i am having a problem digesting this circuit. This wire seems wierd and i do not understand how the current flows through this circuit. if we suppose all of them resistors have the same resistance and one needs to calculate the total resistance then how do you go ahead since this wire is blocking me from moving

to the next step:

  1. i thought maybe we can collapse the wire since the points on both ends have the same potential

  2. an infinite electric current goes thru this wire. but where do you go from here?

enter image description here Thanks

  • $\begingroup$ I am very sorry to cause confusion. I today confirmed my answer with a teacher and my answer is wrong. $\endgroup$ Commented Aug 1, 2020 at 8:20
  • $\begingroup$ If you want to understand what was wrong in my answer chat.stackexchange.com/transcript/message/55122158#55122158 read the conversation after it. $\endgroup$ Commented Aug 1, 2020 at 8:33
  • $\begingroup$ thank you very much $\endgroup$
    – Mad
    Commented Aug 1, 2020 at 9:00

2 Answers 2


You need to draw the diagramme nicely (there's a few way... draw it such that you understand the network) and then apply Kirchhoff's law to each separately and calculate the currents.

networks. Left with wire, right without wire

I labelled the unlabeled resistance in your network with 'e' and assume that on the lhs of your diagramme you have the actual voltage source (and that it's not a capacity).

comparing these two diagrammes you see that the addition of the wire will change your network such that you have three parallel wires to consider which share the voltage from the source.

In the original configuration all current passes through resistance 'a' and then branches into two, resistance 'c' and yet another branchy construct where 'b' acts as common resistance to two branches.

  • $\begingroup$ I think you drew with wire structure wrong. $\endgroup$ Commented Jul 31, 2020 at 12:10
  • 1
    $\begingroup$ i.sstatic.net/fyDPQ.png this is what my diagram looks like. It's the diagram with wire one . $\endgroup$ Commented Jul 31, 2020 at 12:14
  • 2
    $\begingroup$ I think I drew that correctly and that the left side of my graphics is electrically equivalent - but easier to understand. And that is the whole point of re-drawing it. Redraw it such, that you see the structure $\endgroup$ Commented Jul 31, 2020 at 12:18
  • $\begingroup$ well.. so whos right? $\endgroup$
    – Mad
    Commented Jul 31, 2020 at 12:58
  • 1
    $\begingroup$ It's useful to distinguish coil and resistors. Should L not be a coil, it is uncomon naming $\endgroup$ Commented Aug 1, 2020 at 10:24

You should indeed try to redraw the circuit without this branch, i.e. by collapsing the points at both ends, as you suggested in (i). Concerning (ii), the electric current is NOT infinite. At the left point, the current I is divided in $I_1$ in the wire and $I_2$ going into $a$. You should solve the whole set of Kirchoff equations to find $I_1$ and $I_2$.

  • $\begingroup$ I redrew it. so in a there is no current but at junction point where the wire connects the other resistors the current will be split and in b and c flows also a current. this means for the calculation of the total resistance we do not consider a but all the others. right? in this case b and c will be connected in series as well as d and L if we sum them then add the sums with the one in the middle in parallel (thus inverse) we get the total resistance right? $\endgroup$
    – Mad
    Commented Jul 31, 2020 at 11:10
  • $\begingroup$ The diagrams drawn by @planetmaker is correct. My conclusion about what Kirchoff equations would give you was wrong. There is indeed a current through $a$ and $b$. $\endgroup$
    – Christophe
    Commented Jul 31, 2020 at 14:09

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