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.

enter image description here


closed as off-topic by John Rennie, Jon Custer, GiorgioP, stafusa, ZeroTheHero Jun 29 at 11:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, Jon Custer, GiorgioP, stafusa, ZeroTheHero
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Each branch will contribute 1A. I + I = x and x = 2A and I=I. $\endgroup$ – PhysicsDave Jun 1 at 17:08

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Not the answer you're looking for? Browse other questions tagged or ask your own question.