis
  • Learn advanced techniques for solving circuits with multiple unknowns
  • Explore the use of simulation tools like LTspice for circuit analysis
  • Review examples of complex circuits using Kirchhoff's laws
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing complex resistor networks will benefit from this discussion.

valhakla
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Homework Statement
Help me to find the total resistance between two points A and B. I have tried to find the total resistance using the formulas Rparralel1 = 2R * 2R /2R+2R, Rparallel2 = 6R *3R / 6R + 3R, and R total = Rparralel1 + Rparralel2 + 2R, the answer was incorrect. Can you tell me where is my mistake, and how can I determine parallel and series resistors to find the total resistance in such circuits?
Relevant Equations
R_parallel = R1 * R2 / R1 + R2, R_series = R1 + R2
Screenshot 2023-05-31 at 17.59.50.png
 
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valhakla said:
Rparralel1 = 2R * 2R /2R+2R,
The two R2 on the left are not parallel. To count as parallel, the two resistors must be directly connected to each other, with no intervening resistance, at each end.
 
I should add to what @haruspex noted that if two resistors are not parallel, this does not mean that they must be in series. It is entirely possible to have circuits where there are no parallel or series combinations.
 
@valhakla What circuit analysis methods have you learned? Have you covered KCL, KVL? How about mesh and nodal analysis? Perhaps delta-Y and y-delta transforms?
 
gneill said:
@valhakla What circuit analysis methods have you learned? Have you covered KCL, KVL? How about mesh and nodal analysis? Perhaps delta-Y and y-delta transforms
Yes, I have tried to apply Kirchkhoff's and loop rule for this ciruit, but it didn't helped, as it gives only the values of the current flowing through particular resistor not the total resistance between points A and B. Regarding the transforms, I am not familiar with that yet.
 
haruspex said:
The two R2 on the left are not parallel. To count as parallel, the two resistors must be directly connected to each other, with no intervening resistance, at each end.
I understood. So are there any algorithms to approach such circuits? This is my first time encountering this.
 
valhakla said:
Yes, I have tried to apply Kirchkhoff's and loop rule for this ciruit
That’s enough to solve the problem. If that’s all you have covered in class so far, I’d guess that’s what you are meant to use.

Add a battery (emf ##= E##) between A and B so you have a complete circuit.

Analyse the circuit using Kirchhoff’s first law (the ‘current rule’) and second law (the ‘loop rule’). You can find the total current, ##I##, entering A (or leaving B) in terms of ##E## and ##R##.

The resistance between A and B is then ##\frac EI##. ##E## should cancel and the answer will be some multiple of ##R##.

It’s messy because you have a lot of unknown currents.

Note the circuit is ‘linear’. Without loss of generality, you can consider the case with ##R=1 \Omega## and take the battery’s emf as some convenient arbitrary value (e.g. ##E =10V##). The final answer will be a multiple of ##R##, not ohms. However, if you are not comfortable with this ‘shortcut’ you will need to work with the symbols ##R## and ##E##.

Post your working if you need further help.

Also, note spelling: Kirchhoff.
 
  • #10
Steve4Physics said:
Add a battery (emf ##= E##) between A and B so you have a complete circuit.
I tried to add 10V battery, but my final answer wasn't even close. However the delta-wye transformation method helped me to finally solve the problem.
 
  • #11
valhakla said:
I tried to add 10V battery, but my final answer wasn't even close.
The risk of algebra/arithmetic errors is high with 5 unknowns! But it will work.

valhakla said:
However the delta-wye transformation method helped me to finally solve the problem.
It's a good choice if you are allowed to use that method.
 

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