Find equivalent resistance R in the circuit

In summary: It can help simplify more complex circuits. Just be careful with the algebra and make sure to include the delta-wye transformation in your final answer. In summary, the discussion covers the concept of parallel resistors and how they must be directly connected to each other with no intervening resistance. The use of various circuit analysis methods is also mentioned, including Kirchhoff's laws and delta-wye transformations. The conversation also offers a helpful resource for further understanding. Ultimately, the delta-wye transformation method is recommended for solving the given problem due to its ability to simplify complex circuits.
  • #1
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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  • #2
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.
 
  • #3
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.
 
  • #4
@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?
 
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  • #5
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.
 
  • #6
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.
 
  • #8
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.
 
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  • #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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FAQ: Find equivalent resistance R in the circuit

1. How do you find the equivalent resistance in a series circuit?

To find the equivalent resistance in a series circuit, simply add up the resistance values of all the resistors. The formula is \( R_{eq} = R_1 + R_2 + R_3 + \ldots + R_n \).

2. How do you find the equivalent resistance in a parallel circuit?

To find the equivalent resistance in a parallel circuit, you use the reciprocal sum of the resistances. The formula is \( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots + \frac{1}{R_n} \). After calculating the reciprocal sum, take the reciprocal of the result to get the equivalent resistance.

3. What is the equivalent resistance of resistors in a combination of series and parallel circuits?

For a combination of series and parallel circuits, you need to break down the circuit into simpler series and parallel parts. Calculate the equivalent resistance for each part separately and then combine them step-by-step, either in series or parallel, as required by the circuit configuration.

4. How do you handle circuits with both series and parallel resistors?

To handle circuits with both series and parallel resistors, first identify and solve the simplest series or parallel sections of the circuit. Replace those sections with their equivalent resistances. Continue simplifying the circuit step-by-step until you have a single equivalent resistance for the entire circuit.

5. Can the equivalent resistance be greater than the largest individual resistor in the circuit?

In a series circuit, the equivalent resistance is always greater than the largest individual resistor because resistances add up. In a parallel circuit, the equivalent resistance is always less than the smallest individual resistor because the reciprocal sum of resistances results in a smaller value.

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