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schematic

simulate this circuit – Schematic created using CircuitLab

The voltage across R ohm should be always V volts for every single value of R except R=0 and at that resistance the current through R needs to be infinite to get rid of all energy and reduce the potential energy to become same with the negative terminal (or 0 joule).

So, at R=0, voltmeter reading will be, V = IR = infinity x 0 = 0. But isn't it true that if all the resistance are connected in parallel with a voltage source the voltages across all resistors should be same? Then why at R=0, voltmeter will read 0 volts and not v volts? I mean for as small value of R as you can think, voltmeter will read V volts then why not for R=0?

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    \$\begingroup\$ You are discussing ideal components which don't exist. \$\endgroup\$
    – Transistor
    Jan 9 '20 at 23:12
  • \$\begingroup\$ If you draw this circuit with R=0 ohms and V = 5 V (for example), it doesn't make any more sense than writing the mathematical equation 5 = 0. What will the voltmeter read in the comparable physical situation? It depends how close to 0 the resistance really is, and how close to an ideal voltage source the source really is. Trying to talk about it in the abstract (an ideal voltage source and an ideal 0 ohm resistor) simply isn't sensible. \$\endgroup\$
    – The Photon
    Jan 9 '20 at 23:17
  • \$\begingroup\$ Hmm but even with non ideal components if I short circuit that R part with a wire and the measure voltage cross it's terminals I read voltage close to 0 volts but not 0. But why is it so? the shorting wire has a some resistance and though it is extremely low, I should read V volts across it. \$\endgroup\$ Jan 9 '20 at 23:17
  • \$\begingroup\$ Related: Different and opposing voltage sources? \$\endgroup\$
    – The Photon
    Jan 9 '20 at 23:18
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    \$\begingroup\$ @AbhirupBakshi I'll take you seriously. Conductance \$G=\frac{1}{R}\$. So \$I_R=V\cdot G\$. The voltage measured by the meter is \$V_{_\text{METER}}=I_R\cdot R=V\cdot\frac1{R}\cdot R\$. Tell me, how does the following mathematical expression behave?$$V_{_\text{METER}}=\lim_{R\to0^+} \left[V\cdot\left(\frac1{R}\right)\cdot R\right]$$It's valid for all values of \$R\$ except at \$R=0\$, when the result is \$V\cdot\frac00\$ (I'm sure you know that \$\frac00\$ is mathematically indeterminate.) Note this is a purely mathematical question. Yes? \$\endgroup\$
    – jonk
    Jan 10 '20 at 4:28
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Your hypothetical circuit is invalid...the normal rules of circuit analysis can not be used.

The definition of a short circuit is a path that has zero voltage across it regardless of current. Our definition of an ideal voltage source is an element that has some constant voltage (nonzero in your example) across it regardless of the current through it. Our definition of parallel elements requires that both elements have the same voltage across them.

So, if you connect an ideal short circuit in parallel with an ideal, non-zero voltage source then you have violated one of these definitions. It's a nonsensical circuit. Don't try to make sense of it.

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  • \$\begingroup\$ But short circuit means no resistance electrical path between an voltage source ideally. But lets just we forget about the short circuit thing and ideal components. In real world, I have a 12v supply and if I connect any resistor (trying to say high resistance than a wire) with it I'm perfectly reading 12 volts across it. But if I connect a wire instead of a resistor I'm getting very low voltage (almost 0.8v) across it and not 12 volts. Why so? \$\endgroup\$ Jan 9 '20 at 23:59
  • \$\begingroup\$ @Abhirup Because you are exceeding the capability of the power supply. If a 12V supply has a current limit of 1 amp, then, yes, its voltage will be constant as long as the load resistance is more than 12/1 or 12 ohms. Once the load resistance drops below 12 ohms, the supply will, depending on its design, reduce is voltage to keep the current at 1 amp or turn itself off. In any case, it will not supply 12 V across any load as you seem to think it should. \$\endgroup\$
    – Barry
    Jan 10 '20 at 0:02
  • \$\begingroup\$ actually its a transformer (220v to 12v) not psu. But I thought about that and just realized what was happening that the total resistance of my transformer and all the wires except the wire which I'v used to short circuit, is much greater than that short circuiting wire. that's why almost all voltage was dropping across those wires and a little amount was dropping in in that short circuited wire. and in the ideal wire scenario if the source would have short circuited the I would read 0v as that infinite current should have been needed to reduce the potential from 12v to 0v. \$\endgroup\$ Jan 10 '20 at 0:16
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I remember having similar confusion learning about analog electronics the first time. The book and my calculator would show some value, and the real-world circuit on a breadboard would be a little different.

It's convenient to think of those black lines on a schematic as a perfect conductor, a perfect wire. But in reality, even the wires connecting components together have some small resistance. It doesn't matter if they're "oxygen-free copper" or solid gold or even a very expensive super-conductor chilled to extremely low temperatures: they will all have some small resistance.

If somehow you could achieve a perfect "zero ohm" resistance, the current would necessarily have to approach infinity, which is at best an abstract concept. But let's say you found the world's most hefty power supply. You would only be finding ways to vaporize your low resistance wires and components...

Current equals voltage divided by resistance. If resistance is zero, then you're trying to divide by zero, which is an error.* 1 volt across 1 Ω is 1 amp. Across 1mΩ would be 1 kA. Across 1µΩ would be 1 MA. The current is inversely proportional to resistance. In the real world you encounter limits of equipment and components well before these hypothetical large current values occur.

* Unless you are Chuck Norris. :)

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  • \$\begingroup\$ Yup but if I measure the voltage across that 1µΩ I should get the almost same voltage I applied to it. If I applied V volts I should read basically V volts across it. now this should be true even with smaller resistances as the current is also increases as the resistance drops. But I'm not getting that V volts. If I replace that R part with a wire which have extremely small resistance the voltage across it is I'm reading is zero point something but not V(in my case 12v). the wire has a resistance so why is it happening? It's not happening with high resistances like with resistors. \$\endgroup\$ Jan 9 '20 at 23:45
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    \$\begingroup\$ @Abhirup As long as the load requires less current than the rating of the supply, the supply will output its rated voltage. As you have observed, that is the case for high load resistances. However, if you load the supply with a resistance that requires more current than the rating of the supply, the output voltage, as you also have observed, will drop. This is the nature of real world power supplies. They are only able to supply a given amount of current unlike an ideal supply which, by definition, can supply as much current as it takes to keep the voltage constant. \$\endgroup\$
    – Barry
    Jan 10 '20 at 0:11
  • \$\begingroup\$ actually its a transformer (220v to 12v) not psu. But I thought about that and just realized what was happening that the total resistance of my transformer and all the wires except the wire which I'v used to short circuit, is much greater than that short circuiting wire. that's why almost all voltage was dropping across those wires and a little amount was dropping in in that short circuited wire. and in the ideal wire scenario if the source would have short circuited the I would read 0v as that infinite current should have been needed to reduce the potential from 12v to 0v. \$\endgroup\$ Jan 10 '20 at 0:18
  • \$\begingroup\$ @AbhirupBakshi As Barry said, the voltage on the power supply will drop when the current being pulled exceeds its rating. Either that or it will fail in some fashion. \$\endgroup\$
    – JYelton
    Jan 10 '20 at 0:57
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In comments you wrote,

Hmm but even with non ideal components if I short circuit that R part with a wire and the measure voltage cross it's terminals I read voltage close to 0 volts but not 0. But why is it so? the shorting wire has a some resistance and though it is extremely low, I should read V volts across it.

Because your real-world voltage source isn't an ideal voltage source. Also the wires connecting the voltage source to the resistor aren't ideal 0-ohm wires.

A benchtop voltage source has some maximum compliance current. If you try to draw more current from it, it might switch to a low-voltage output mode (crowbar circuit), or switch to a constant-current mode, or blow a fuse, depending on how its designed.

A battery has energy loss mechanisms that act as if there were a small resistor inside between the "ideal battery" and the visible terminals. As you draw more current from it, the voltage across its terminals drops.

Etc.

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  • \$\begingroup\$ yea but in real world if I short circuit with a wire which have a resistance, though extremely low but it have that, I'm reading very low voltage across it and not V volts(12v in my case). \$\endgroup\$ Jan 9 '20 at 23:50
  • \$\begingroup\$ @Abhirup You are not reading what we are writing. Your power supply is not ideal. It can't supply more current that its rating. Your low resistance load requires more current than the supply can output. Something has to give and what happens, as you have observed, is that the supply voltage will decrease. Real world supplies are not ideal and cannot drive any load. Once the load causes the supply to need to output more than its rated current, its output voltage will fall. \$\endgroup\$
    – Barry
    Jan 10 '20 at 0:06
  • \$\begingroup\$ actually its a transformer (220v to 12v) not psu. But I thought about that and just realized what was happening that the total resistance of my transformer and all the wires except the wire which I'v used to short circuit, is much greater than that short circuiting wire. that's why almost all voltage was dropping across those wires and a little amount was dropping in in that short circuited wire. and in the ideal wire scenario if the source would have short circuited the I would read 0v as that infinite current should have been needed to reduce the potential from 12v to 0v. \$\endgroup\$ Jan 10 '20 at 0:13

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