Equivalent resistance between two points

In summary: The upper part is then ## R_{upper}= 45 \Omega || 45 \Omega +15 \Omega ##.In summary, in order to solve this problem, you need to find symmetry in the voltage between A and B, find the voltage between C and D, and then use Ohm's law to find the resistance between the upper and lower parts.
  • #1
Jahnavi
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Homework Statement


Eq resistance.jpg


Homework Equations

The Attempt at a Solution



From symmetry , C and D are at the same potential . So , resistance between them can be removed . This leaves us with three parallel branches between A and B i.e 45 Ω || 45 Ω || 15Ω .

The net resistance will be 9 Ω .

But this is not an option .
 

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  • #2
C and D are not symmetrical.
D is "closer" to A and C is closer to B.

On the other hand, it doesn't matter whether the crossing point in the middle represents an electrical connection or not.
 
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  • #3
I should have been more careful . I realize my mistake .

Thanks !
 
  • #4
Hello. :) A good problem. I figured out how to do it. You are on the right track to use symmetry, but ## C ## and ## D ## are not the same voltage Instead, if you let ## V_A=V ## , and ## V_B=0 ##, ## V_C ## and ## V_D ## will be related by symmetry, but they won't be equal. I can give you a hint that should be helpful: By symmetry, what can you say about the voltage of the point in the middle? There is one more trick you need to see, once you get this part. ## \\ ## Edit: I see @.Scott got the solution already.
 
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  • #5
Hi ,

I have got the correct answer . But I am still interested in discussing this , as I often get confused in these problems .

Charles Link said:
I can give you a hint that should be helpful: By symmetry, what can you say about the voltage of the point in the middle?

Half of the voltage between A and B ?
 
  • #6
Jahnavi said:
Hi ,

I have got the correct answer . But I am still interested in discussing this , as I often get confused in these problems .
Half of the voltage between A and B ?
Yes. And @.Scott gave the additional part that you need: You can separate the circuit at the middle, with the upper part in one loop with C and D, and the lower part in another loop with A and B. ## \\ ## Edit: The reason is by symmetry: the current going toward the middle from C must be the same as the current coming out of the middle and going to D, etc., because ## V_C-V_{Middle}=V_{Middle}-V_D ##.
 
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  • #7
Charles Link said:
The reason is by symmetry: the current going toward the middle from C must be the same as the current coming out of the middle and going to D, etc...

How does middle point being at half the potential helps us in deciding this ?
 
  • #8
Jahnavi said:
How does middle point being at half the potential helps us in deciding this ?
See my "Edit:" additions to the previous post.## \\ ## Once you see that the current into the middle from C is the current that comes out of the middle to D, (and the same with that coming from A and into the middle, and that coming out of the middle and going to B), the middle connection is not necessary, because no current crosses the middle between the upper and lower halves.
 
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  • #9
Charles Link said:
because ## V_C-V_{Middle}=V_{Middle}-V_D ##.

This is the key point :smile: . I am thinking more on this specifically .

I am curious to know as to how you found symmetry in this circuit .

As far as I am concerned , I drew a line perpendicular to the line joining AB and found that the left and right parts are mirror images of each other . C is mirror image of D .

I hope this is right .

What did you do to find symmetry ?
 
  • #10
Jahnavi said:
This is the key point :smile: . I am thinking more on this specifically .

I am curious to know as to how you found symmetry in this circuit .

As far as I am concerned , I drew a line perpendicular to the line joining AB and found that the left and right parts are mirror images of each other . C is mirror image of D .

I hope this is right .

What did you do to find symmetry ?
I essentially did the same thing as drawing a vertical line through the middle, with symmetry on both sides of the line.## \\ ## And when I saw your OP of setting ## V_C=V_D ##, although it was incorrect, it did get me looking for a symmetry about those two points... Notice also that ## V_A-V_{Middle}=V_{Middle}-V_B ##. ## \\ ## And, in addition, notice the resistance of the lower part is then ## R_{lower}= 15 \Omega || (15 \Omega +15 \Omega) ##, and the upper part is also a simple resistor configuration. ## \\ ## This problem reminded me of a related one I saw a number of years ago: Take a cube with ## R=1 \Omega ## on each of the 12 edges. Find the equivalent resistance between diagonal points of the diagonal that passes through the center of the cube (it would be easier to draw a picture). This one also has symmetry in it, but it's a more complex symmetry. (And maybe it would be good to give a hint on this, because when I solved it, I had already learned about a symmetry of a cube from solid state physics courses: About the diagonal that passes through the center, there is a 3-fold axis of symmetry: You can rotate the cube 120 degrees about this axis and you get the same thing. That means the 3 edges coming from what would be point A in this cube are equivalent. Similarly, the 3 edges going into point B are equivalent). And it came up in a google of it:
 
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  • #11
I will have to admit that my reasoning in OP was nonsense . Even though after drawing the line perpendicular to AB and finding that left and right parts are mirror images , I thought that the points C and D were symmetric with respect to points A and B , which is wrong .This is why I wrote in the OP that C and D are equipotential .

The correct thing is that C and D are symmetric with respect to the middle line drawn .

Right ?

 
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  • #12
Jahnavi said:
I will have to admit that my reasoning in OP was nonsense . Even though after drawing the line perpendicular to AB and finding that left and right parts are mirror images , I thought that the points C and D were symmetric with respect to points A and B , which is wrong .This is why I wrote in the OP that C and D are equipotential .

The correct thing is that C and D are symmetric with respect to the middle line drawn .

Right ?
I think the correct term for the relation between their potentials, if you let the middle be potential zero, is the potentials of ## C ## and ## D ## are basically "antisymmetric" about the middle. If ## V_C=+V ##, with ## V_{middle}=0 ##, then ## V_D=-V ##.
 
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  • #13
Wow ! This is such a nice discussion . I get to learn so much from you .

Thank you so much :smile:
 
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What is equivalent resistance between two points?

The equivalent resistance between two points is the total resistance that an electrical current would encounter when traveling from one point to another. It takes into account all of the resistors, wires, and other components in the circuit.

How do you calculate equivalent resistance?

Equivalent resistance can be calculated using Ohm's Law, which states that resistance is equal to voltage divided by current. For a series circuit, the equivalent resistance is the sum of all individual resistances. For a parallel circuit, the equivalent resistance is equal to the reciprocal of the sum of the reciprocals of each individual resistance.

What is the difference between series and parallel circuits?

In a series circuit, all of the components are connected in a single loop, so the same current flows through each component. In a parallel circuit, the components are connected in branches, so the current is divided between them. This results in different equivalent resistance calculations for each type of circuit.

Why is equivalent resistance important?

Equivalent resistance is important because it helps us understand how components in a circuit affect the flow of electrical current. It also allows us to simplify complex circuits into a single, equivalent resistance, making calculations and analysis easier.

What are some real-world applications of equivalent resistance?

Equivalent resistance is used in a variety of electrical engineering and physics applications, such as designing circuits for electronic devices, calculating power consumption and voltage drops, and analyzing the behavior of complex systems. It is also important in understanding and troubleshooting issues in electrical networks and systems.

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