# Confused on how to work out the total resistance HELP!

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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!

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!

Last edited by Yatayyat; 2 years ago

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#2

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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!

**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!

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(Original post by

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.

**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.

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**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.

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.

Last edited by Yatayyat; 2 years ago

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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?

**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?

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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(Original post by

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

**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.Is there not meant to be 3 journeys i.e 3 routes to go from x to y?

Last edited by Yatayyat; 2 years ago

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#7

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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?

**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?

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.

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#8

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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.

**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.

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(Original post by

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.

**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.

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(Original post by

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.

**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.

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#11

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Thank you this makes sense, the diagrams helped to visualise it.

**Yatayyat**)Thank you this makes sense, the diagrams helped to visualise it.

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#12

**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

Thank you this makes sense, the diagrams helped to visualise it.

**Yatayyat**)Thank you this makes sense, the diagrams helped to visualise it.

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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(Original post by

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:

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.

**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:

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.

Last edited by Yatayyat; 2 years ago

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#14

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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.

**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.

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#15

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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