Confused on how to work out the total resistance HELP!

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Yatayyat
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Attachment 789874How would I work out the effective resistance between X and Y.

This network is arranged in some square formation and I don't see how I would find the resistance between X and Y? I

Any help would be really appreciated. Thanks a lot!
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Pangol
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(Original post by Yatayyat)
Attachment 789874How would I work out the effective resistance between X and Y.

This network is arranged in some square formation and I don't see how I would find the resistance between X and Y? I

Any help would be really appreciated. Thanks a lot!
It's the way that it is drawn that is making it tricky. You can think of the "bottom" route from X to Y as two resistors in parallel (on the right), in series with the resistor on the left. Work out the total resistance of this combination. Then this single resistance is in parallel with the resistor in the "top" route from X to Y.
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Yatayyat
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(Original post by Pangol)
It's the way that it is drawn that is making it tricky. You can think of the "bottom" route from X to Y as two resistors in parallel (on the right), in series with the resistor on the left. Work out the total resistance of this combination. Then this single resistance is in parallel with the resistor in the "top" route from X to Y.
I'm getting it from where you're heading at, but why is it when we have the the total resistance of the first combination that we take that that single resistance to be in parallel with the "top" route? Why is it wrong for me to assume that the single resistance is in fact connected in series with the "top" resistor?
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Yatayyat
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(Original post by Pangol)
It's the way that it is drawn that is making it tricky. You can think of the "bottom" route from X to Y as two resistors in parallel (on the right), in series with the resistor on the left. Work out the total resistance of this combination. Then this single resistance is in parallel with the resistor in the "top" route from X to Y.
Following from what you have said

Here is my working:

1) Two resistors that are seen at the bottom right of the square can be thought of as resistors connected in parallel, so their total resistance is '(1/2 +1/2)^-1 = 1 ohm.'

2) Then that parallel part is in in series with the left resistor like you have said so then I add up those two resistances giving '1 + 2 = 3 ohms'

3) Now when I have that combination reduced to 3 ohms it is then parallel to another 2 ohm resistor, so again the total resistance between those two are '(1/3 + 1/2)^-1 = 1.2 ohms'

This is the answer I have got now.

The only thing that I don't seem to get is why the whole combination is then parallel to the "top" resistor and not in series, which is step 3 that I labelled above.
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Pangol
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(Original post by Yatayyat)
I'm getting it from where you're heading at, but why is it when we have the the total resistance of the first combination that we take that that single resistance to be in parallel with the "top" route? Why is it wrong for me to assume that the single resistance is in fact connected in series with the "top" resistor?
If you think of "series" as meaning "one after the other", then the journey from X to Y does not involve doing the top branch and then the bottom part.

Similarly, if you think of "parallel" as meaning "side by side", then the journey from X to Y does involve the top and bottom branches doing this journey next to each other.
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Yatayyat
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(Original post by Pangol)
If you think of "series" as meaning "one after the other", then the journey from X to Y does not involve doing the top branch and then the bottom part.

Similarly, if you think of "parallel" as meaning "side by side", then the journey from X to Y does involve the top and bottom branches doing this journey next to each other.
Sorry, I’m still a bit confused why do you mean when you say journey from x to y? I’m struggling a bit on how to picture all of this.

Is there not meant to be 3 journeys i.e 3 routes to go from x to y?
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Pangol
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(Original post by Yatayyat)
Sorry, I’m still a bit confused why do you mean when you say journey from x to y? I’m struggling a bit on how to picture all of this.

Is there meant to be 3 journeys i.e 3 routes to go from x to y?
I'm sure we'd all agree that the first picture below is two resistors in series, and the second is two resistors in parallel.

The third picture is the one in the question, just drawn in a different way. Looked at like this, it should be clear that the single resistor on the top and the network on the bottom are in parallel with each other.

Name:  resistors.jpg
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Size:  138.1 KB
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RogerOxon
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(Original post by Yatayyat)
Following from what you have said

Here is my working:

1) Two resistors that are seen at the bottom right of the square can be thought of as resistors connected in parallel, so their total resistance is '(1/2 +1/2)^-1 = 1 ohm.'

2) Then that parallel part is in in series with the left resistor like you have said so then I add up those two resistances giving '1 + 2 = 3 ohms'

3) Now when I have that combination reduced to 3 ohms it is then parallel to another 2 ohm resistor, so again the total resistance between those two are '(1/3 + 1/2)^-1 = 1.2 ohms'

This is the answer I have got now.

The only thing that I don't seem to get is why the whole combination is then parallel to the "top" resistor and not in series, which is step 3 that I labelled above.
1.2 Ohm is correct. The 2 Ohm resistor is in parallel with the rest because current does not need to go through any other resistor to travel from X to Y.
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Yatayyat
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(Original post by Pangol)
I'm sure we'd all agree that the first picture below is two resistors in series, and the second is two resistors in parallel.

The third picture is the one in the question, just drawn in a different way. Looked at like this, it should be clear that the single resistor on the top and the network on the bottom are in parallel with each other.

Name:  resistors.jpg
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Thank you this makes sense, the diagrams helped to visualise it.
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Yatayyat
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(Original post by RogerOxon)
1.2 Ohm is correct. The 2 Ohm resistor is in parallel with the rest because current does not need to go through any other resistor to travel from X to Y.
Oh I see, yep I agree with that
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Joinedup
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(Original post by Yatayyat)
Thank you this makes sense, the diagrams helped to visualise it.
fwiw if you run into a weird circuit diagram under exam conditions it might help to do a sketch, redrawing it into a more conventional and less baffling form like that. for a mcq it might be something to come back to in any spare time at the end of the exam
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Eimmanuel
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(Original post by Yatayyat)
Attachment 789874How would I work out the effective resistance between X and Y.

This network is arranged in some square formation and I don't see how I would find the resistance between X and Y? I

Any help would be really appreciated. Thanks a lot!

(Original post by Yatayyat)
Thank you this makes sense, the diagrams helped to visualise it.
As you already know that we can redraw the arrangement of resistors but the question is usually how can we do so. A trick that some students find it useful is as follow:
Name:  resistor_network_02 (3)_LI.jpg
Views: 205
Size:  62.6 KB

Imagine that a current is split at junction X.

When current split, this usually implies that the resistors will be in a parallel setup.

As the current (blue colour) further split at Z, the final two resistors are parallel to each other.

By following the “flow” of current, you can get the drawing as shown by Pangol.
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Yatayyat
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(Original post by Eimmanuel)
As you already know that we can redraw the arrangement of resistors but the question is usually how can we do so. A trick that some students find it useful is as follow:
Name:  resistor_network_02 (3)_LI.jpg
Views: 205
Size:  62.6 KB

Imagine that a current is split at junction X.

When current split, this usually implies that the resistors will be in a parallel setup.

As the current (blue colour) further split at Z, the final two resistors are parallel to each other.

By following the “flow” of current, you can get the drawing as shown by Pangol.
Thank this trick seems really handy to know to quickly spot which resistors seem to be parallel to each other by the current splitting. The square formation just really threw off there before.
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Eimmanuel
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(Original post by Yatayyat)
Thank this trick seems really handy to know to quickly spot which resistors seem to be parallel to each other by the current splitting. The square formation just really threw off there before.
Glad that you find it useful.
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Aykan124
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In that case would the resistance from y-x be different as the current would split into the 3 different parts? And if not why wouldn't it be different?
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