Below, there is a sequence of images showing how the original diagram gets transformed into the diagram in section (c), by applying these simple principles:

The key point to remember here is that any two points that are directly connected by a wire are electrically the same point.

So first ignore any wires that just ‘stick out’ from the points A, B, C and D. They don’t add any information.

Then notice that B and C are connected by a wire and so in electrical terms B and C are the same point. So erase the point C and redraw C at the same position as B.

Then imagine that you can pick up the point D and move it to the left of the point A without breaking any of the connections.

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Hi. For part (c), why is the answer different to part (a)? Either way, you’re finding the total resistance in the circuit, right, as all resistors are accounted for? So shouldn’t the answer be the same for both?

The answer to (c) is different from the answer to (a) because point A is a different point from point D. Different pairs of points in a circuit generally have different resistances between them. The argument “Either way, you’re finding the total resistance in the circuit, right, as all resistors are accounted for?” just isn’t valid.

Try to go through the answer, applying Ohm’s law, and you should understand how it all works.

Thank you, and sorry for the late response.

Hi,

Can you please explain why two points connected by a wire are electriccally the same point? Surely b has more current going through it than c does. furthermore on is on a junction and the other is not. Are their apparent ressitances the same? I don’t quite understand.

Please you you explain? much appriciated. (or is there anything I should read up on?)

That’s a good question. But it’s important to remember that here we are talking about a schematic, and the question is about how we can simplify the schematic to work out resistor values. What I mean here by ‘B and C are electrically the same point’ is that that if I were to add another wire joining B and C, then the behaviour of the circuit would not change in any way. If this is true then when simplifying the circuit diagram I am entitled to slide C down to be where B is, or even swap C and B.

I recommend that you satisfy yourself that the circuit really would be the same if you were to add a wire between B and C, connect the voltage source between A and C instead of A and B, and so on. Once you have played around with solving the circuits you’ll see why your question isn’t something to worry about, even though it is a perfectly sensible observation.

Hello,

I have tried your advice; and it seems to work very well. It seems that the current between AC and AB is the same after all. Can I ask if this trick is applicable to every circuit problem? And is there in any circumstance where I should be wary of using this? Complex circuit problems seem to be confusing for us all because of the way we are originally taught. (extreme narrow set of problems leading to oversimplification and a false sense of understanding. I will be looking for resources online to practice this skill

Glad to hear it. The principle

“‘B and C are electrically the same point’ if and only if adding another wire joining B and C does not change the behaviour of the circuit in any way”is always true.Hi, would you be able to explain why, in part b) how you get Vbd? Thank you!

Don’t worry, I have just seen your explanation in the comments!

Hi, I understand part a and c – but with part b I don’t understand why R is divided by 3/2 R. I understand the use of R= V/I and P = V^2/R but i would have thought you added the total resistance between the points.

Thank you for your solutions and hope you can help me!

When there is a voltage V between A and B, there is some current I that flows through the point D, and this is the current in the resistor between B and D.

We can work out the value of the current I: the resistance between A and D is R/2 (two resistors R connected in parallel), and the resistance between D and B is R, so the resistance along the path through D between A and B is 3R/2; therefore the current I is equal to V/(3R/2).

Given the value for I we can work out that the voltage across the resistor between B and D is RV/(3R/2), which equals 2V/3.

do you have any way to practice redrawing circuits like in the last part here im not sure how you really go about it?

how can i learn this?

One way could be to try simplifying the examples discussed in a web page like this one: http://www.allaboutcircuits.com/textbook/direct-current/chpt-7/re-drawing-complex-schematics/. I found this page with a web search for “complex resistor circuits”, and this turns up quite a few other links. Failing that, just start drawing messed up combinations of resistors yourself and then try and simplify them. After a bit of practice you should find it easy.

Can you explain to me your thought process as you resketched the circuit?

I don’t get the two resistors and the other one parallel between a and c on the redrawn graph

OK, here’s a way of redrawing the circuit.

The key point to remember here is that any two points that are directly connected by a wire are electrically the same point.

So first ignore any wires that just ‘stick out’ from the points A, B, C and D. They don’t add any information.

Then notice that B and C are connected by a wire and so in electrical terms B and C are the same point. So erase the point C and redraw C at the same position as B.

Then imagine that you can pick up the point D and move it to the left of the point A without breaking any of the connections.

After doing those steps, you should have a diagram that is the same as the one I have drawn.

when i redraw it like that what happens is the resistor between B and D goes in the same line that A and b are and the parallel compenent in your diagram theyre parallel to AB but when i move d to the left i move that whole bottom like to the left of A and that happens

Thank you for your time

I’m having trouble visualizing that. Why not grab a photo of your answer and upload it to a site like tinypic.com, then post the link here.

I cant seem to reply to your comment so i hope replying here will be ok

if i transfer the bottom circuit beside A i get this, maybe the key word you used was not breaking any of the connections but im having a hard time visualizing that

I’m away from home at the moment but I’ll scan step-by-step pictures in a few days.

Hi are you home yet?

Yes — I have updated the answer with some more explanation. I hope it helps.

Thank you that was great!

Do you offer paid classes online?

No problem, glad it helped. I’m afraid I don’t do any paid-for classes. But I’m happy to answer any questions that you want to ask here for no charge.