Equivalent Resistance of Figure

In summary, the conversation is discussing the calculation of equivalent resistance in a circuit with two resistors and a direct cable connecting the two sides. The formula used is (1/Req) = (1/Rv1) + (1/Rv2), and the solution involves getting rid of Rv1 and Rv2 using the image in #4. The final equivalent resistance is 0 Ohm.
  • #1
Yealtas

Homework Statement


https://imgur.com/a/aVCeU

Homework Equations


(1/Req) = (1/Rv1) +(1/Rv2)
Req = Rv1 + Rv2

The Attempt at a Solution


I'm not sure which formula is the correct one to use.So..
I'm absolutely stuck. I tried downloading a bunch of circuit board simulators, but I couldn't find an easy one to use. And googling yields 0 results.

Which formula should I use for the equivalent resistance, and why?

Thank you very much. :)
 
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  • #2
There is a direct cable from one side to the other. What do you expect?
 
  • #3
I would say electrons choose the path of least resistance, and therefor never even attempt to go through the resistances. But I know that's extremely wrong, so I won't say it. In all honesty man, I have no idea.
 
  • #4
Yealtas said:
I would say electrons choose the path of least resistance, and therefor never even attempt to go through the resistances.
Correct.

As image:

resistanceisfutile.gif
 

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  • #5
mfb said:
Correct.

As image:

View attachment 213652

Hahaha, I want that as a shirt now to be honest!

Alright, i'll give it a better shot tho:

Let's call the wire "Rv3". Rv3 should have 0 resistance. You'd then calculate the equivalent resistance of Rv1 and Rv3 (With the equation for parallel resistances). Let's call this answer Rv4, so that finally Req = Rv2 + Rv4..?

edit: Rv4 would equal Rv1, so that Req is simply Rv1 + Rv2..?
 
  • #6
What is Rv4?

There is a cable parallel to R1 and one parallel to R2.
 
  • #7
I called the cable in the middle Rv3. Rv3 = 0 (so all current can flow freely). Then I call Rv4 the equivalent resistance of Rv1 and Rv3 (1/Rv4) = (1/Rv1) + (1/Rv3).

I just now realize that what I said earlier doesn't make sense. If Rv3 was infinite I could say that (1/Rv4) = (1/Rv1), but Rv3 is 0...

I have a feeling that Req = Rv1 + Rv2, but I can't explain why. And just the plain answer doesn't satisfy me. I need a method :/
 
  • #8
Yealtas said:
And just the plain answer doesn't satisfy me. I need a method :/
Think about the image in #4.
Although your statement about current taking the path of least resistance is not accurate, your conclusion below is correct (since the wires are assumed to be ideal).
Yealtas said:
and therefor never even attempt to go through the resistances
So what does that tell you about the resistance between the two points?
 
  • #9
cnh1995 said:
Think about the image in #4.
Although your statement about current taking the path of least resistance is not accurate, your conclusion below is correct (since the wires are assumed to be ideal).

So what does that tell you about the resistance between the two points?

Sorry, I don't quite understand what you mean. Do you mean the resistance of the wire in the middle..?
 
  • #10
Yealtas said:
Sorry, I don't quite understand what you mean. Do you mean the resistance of the wire in the middle..?
How many paths can you find to go from the left terminal to the right terminal? Which of them is the path of "least" resistance? How much is that least resistance?
 
  • #11
Yealtas said:
Hahaha, I want that as a shirt now to be honest!
It exists as shirt in many places.

If you understand the circuit in #4, you can use it to get rid of R1. And then you can use it to get rid of R2. What is left?
 
  • #12
cnh1995 said:
How many paths can you find to go from the left terminal to the right terminal? Which of them is the path of "least" resistance? How much is that least resistance?
cnh1995 said:
How many paths can you find to go from the left terminal to the right terminal? Which of them is the path of "least" resistance? How much is that least resistance?

4 paths. 2 at the start, and then each path has another 2 'sub-paths'. the least resistance is the path through the wire alone.
 
  • #13
Yealtas said:
the least resistance is the path through the wire alone.
Yes, and how much is that resistance?
 
  • #14
mfb said:
It exists as shirt in many places.

If you understand the circuit in #4, you can use it to get rid of R1. And then you can use it to get rid of R2. What is left?
See, i don't understand how i can get rid of Rv1 with #4. I'm still stuck on this problem.

The figure is basically just #4 twice. Any more help please?
 
  • #15
cnh1995 said:
Yes, and how much is that resistance?
Speaking in term of ideal circumstances; 0 Ohm.
 
  • #16
Yealtas said:
Speaking in term of ideal circumstances; 0 Ohm.
Right.
 
  • #17
.
 
  • #18
cnh1995 said:
Right.
Would the equivalent resistance of the Figure simply be 0..?
 
  • #19
Yealtas said:
Would the equivalent resistance of the Figure simply be 0..?
Yes.
 
  • #20
cnh1995 said:
Yes.
Thanks for the help Sheldon.

:)
 

1. What is equivalent resistance?

Equivalent resistance is the total resistance of a circuit when all individual resistances are combined into one single resistor.

2. How do you calculate equivalent resistance in a series circuit?

In a series circuit, equivalent resistance is calculated by adding all the individual resistances together.

3. How do you calculate equivalent resistance in a parallel circuit?

In a parallel circuit, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of all the individual resistances. This can be represented by the formula: 1/Req = 1/R1 + 1/R2 + 1/R3 + ...

4. What is the difference between series and parallel circuits in terms of equivalent resistance?

In a series circuit, equivalent resistance is always greater than any individual resistance, whereas in a parallel circuit, equivalent resistance is always less than any individual resistance.

5. How does the arrangement of resistors affect the equivalent resistance?

The arrangement of resistors can greatly affect the equivalent resistance of a circuit. In a series circuit, resistors are arranged one after the other, resulting in a cumulative increase in resistance. In a parallel circuit, resistors are arranged side by side, providing multiple paths for current to flow, resulting in a decrease in resistance.

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