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I want to calculate the current flowing in the node A of the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

This can be approached in several ways.

  1. The tension in a short circuit is 0 but what about the current ? By Ohm's law \$I = {V \over R} \$ if R approach 0 and \$V > 0\$ then I approach infinity. If \$V = 0\$ as in a short circuit then this lead to \${0\over 0}\$. Both forms has no meaning in mathematics and are undefined.

  2. From the parallel resistors formuala, a resistor in parallel with a short circuit is equivalent to a short circuit. So if the entire circuit is equivalent to a shorted voltage source. The voltage provided by the source is \$V > 0\$ but in a short circuit \$V = 0\$. So again an undefined rational number.

  3. By KCL the sum of the currents in node B must be 0. Maybe we can consider V1 and R1 in series if the short circuit is ignored. If that's the case the sum of the currents is something like \$I - I + I_s = 0 \Rightarrow I_s = 0 \$. Is it a valid reasoning ?

How can I solve this problem ?

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  • \$\begingroup\$ Some people consider it in form U=R×I so when R is 0 due to short circuit, voltage over the load R is also 0 regardless of current flowing or not. Also the form I=U/R can be considered, because current can't be defined when R is 0. It is just not practical to actually calculate these beyond mental exercise. \$\endgroup\$ – Justme Jan 11 at 12:26
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    \$\begingroup\$ We always encounter this kind of contradiction if we have to deal with the ideal voltage/current source and ideal wires. In the real world, this is impossible so why you bother about this? \$\endgroup\$ – G36 Jan 11 at 12:27
  • \$\begingroup\$ @G36 Because when I apply the superposition theorem I end up with this kind of circuit quite often. The voltage source might be a current source but the contradictions came up anyway. \$\endgroup\$ – Bemipefe Jan 11 at 14:15
  • \$\begingroup\$ What if the circuit has another resistor (let's say R2) between B and V1 ? I believe that in this case the circuit has a valid solution. R1 is replaced by the short circuit and the circuit become a loop with R2 and V1 in series. Correct ? \$\endgroup\$ – Bemipefe Jan 11 at 14:34
  • \$\begingroup\$ Can you show the full circuit? And remember that current sources will likes to works in a short circuit. And yes, you will have a valid circuit if you add a resistor between B and V1. \$\endgroup\$ – G36 Jan 11 at 15:19
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The circuit you present is invalid and can not be analyzed using our conventional rules and laws.

An ideal, non-zero, voltage source must have a constant, non-zero voltage across it. An ideal short-circuit must have zero volts across. When you connect two elements in parallel they must have the same voltage across them.

Your circuit must violate one of these rules. Therefore, your circuit is invalid and any analysis you try to do will be nonsense.

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If V1 is an ideal voltage source, its voltage will not drop even at infinite current. The current at A will be infinite, and V1 will be delivering infinite power at its rated voltage.

There's not much point in considering shorted ideal voltage sources, though.

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  • \$\begingroup\$ Your answer is not correct. Calculating the current using a resistance of 0 Ohm results in an equation with 1/0 which is undefined, and not equal to infinity, this is a very important differentiation. \$\endgroup\$ – Vinzent Jan 11 at 12:18
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    \$\begingroup\$ @Vinzent While you are strictly correct, one could say that as the resistance of the short circuit approaches zero, the limit of the current will be infinite. As engineers we usually make that leap implicitly. \$\endgroup\$ – Elliot Alderson Jan 11 at 12:59
  • \$\begingroup\$ @ElliotAlderson "one could say that as the resistance of the short circuit approaches zero" No, one couldn't. Because that is not the circuit diagram that he has drawn, he didn't draw a circuit with a resistor, and then say the value of this resistor is approaching zero, he drew a circuit with a straight connection, and connections in a circuit diagram are not resistances with values approaching zero, they have resistance strictly equal to zero. while you can always insert a resistor with value approaching zero, we have to be able to agree what we mean when we draw a circuit diagram. \$\endgroup\$ – Vinzent Jan 11 at 13:07
  • \$\begingroup\$ @ElliotAlderson I considder this a very important point, as there is a very well known video on youtube with a very famous retired professor Walter Lewin, the video is called "Kirchhoff's Loop Rule Is For The Birds" in this video he argues that you can't trust Kirchhoff's laws, by implying coupled inductances to traces in his circuit diagram, and then concluding that if the diagram is subjected to a changing magnetic field then Kirchhoff's laws don't hold, this is simply bullshit. That is why I am so reluctant to considder connections anything but strictly ideal in all regards. \$\endgroup\$ – Vinzent Jan 11 at 13:12
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    \$\begingroup\$ @Vinzent Calm down. For engineers it is useful and helpful to have an intuition of how a circuit would behave if the there is an almost ideal short circuit...a very small resistance...because that's how the real world works. Saying that the current is undefined provides no such intuition, while saying that the current approaches infinity does provide that intuition. \$\endgroup\$ – Elliot Alderson Jan 11 at 13:16
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The current is undefined

In circuit diagrams voltage sources are ideal, and connections have zero ohm resistance.

meaning this:

schematic

simulate this circuit – Schematic created using CircuitLab

Equates to this:

schematic

simulate this circuit

And according to Ohm's law we have;

$$I=\frac{V}{R}$$

Which in this case is;

$$I=\frac{V}{0}=undefined$$

And dividing by zero is not allowed, it is what we call "undefined".

So the answer to you question is that the current is undefined

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    \$\begingroup\$ You can avoid the "schematics for the deaf" effect by placing a small dot over on the right to force the rendered image to the 640 pixel wide SE image width. Alternatively add an 'l', 'm' or 's' (large medium or small) before the '.png' on the CircuitLab image link (but you'll have to remove it if you need to edit the schematic). \$\endgroup\$ – Transistor Jan 11 at 13:28
  • \$\begingroup\$ @Transistor Thanks, that helps make it a bit more readable (: \$\endgroup\$ – Vinzent Jan 11 at 13:31
  • \$\begingroup\$ @Transistor shud hold a live show'N tell on tricks to make this better \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 11 at 19:24

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