Basic resistor question

  • Thread starter nhmllr
  • Start date
  • #1
185
1
Are the following Req's between the two nodes equivalent? I'm pretty sure the first and seconds ones are, sort of sure the third is, and a little sure the fourth one is. The r's are the resistors. I'm sorry that the pictures are so crude, I couldn't attach my MS paint picture's jpeg or png
Bonus Question: Why does Req stand for Total Resistance??? Nobody seems to know...


1.
o------------
.....|.........|
.....r.........r
.....|.........|
o------------

.....-----r---
.....|.........|
o---|.........|----o
.....|.........|
.....----r-----

2.
o--------------------
.....|.........|.........|
.....r.........r.........r
.....|.........|.........|
o--------------------

.....-----r---
.....|.........|
o---|---r---|----o
.....|.........|
.....----r-----

3.
.....----r-------r-----
.....|.........|.........|
o---|.........|.........|---o
.....|.........|.........|
.....----r-------r-----

.....----r----......---r-----
.....|.........|.....|.........|
o---|.........|----|.........|---o
.....----r----......---r-----

4.
.....----r-------r-----
.....|.........|.........|
o---|.........r.........|---o
.....|.........|.........|
.....----r-------r-----

.....----r----......---r-----
.....|.........|.....|.........|
o---|.........|-r--|.........|---o
.....|.........|.....|.........|
.....----r----......---r-----
 
Last edited:

Answers and Replies

  • #2
Integral
Staff Emeritus
Science Advisor
Gold Member
7,201
56
Req stands for equivalent resistance.

Please share more of your thoughts on your problems.
 
  • #3
754
1
It may help you to look at it this way:
Re-label the resistors with unique identifiers (R1, R2, etc.)
Determine the minimum path or paths from one node to the other.
Which resistors are passed through in each case?
Use that information to determine if the 2 circuits are equivalent.

Example:
In #1, you can get from one node to the other by passing through either R1 or R2.
This holds true for the both circuits, so they are equivalent.


In #3, let's relabel the resistors in the 1st circuit so that the upper left resistor is R1, the upper right one is R2, the lower left one is R3, and the lower right one is R4. Now, notice that there are four ways to traverse from one node to the other (without looping back or re-using a resistor): R1 to R2, R1 to R4, R3 to R2, and R3 to R4. Keeping the same labeling convention for the 2nd circuit, you can see that the same 4 routes get you from one node to the other. Therefore, the circuits are equivalent.

Now, try this with problem #4.
 
  • #4
185
1
So the 4th one isn't equivalent... Then how do I find the Req?

.....----r1-------r2--
.....|.........|.........|
o---|.........r3.......|---o
.....|.........|.........|
.....----r4-------r5--

Would it be the same as this?

.....|-----r1-----r2----|
o---|--r1----r3---r5--|---o
.....|-----r4-----r5----|
.....|--r4----r3---r2--|

ugh that's annoying...
 
Last edited:
  • #5
185
1
Also by that logic, this (assuming the second node is not on the intersection)

------r-------
...|........|
...----o---

could be replaced with this?

---r---o---r-----

Just curious...
 
Last edited:
  • #6
berkeman
Mentor
59,469
9,593
Also by that logic, this (assuming the second node is not on the intersection)

------r-------
...|........|
...----o---

could be replaced with this?

---r---o---r-----

Just curious...

What's "o"? Another resistor? Whatever it is, it's in parallel with "r", and you would need to use the parallel combination formula to combine them.
 
  • #7
berkeman
Mentor
59,469
9,593
So the 4th one isn't equivalent... Then how do I find the Req?

.....----r1-------r2--
.....|.........|.........|
o---|.........r3.......|---o
.....|.........|.........|
.....----r4-------r5--

Would it be the same as this?

.....|-----r1-----r2----|
o---|--r1----r3---r5--|---o
.....|-----r4-----r5----|
.....|--r4----r3---r2--|

ugh that's annoying...

Yes, trying to draw resistors like that is annoying. You could use "code" tags in order to force a non-proportional font, or you could just draw them in Visio and convert to PDF.

And you are correct that #4 is not equivalent. I'm not sure there's an easy way to find the equivalent resistance (there may be some trick), but in general, I just write a couple KCL equations for the network and solve.
 
  • #8
185
1
My bad- the o is just a node
 
  • #9
185
1
So both my first and second attempts at trying to find an equivalent layout are both wrong?
 
  • #10
berkeman
Mentor
59,469
9,593
So both my first and second attempts at trying to find an equivalent layout are both wrong?

I could be missing something, but I don't think they are right.

Are you familiar with how to write Kirchoff's Current Law (KCL) equations? That's the sure-fire way to figure out the equivalent resistance if it doesn't look like the resistor structure will simplify with traditional series/parallel combinations.
 
  • #11
2,967
5
I could be missing something, but I don't think they are right.

Are you familiar with how to write Kirchoff's Current Law (KCL) equations? That's the sure-fire way to figure out the equivalent resistance if it doesn't look like the resistor structure will simplify with traditional series/parallel combinations.

No, please don't follow this advice. It is a painful exercise for a one-line problem.
 
  • #12
berkeman
Mentor
59,469
9,593
No, please don't follow this advice. It is a painful exercise for a one-line problem.

Is there a trick when all 5 resistors are different?

EDIT -- Certainly if all 5 resistors are the same value, there is an easy simplification. But I didn't think we were making that assumption in this thread....?
 
  • #13
2,967
5
Is there a trick when all 5 resistors are different?

EDIT -- Certainly if all 5 resistors are the same value, there is an easy simplification. But I didn't think we were making that assumption in this thread....?

I am sorry, I thought they have some symmetry. However, if they don't, maybe he can use the triangle(polygon)/star conversion. It still relies on KCLs, though.

EDIT:

Link: http://en.wikipedia.org/wiki/Y-Δ_transform
 
  • #14
berkeman
Mentor
59,469
9,593
I am sorry, I thought they have some symmetry. However, if they don't, maybe he can use the triangle(polygon)/star conversion. It still relies on KCLs, though.

EDIT:

Link: http://en.wikipedia.org/wiki/Y-Δ_transform

No worries, and actually, looking back at Post #1, they *are* all labelled the same. Maybe it was us "helpers" that generalized the problem for the OP...?:tongue2:

Hey nhmllr -- were you supposed to assume that all the "r" values were the same? That definitely simplifies configuration #4. Quiz Question -- Why does it?
 
  • #15
754
1
So the 4th one isn't equivalent... Then how do I find the Req?

.....----r1-------r2--
.....|.........|.........|
o---|.........r3.......|---o
.....|.........|.........|
.....----r4-------r5--

Would it be the same as this?

.....|-----r1-----r2----|
o---|--r1----r3---r5--|---o
.....|-----r4-----r5----|
.....|--r4----r3---r2--|

ugh that's annoying...

No, that doesn't work. You can't convert like this since your paths utilize R3 more than once. That's the flag telling you that there's something else going on here.

As Dickfore mentioned, you'll need to use "triangle-star conversion" (better known as "delta-wye conversion"). Google that ... there's plenty of info out there on how to do it.

To summarize though, problems 1, 2 & 3 all have equivalent circuits (regardless of resistor values). Problem 4 however, does NOT show equivalent circuits.
 
  • #16
754
1
Note that this network
Code:
    ---- r1 -------- r2 ----
    |          |           |
o---|          r3          |---o
    |          |           |
    ---- r4 -------- r5 ----
is equivalent to these 2 networks in series:

Code:
    ---- r1 --------o          o---- r2 ----
    |          |                           |
o---|          r3                          |---o
    |          |                           |
    ---- r4 --------o          o---- r5 ----

The network on the left is in a delta configuration and will need to be converted into an equivalent wye configuration of the form

Code:
           ----o
           |
           s1
           |
o--- s2 ---|
           |
           s3
           |
           ----o

The "S's" represent resistors with values different than those in the delta configuration.

The original network can then be shown as:
Code:
           ------ r2 ----
           |            |
           s1           |
           |            |
o--- s2 ---|            |----o
           |            |
           s3           |
           |            |
           ------ r5 ----

which is the same as
Code:
            ---- s1 ---- r2 ----
           |                   |
o--- s2 ---|                   |----o
           |                   |
            ---- s3 ---- r5 ----
 
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  • #17
205
0
So the 4th one isn't equivalent... Then how do I find the Req?

.....----r1-------r2--
.....|.........|.........|
o---|.........r3.......|---o
.....|.........|.........|
.....----r4-------r5--

...

If (and only if) all the resistors have equal value r, the bridge circuit simplifies very easily. Can you see what the total would be?

Hint, imagine a voltage applied across the network: what then would be the voltage across the central resistor R3?

Oops. Berkemann has already pointed this out. Sorry!
 
Last edited:
  • #18
2,967
5
If (and only if) all the resistors have equal value r, the bridge circuit simplifies very easily. Can you see what the total would be?

Hint, imagine a voltage applied across the network: what then would be the voltage across the central resistor R3?

Oops. Berkemann has already pointed this out. Sorry!

The condition denoted in red is not necessary for that to be possible.
 
  • #19
205
0
The condition denoted in red is not necessary for that to be possible.

Dead right too - I must have been asleep. Any balance condition would do the trick.
 

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