This is not a homework problem. I posted this specific question as it would be easy for me to explain. If a current flows through the wire it encounters a short circuit 1 and after travelling through the short circuit 1, it again enters the main wire. Then it once again encounters a short circuit 2. So does the current enter the short circuit 2 or it just move on to the end. If it goes to the short circuit 2 it will be travelling in a backward direction. Is it possible and if it is, will it not be stuck in a loop forever as once again it would once again encounter short circuit 1 and go backwards once again and then the cycle would repeat. Just explain the concept. No need for answer. By the way it is 37.5 ohms enter image description here


If you don't assume that it doesnt go through the resistors in between then answer doesnt come

Edit 2

I guess whole current doesn't go through a single branch after going through short circuit 1. If I'm correct then how will it be distributed?Applying kvl i came to know that all current will not go to short circuit 1. If I'm correct them how is this possible??


closed as off-topic by G. Smith, Jon Custer, Yashas, John Rennie, ZeroTheHero Jun 6 at 11:03

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  • $\begingroup$ Hint: these kind of problems are meant to trick you by drawing the schematic in a confusing way. You can redraw the schematic so that you shorten the lines marked "I" and "II" and it will be much more clear which resistors are in series and which in parallel. $\endgroup$ – The Photon Jun 1 at 16:03
  • $\begingroup$ My concern is if i let it to be like this only then what will happen to current not any other trick to reduce the circuit as i have already got the desired answer by applying it. I want a proper logic for the current in this figure $\endgroup$ – Satwik Jun 1 at 16:08
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    $\begingroup$ I think the best way to analyze this problem and solve for the current is by using Kirchoff's Laws. $\endgroup$ – Tausif Hossain Jun 1 at 16:10
  • $\begingroup$ Yes i considered that option but applying that doesn't give us current in the wires that short circuit as they have no resistance $\endgroup$ – Satwik Jun 1 at 16:16
  • $\begingroup$ Yes redraw it so the wires do to cross and it will be more obvious. In this problem there is no negative current. $\endgroup$ – PhysicsDave Jun 1 at 17:12

The currents circulate like in the figure. There are no infinite loops or anything else out of the ordinary. If you apply some specific voltage you can calculate the values of the currents, with or without redrawing the circuit.

enter image description here

  • $\begingroup$ So that means these seeming short circuits aren't short circuits after all. $\endgroup$ – Satwik Jun 2 at 11:40
  • $\begingroup$ physics.stackexchange.com/questions/346650/…. This link addresses the same kind of problem and supports the answer of nasu that these aren't short circuits. For more understanding you can visit this question $\endgroup$ – Satwik Jun 2 at 11:44

enter image description here As it has been pointed out already , you just need to simplify the circuit , the wires 1 and 2 have no resistance in them so the ends of those wires will be at same potential , you can , for your circuit interpret this as the two ends being the same . Following the above route , you can simplify your circuit easily.

  • $\begingroup$ I just read you do not want a simplification, let me think of something. $\endgroup$ – ADITYA PRAKASH Jun 1 at 18:03

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