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I'm having hard times understanding why when you combine 20 and 30 resistances which are in parallel, and then combine 60 and 40 (which are also in parallel) ohms resistances you will get two parallel resistances? Aren't they supposed to be in series when you combine 20 and 30, 60 and 40?

And if the question was asking the equivalent resistance between terminal b and a but not a and b, would it be just the equivalent resistance of 60 and 40 ohms resistances?

Also, if there was no wire in the middle, would 20 ohms resistance be in series with 60 ohms resistance as well as 30 ohms be in series with 40 ohms resistances?

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This is a very good question and you are not alone in finding it difficult to decide which resistors are in series and which resistors are in parallel.

So go back one step and think about the reason that you want to know if resistors are in parallel or in series.

The reason is that if you can identify series and parallel resistors you have formulae which enable you to find the equivalent resistance.

So go back another step and think about the assumptions which were made when the series and parallel formulae for equivalent resistance were derived.

  • For the resistors in series the crucial condition was that the current through each of the resistors is the same.
  • For resistors in parallel the crucial condition is that the voltage across the resistors was the same.

I have modified your circuit and the as follows:

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So looking at the left hand circuit can it be decided whether or not the 20 Ω and the 60 Ω are in series?
Because the current through the 20 Ω resistor is not the same as the current through the 60 Ω resistor you cannot consider these two resistors to be in series.
So for these two resistors the series equivalent formula should not be used.

Are the 20 Ω and the 30 Ω in parallel?
The answer is “yes” because the voltage across them is the same and so you can use the parallel equivalent formula with them.

What about if the connection across the middle was removed as in the middle circuit?
Then for the 20 Ω and 60 Ω resistors the series equivalent formula can be used because the current passing through them is the same.

Now suppose that instead of the connection across the middle there was a 4 Ω resistor as in the right hand diagram.

Are the 20 Ω and 60 Ω resistors in series?
“No”, because the current through them is not necessarily the same.

Is the 30 Ω resistor in parallel with the 20 Ω and 4 Ω resistor combination?
At first glance it looks like they are because the voltage across the 30 Ω resistor is the same as the voltage across the 20 Ω and 4 Ω resistor combination.

Here one must go one step back and ask:
“How am I going to use my parallel equivalent formula?”
The answer is that first one must find the equivalent resistor for the 20 Ω and 4 Ω resistor combination.

Can that be done?
“No”, because the current through them is not necessarily the same.
That means that the series equivalent formula cannot be used which in turns means the parallel equivalent formula cannot be used.

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  • $\begingroup$ Thank you ! You clarified a lot of things for me ! Just quick two question: 1) In the third circuit, none of the resistors are in parallel and also none of them are in series, right? 2) The second circuit, if the current goes from a to b, will that mean that the polarities of all resistors will be "+ -" ? $\endgroup$ – Student Feb 19 '16 at 1:09
  • $\begingroup$ 1) You are correct. 2) positive at the end nearest $a$ because for a resistor the current flows from the high potential to the low potential. $\endgroup$ – Farcher Feb 19 '16 at 6:13
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If $1/R1=1/20+1/30$ and $1/R2=1/60+1/40$, the branch is converted to R1, R2 in series, with a tap to terminal a between them.

The conductor connecting running across from terminal a can be collapsed to a single point; it simply indicates the common voltage of the two points. If it were removed the circuit would be quite different - and you are correct, the 20, 60 branch would be in series, but the 30, 40 branch would not be in series due to the tap.

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  • $\begingroup$ so when 20 and 30 ohms, and 60 and 40 ohms resistors are combined into R1 and R2, they won't be in series only because of a terminal between them? $\endgroup$ – Student Feb 18 '16 at 3:28
  • $\begingroup$ Their still in series, but you cannot combine them due to the terminal. If you short the terminals a, b they are no longer in series ... but if nothing is attached, then they are in series. Sometimes you will be adding another complete circuit off of the two terminals. So you have to retain them as part of the topology of the equivalent circuit. $\endgroup$ – Peter Diehr Feb 18 '16 at 3:40
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I'm having hard times understanding why when you combine 20 and 30 resistances which are in parallel, and then combine 60 and 40 (which are also in parallel) ohms resistances you will get two parallel resistances? Aren't they supposed to be in series when you combine 20 and 30, 60 and 40?

The $20\,\Omega$ and $30\,\Omega$ are indeed parallel to each other. Also, $60\,\Omega$ and $40\,\Omega$ are in parallel to each other. these two equivalent resistances, however, are not in parallel with each other because that are not touching/connected to each other at both ends. You need to figure out what to do with the $4\,\Omega$. Draw a picture after each combination so that your mind doesn't make incorrect assumptions about the actual calculations.

And if the question was asking the equivalent resistance between terminal b and a but not a and b, would it be just the equivalent resistance of 60 and 40 ohms resistances?

There's absolutely no difference. Choosing two points establishes the resistance path for a passive circuit (no sources). The order of the names means nothing.

Also, if there was no wire in the middle, would 20 ohms resistance be in series with 60 ohms resistance as well as 30 ohms be in series with 40 ohms resistances?

$20\,\Omega$ and $60\,\Omega$ would be in series, but $30\,\Omega$ and $40\,\Omega$ are not, because of terminal a.

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