Equilavent Resistance of a Triangular Circuit

In summary: Unless a problem explicitly tells you to solve by a certain method it should be fair game to use these transformations.In summary, the student is trying to find the equivalent resistance between the points a and c, but is having trouble due to the resistance between a and b. There is some symmetry that the student can exploit for a-c, but not for a-b. There is also a potential difference on R' that should be zero.
  • #1
Dilemma
15
1
Hello,

Circuit:
mage=https%3A%2F%2Fphysics.knox.edu%2FOnlineHW%2Fphys130a%2Fspring%2Fhomework%2F09%2F02%2FP26_36.jpg


1. Homework Statement

Determine the net resistance between points a and c and a and b. Assume R' ≠ R.

Homework Equations



ε - IR = 0
V = IR

The Attempt at a Solution



http://i.imgur.com/Y2KPI20.jpg

I applied an emf between the points a and c, and tried to solve using Kirchhoff's rules. The question suggests me to use symmetry at junctions but I can not see any.

Thanks in advance.
 
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  • #2
Hi Di, :welcome:,

From the picture it is not clear what you are doing. A few words might help. What do you mean when you write V = (a+b+c) Req ? To me a is the leftmost point in the circuit and you can't use a symbol twice.

There is no symmetry to exploit for the resistance between points a and c , but there is for the resistance between a and b.
 
  • #3
BvU said:
There is no symmetry to exploit for the resistance between points a and c

Yes there is some symmetry you can exploit for a-c and it allows you to remove a resistor.
 
  • #4
CWatters said:
Yes there is some symmetry you can exploit for a-c and it allows you to remove a resistor.

Ignore that. I didn't spot the R' not equal to R.
 
  • #5
Thank you so much for your replies. a, b, and c denote the currents passing through each resistor. I'm using Kirchhoff's idea of "The directed sum of the electrical potential differences (voltage) around any closed network is zero "(Wiki). Therefore ε - Ia+b+cReq = 0

I noticed the so-called symmetry between the points a and b, though I can not utilize that because of the resistance R'.

If this question were not to include an hint, I would not ask this question and try to solve the mathematical equations that I discovered. The hint makes me wonder if I have been following the wrong path from the beginning.

I should also mentioned that I'm currently trying to find the Req between the points a and c.
 
  • #6
Have you investigated whether ΔY or YΔ transformations might be useful here? When the symmetry is right they can be very easy to apply :wink:
 
  • #7
Dilemma said:
I noticed the so-called symmetry between the points a and b, though I can not utilize that because of the resistance R'.
And what voltage difference do you expect over R' in the a-b case ?
 
  • #8
Hello BvU,

Which part of the question are you focusing on? As I said, I'm currently trying to calculate the equivalent resistance between the points a and c.

Since you pointed out, second part of the question seems pretty straightforward, I shouldn't had that part included in the question. The potential difference on R' should be zero and the rest is very easy to calculate.

Dear gneill,

I do not know what those transformations are. I googled it but couldn't find a relevant source.
 
  • #9
Dilemma said:
I do not know what those transformations are. I googled it but couldn't find a relevant source.
They are also called Delta-Y and Y-Delta.
 
  • #10
I believe we will not be covering that in my Phys102 class.
 
  • #11
Dilemma said:
I believe we will not be covering that in my Phys102 class.
Well, it is nevertheless a useful tool for simplifying resistor networks when you run into Y or Δ shaped configurations. In many cases it can allow you to avoid turning to writing and solving KVL and KCL equations, and let's you proceed with standard series and parallel reductions. The transformations are trivial when all the resistors have the same value.

Unless a problem explicitly tells you to solve by a certain method it should be fair game to use these transformations.
 
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Related to Equilavent Resistance of a Triangular Circuit

What is the equivalent resistance of a triangular circuit?

The equivalent resistance of a triangular circuit is the single resistance value that can replace the entire circuit without changing the current or voltage in the circuit.

How do you calculate the equivalent resistance of a triangular circuit?

To calculate the equivalent resistance of a triangular circuit, you can use the formula Req = R1 + R2 + R3, where R1, R2, and R3 are the individual resistances of each side of the triangle. Another method is to use the parallel and series resistance rules to simplify the circuit and determine the equivalent resistance.

What is the purpose of finding the equivalent resistance of a triangular circuit?

The purpose of finding the equivalent resistance of a triangular circuit is to simplify the circuit and make it easier to analyze. It allows us to calculate the total current and voltage in the circuit without having to take into account the individual resistances of each side.

Can the equivalent resistance of a triangular circuit be greater than the sum of its individual resistances?

Yes, the equivalent resistance of a triangular circuit can be greater than the sum of its individual resistances. This can happen when the circuit contains both series and parallel components, which can affect the overall resistance.

How does the shape of a triangular circuit affect its equivalent resistance?

The shape of a triangular circuit does not have a direct effect on its equivalent resistance. However, the arrangement of the components within the circuit can impact the overall resistance. For example, a triangular circuit with all sides in parallel will have a lower equivalent resistance compared to a circuit with all sides in series.

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