I think there are 6 nodes(as I've labelled in the picture),8 branches and 6 loops in the circuit.But which resistors are parallel to each other?R7||R4,R2||R5||R3 + R6 ?I'm really not sure.Can someone give an answer an explain?

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    \$\begingroup\$ Include the picture instead of only a link to it. Also, if you cannot tell which resistors are in series and which are in parallel then you really need to refer back to your book or teacher. You must know the difference between series and parallel, it is really easy and a basic skill and not something we're going to explain to you here. \$\endgroup\$ – Bimpelrekkie Oct 12 '17 at 9:56
  • \$\begingroup\$ @Sook Lim: what's the definition of series? And of parallel? If you find two resistors that fall in one of the two definitions, then they are either in series or in parallel. If two resistors do not fall in any of the two definitions, then they are neither in series, nor in parallel. \$\endgroup\$ – next-hack Oct 12 '17 at 10:00
  • \$\begingroup\$ I just know that in parallel circuits there's more than one path for current to flow.Thus the current is divided but voltage is the same. But the circuit is a bot complex for me to point out which is parallel \$\endgroup\$ – Sook Lim Oct 12 '17 at 10:30
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    \$\begingroup\$ The definition of parallel is not "there is more than one path for current", but that the same voltage is applied to the two bipoles (2-terminal components). These two definitions are not equivalent. \$\endgroup\$ – next-hack Oct 12 '17 at 11:05

In schematics, we can think of resistors and other components as the roads and wires as nodes or forks in those roads.

Note I said wires, not "connections". If you look at the schematic on the left below, you may think it has two connecting nodes. However that is just an illusion created by drawing components orthogonally. The circuit can be drawn equally correctly using the diagonal resistors as shown on the right. Here it is clearer that there is only one connecting node.


simulate this circuit – Schematic created using CircuitLab

Think of a node as a fork in the road...

enter image description here

Current will split between each path at a fork, but that does not necessarily make the roads parallel.

If you have a fork in the road, and both paths ONLY lead you to the same place, then it wont matter which fork you take. The paths are logically parallel.

One may be a harder path and go much further, but they are still logically parallel.

This can be confusing in some schematics, however, it is important to look for forks and paths that start and end at the same point with no intermediate forks on either path that lead you to a different destination.

Look at the schematic on the left below


simulate this circuit

You may now see that there are two nodes A, and B. At fork A, you can go down either the R1 path OR the R2 path and you will end up at node B. The opposite is also true, though maybe a little harder to see.

Once you see that you can quickly rearrange your circuit to show that R2 and R1 are in fact in parallel as shown.

Note I personally draw resistors in parallel as I have indicated on the schematic on the right above just to make it clear to whomever is reading it that I intended them to be thought of as in parallel.

Of course, once you make a combination like that you have to then review the rest of the schematic to see that other paths you can now simplify.

Ultimately, rearranging your schematic into a logical forking flow makes the task a lot clearer.

enter image description here

  • \$\begingroup\$ So I'm sure R7||R4.Are the others all in series? \$\endgroup\$ – Sook Lim Oct 12 '17 at 19:22
  • \$\begingroup\$ @SookLim take it step by step! \$\endgroup\$ – Trevor_G Oct 12 '17 at 19:29
  • \$\begingroup\$ If nodes A and B are one node and nodes D and e are one node,then the only two resistors that share common nodes is R7 and R4 though \$\endgroup\$ – Sook Lim Oct 12 '17 at 20:25
  • \$\begingroup\$ @SookLim individually yes, but looking at the chart above... can you see R5||(R3+R6) etc. \$\endgroup\$ – Trevor_G Oct 12 '17 at 20:26
  • \$\begingroup\$ Then R2 ||(R5+R3+R6)? \$\endgroup\$ – Sook Lim Oct 12 '17 at 20:56
I think there are 6 nodes

No, there are not. Look more carefully. Any places that are connected to each other by just a line are the same node.

This kind of assignment is deliberately drawn to obfuscate the circuit. That's to make you think what is really connected to what and how, and to get you used to looking at crappy schematics. For now, start such problems by redrawing the schematic. Have power voltages descending down the page, and logical flow left to right.

In your case there is no logical flow. Start by considering A high voltage, B ground, and redraw accordingly. In your case, all the resistors should be vertical.

For further help, post the properly redrawn schematic. Of course once you do that, you may see parallel and series combinations popping out at you and you won't need additional help. That's fine, but don't come back asking more about this problem until you have redrawn the schematic with proper easily-readable layout.

  • \$\begingroup\$ Wait so nodes A and B are actually one node so it's 5 nodes right? \$\endgroup\$ – Sook Lim Oct 12 '17 at 14:12
  • \$\begingroup\$ "so nodes A and B are actually one node". Yes. "so it's 5 nodes". No. Look more closely. \$\endgroup\$ – Olin Lathrop Oct 12 '17 at 14:15
  • \$\begingroup\$ D and C are a node as well? \$\endgroup\$ – Sook Lim Oct 12 '17 at 14:49
  • \$\begingroup\$ D and e* I mean \$\endgroup\$ – Sook Lim Oct 12 '17 at 20:21

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