A strange electrical circuit - resistance calculation

AI Thread Summary
The discussion focuses on calculating the total resistance of a complex electrical circuit involving a circular arrangement of resistors. Participants suggest using Y-Delta transformations and Kirchhoff's laws for analysis. Observations about the resistor ratios between nodes A and B lead to insights about potential differences across certain resistors, particularly a 5Ω resistor, which can be treated as a short circuit if both ends are at the same potential. The conversation emphasizes the importance of recognizing symmetrical properties in circuits to simplify calculations. Overall, the thread provides guidance on approaching resistance calculations in intricate circuits.
KayEl9
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Hello. So I've got an electrical circuit on page 4 of this: http://www.iuventa.sk/files/documents/2_olympiady/fo/54.%20rocnik%202012-2013/ulohy%20a%20riesenia/skolske%20kolo/f54skcul13.pdf and I am to calculate the total resistance of the circuit, but I have no idea on how to solve that strange, circular group of resistors. Could anyone help me? I don't want the solution, just a little clue on what to do when the resistors are grouped like this. Thanks a lot.
 
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Look up Y-Delta transformations. Otherwise, use Kirchhoff's laws or mesh or nodal analysis to find the current that the battery supplies and from that the resistance of its load.

EDIT: I just noticed the ratios of the resistors in the group between nodes A and B. What can you say about the potentials at the nodes at either end of the 5.0Ω resistor?
 
gneill said:
EDIT: I just noticed the ratios of the resistors in the group between nodes A and B. What can you say about the potentials at the nodes at either end of the 5.0Ω resistor?
Well spotted. That's an observation which can only come from experience! (it's the sort of exam question which I hated.)
 
NascentOxygen said:
Well spotted. That's an observation which can only come from experience! (it's the sort of exam question which I hated.)

Thanks! :smile: I find it's usually worth looking for an "out" when there's a Δ-Y or Y-Δ "opportunity" in the offing...
 
Thanks, that's interesting. I'm just still not sure how to use that to solve the circuit…
 
The sub-circuit between points A and B includes two potential dividers. The resistors make them practically equivalent. As gneill suggested, look at the voltages at each end of that 5 Ohm resistor...
 
So does that mean that resistance on both branches of the A-B part of the circuit will be the same?
 
KayEl9 said:
So does that mean that resistance on both branches of the A-B part of the circuit will be the same?

No, not that. What do you notice about the resistor ratios in both branches?
 
KayEl9 said:
So does that mean that resistance on both branches of the A-B part of the circuit will be the same?
Name the resistor values that you see comprising the two potential dividers we refer to.
 
  • #10
Yeah, I noticed they are 1:2.

However, following a consultation with a friend who studied electrotechnics, I managed to do the delta-y transformation and calculate the total resistance, which appears to be 10/3Ω. Now, the next part of the task is to calculate the currents I1 through I6. So my current question is: how do I calculate the currents of the original resistors using the delta-y transformation? Thanks.
 
  • #11
I think that perhaps something went awry with your Delta-Y transformation. I'm not seeing 10/3 Ω as the equivalent resistance.

So you've noticed that the ratio of the resistors in the two parallel branches are 1:2. Suppose that there is some potential across the nodes AB. What can you say about the potentials at either end of the 5Ω resistor?
 
  • #12
Well... do I conclude right that they are the same?

Gotta look to my physics notebook from last year as I don't remember the relation between electrical potential, current and voltage. Thanks for help so far.

By the way, if I wanted to solve it using the delta-y transformation (provided I do it right, which I didn't the first time, apparently), how do I calculate the current on each of the original resistors that I delta-y transformed?
 
  • #13
KayEl9 said:
Well... do I conclude right that they are the same?
Yes, they would be the same. If both ends of the 5Ω resistor are at the same potential, what is the current though it?
Gotta look to my physics notebook from last year as I don't remember the relation between electrical potential, current and voltage. Thanks for help so far.

By the way, if I wanted to solve it using the delta-y transformation (provided I do it right, which I didn't the first time, apparently), how do I calculate the current on each of the original resistors that I delta-y transformed?
You can't calculate the currents through resistors that you've transformed away, so you have to return to the original configuration.

You would first find the current supplied by the battery into the equivalent resistance, then the potential drop across the 2.4Ω resistor (potential BC). That in turn would give you the potential drop from A to B. Then you'd have to analyze the sub-circuit between A and B by other means (maybe mesh or nodal analysis).

Trust me, going by way of the fortuitous resistor ratio will be simpler overall :wink:
 
  • #14
gneill said:
Yes, they would be the same. If both ends of the 5Ω resistor are at the same potential, what is the current though it?
Well, zero? So should I just calculate as the 5Ω resistor wasn't there and it's just a normal two-branch circuit?
 
  • #15
KayEl9 said:
Well, zero? So should I just calculate as the 5Ω resistor wasn't there and it's just a normal two-branch circuit?

Yes, exactly. Much simpler, no?

When two nodes are at the same potential you are allowed to remove what's between, or connect them together with a wire (short circuit). This can come in very handy in certain circumstances (such as this), and often occurs when a circuit exhibits symmetry.
 
  • #16
Yeah, it surely is much easier. Physics never stops to amaze me… :)

Thanks a lot for all your help, I really appreciate that :)
 
  • #17
KayEl9 said:
Yeah, it surely is much easier. Physics never stops to amaze me… :)

Thanks a lot for all your help, I really appreciate that :)

Glad I could help. Good luck with your studies.
 
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