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Suppose there is a ideal voltage source which can provide infinite power if that is needed in an ideal scenario. Now suppose a resister R is connected across it and the e.m.f is E. So by ohm's law the entire E will fall across R, despite of whatever the value of R is; except for 2 conditions,

  1. If R=0, I=E/0=infinity. So E=IR=infinity*0=0, But if the supply gives E volts out and if the charges dissipates it's Total energy by end of the 0 ohm resistance and with the help of infinite current, then how the voltage across it can be 0 and not E?

  2. If R=infinite, I=E/infinite=0. So, E=IR=0*infinite=0. Again the same thing. How can E be zero when at first E was not zero when supplied?

I know in real world nothing is ideal but here I'm taking an ideal case for calculations. And I also know in case 2 if R=infinite or opened the voltage across it will be E but why the equation is not supporting that? And I'm not sure what will be the voltage in case 1. Is there any asymptote in the E vs R graph which is causing those behaviour in two cases?

Abhirup Bakshi
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    If you want pure mathematics - then you should take in account "infinities" of different orders (or work with $lim$ notation and the involved calculus). Anyway, engineering models do not deal with the extreme cases and call such a cases "non-physical". – Eugene Sh. Jan 15 '20 at 20:37
  • I know but can u say why this is happening? and also what do u mean by infinities of different orders? – Abhirup Bakshi Jan 15 '20 at 20:40
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    Infinity can be of different "size". This is a whole field of mathematics and off-topic here. One thing you should understand - "infinity" is not a number and does not obey the regular arithmetic rules. Again, your calculations might make more sense if you rewrite them in the $lim$ notation. – Eugene Sh. Jan 15 '20 at 20:41
  • For example, you say infinity0=0* - which is not necessarily true. Think of $a=x$ and $b=\frac{1}{x^2}$. Now, if $x=0$ then $a=0$ and $b=\infty$. Right? But what about $ab$? Obviously it is $\frac{1}{x}=\infty$. So here we get $\infty\cdot 0 = \infty$ – Eugene Sh. Jan 15 '20 at 20:56
  • Yea its a strange behaviour. need to search more on that topic. – Abhirup Bakshi Jan 15 '20 at 21:02
  • The "behavior" you mention is mathematically pathological. The branch constituent equations we have are (imprecise) models for real-world impedances that we encounter; you mention a pathological condition that can't even arise with ideal superconductors in practice (since even a type-1 Sc has impedance in the form of series L). – nanofarad Jan 15 '20 at 21:52
  • I know that's why I told it's only in ideal case not real...only for calculations. – Abhirup Bakshi Jan 15 '20 at 21:56
  • Does this answer your question? Different and opposing voltage sources? – The Photon Jan 15 '20 at 23:39
  • This question is like asking what happens when an irresistable force is applied to an immovable object. The answer is you've created a logical contradiction. To call one object an irresistable force you must first know that no immovable object exists, and vice versa. – The Photon Jan 15 '20 at 23:41
  • You can connect an ideal short circuit in parallel with an ideal, non-zero voltage source...that is an invalid circuit because you have violated the definition of parallel elements. And multiplying infinity times zero is equally nonsensical. GIGO – Elliot Alderson Jan 16 '20 at 01:43

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Let's try a practical case.

Given a strong 18650 fully charged battery, Li Ion Cell ESR ~ 0.05 ohm 3.7V

If you had AWG 30 magnet wire coiled up to make a little solenoid and somehow it shorted inside, what would be the resistance , R be to make the hottest wire?

Please Cut wire < 1 second, otherwise ...V^2/0.05=273 Watts is explosive. self-heating energy that demands a safety fuse.

What is the short circuit current?

How long would you have before the battery melts down into a fire or explosive hazard?

Don't even joke about it.

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Tony Stewart EE75
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