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In context to the diagram below , it is quite easy to find the current flowing through section AD , which turns out to be 2A , however , I cannot determine the current in any other branches with Kirchoff's Voltage Law . The branches AB,CD,DE and FA have no resistance across them so whenever I take those branches in any of the the possible loops , the potential drop/rise will always be 0 so the terms involving 'I' never appear and consequently they are indeterminable.So how do I find the magnitude of 'I' ?
In this particular case I can guess than I=1A with junction rule and the symmetry of the circuit - the batteries have same emf, so what if the emfs aren't same ? then the original problem comes up again.

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  • $\begingroup$ Each branch will contribute 1A. I + I = x and x = 2A and I=I. $\endgroup$ – PhysicsDave Jun 1 at 17:08
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As modeled, it's totally undetermined. $I$ could be any finite value, positive or negative, with no limits.

In the real world, it would be determined by the non-ideality of the two voltage sources. They won't both be exactly 10.000000 V, and they will both have non-zero equivalent internal resistance. Exactly how much would depend on how these sources are physically implemented.

what if the emfs aren't same?

Then you have a mathematically contradictory model. Say source BC has 10 V, and source FE has value 7 V. Then your schematic is equivalent to writing the equation

$$10\ {\rm V}=7\ {\rm V}$$

which is simply a mathematical false statement.

In the real world, the contradiction would be resolved by the sources having non-zero equivalent internal resistance.

In comments you asked,

How am I supposed to interpret the mathematical invalidity[?]

The mathematical impossibility tells you your model is non-physical. It can't represent any real physical system.

Something is supposed to happen when I connect the circuit even if the set up is ideal , or not ?

Nothing stopped me from writing down "10 V = 7 V". But there's nothing I can learn from writing that down, since it is simply a nonsensical statement. Same thing with drawing a circuit schematic that puts two ideal voltage sources in parallel. It's a nonsensical drawing that tells you nothing about any real circuit.

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  • $\begingroup$ so maintaining the idealistic setup , no current is supposed to flow through 5-ohm resistor ? How am I supposed to interpret the mathematical invalidity , something is supposed to happen when I connect the circuit even if the set up is ideal , or not ? $\endgroup$ – ADITYA PRAKASH Jun 1 at 16:12
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    $\begingroup$ @ADITYAPRAKASH, no, I'm following the conventions you established in your schematic. $x$ is 2 A. The current flowing through one source is $I$ and through the other source $I-x$ (flowing in the positive terminal of the source). $I$ could be -1000 A, +2,000,000 A, or any other finite value. $\endgroup$ – The Photon Jun 1 at 16:17
  • $\begingroup$ @ADITYAPRAKASH, the mathematical error is telling you your model is non-physical. As I mentioned in the answer, to make your model match a physical system, you could add non-zero internal resistance to your model for the sources. $\endgroup$ – The Photon Jun 1 at 16:18
  • $\begingroup$ Ok I got your point , even if both voltage sources are 10V each 'I' can still not be determined , I thought I could depend on symmetry there and I did not know that two ideal emf sources of different voltage cannot be attached in parallel , just realized. Thank you $\endgroup$ – ADITYA PRAKASH Jun 1 at 16:32

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