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I am still trying to understand the meaning of an ideal current source. To this end, I drew this abstract circuit above. Let's say the current source is ideal, and the voltage symbols you see represent ideal voltmeters, reading the voltage over their respective terminals. Va reads the voltage drop over I, the ideal current source, and Vb over the short circuit wire.

I understand that ideal current sources are supposed to provide a current independent of the voltage across it. That should mean no internal voltage drop in I, due to no internal current and therefore an infinite internal resistance. So, according to Ohm's law (Eq. 1), $$ V_a = IR = 0 \times \infty = 0 $$

Is Eq. 1 correct? Isn't 0 x infinity a bit of a definition problem? Perhaps Va should be "undefined"?

Similarly, for Vb, there should be a non-zero current I traveling through the short circuit, which has 0 resistance. Therefore (Eq. 2):

$$V_b = IR = I \times 0 = 0$$

Is Eq. 2 correct? Is the short circuit current really the non-zero constant I provided by the ideal current source? Perhaps I over the short circuit is infinite, as R = 0, but Vb represents voltage drop, and there is no drop in voltage over a short circuit, so I should remain the non-zero constant provided by the non-zero ideal current source. Does that imply that voltage at each of Vb's terminals is (the constant) infinity?

As you can see, I'm a bit confused.

  • $\begingroup$ Where does the infinity in Eq 1 come from. You have a finite I and zero R. There shouldn't be any infinity in that equation. $\endgroup$ – The Photon Jun 18 '20 at 15:36
  • $\begingroup$ @ThePhoton Since there is no voltage drop over an ideal current source, and no internal current leak, R could be anything really and V could still be 0. But I had understood that it was infinite in the ideal case, since internally, an ideal current source is an open circuit. $\endgroup$ – cccube Jun 19 '20 at 18:23

the voltage symbols you see represent ideal voltmeters, reading the voltage over their respective terminals. Va reads the voltage drop over I, the ideal current source, and Vb over the short circuit wire.

The two voltmeters are connected to the exact same two (actually one) circuit nodes, so they will read the exact same value.

So, according to Ohm's law (Eq. 1), $$ V_a = IR = 0 \times \infty = 0 $$

The current source is connected in parallel with the zero-ohm wire. Therefore the resistance that applies here is the parallel combination of the source's internal resistance and the wire. Since 0 ohms in parallel with any other resistance is 0 ohms, in this equation we should have $R=0\ \Omega$.

Then there is no infinity involved and your question is moot.

It may make it easier to understand this circuit if instead of a short-circuit wire, you place a resistor across the source. Then you can use the mathematical tools of limits to find out what happens if the resistor value approaches zero or increases toward infinity.

For example, the short circuit case can be studied as

$$V=\lim_{R\to 0}IR$$

Since $I$ is a finite constant value, this limit is very easy to evaluate.

The case where the resistance goes to infinity (representing an open circuit rather than a short circuit) is more challenging, and can in fact not be solved. It represents a situation where your model is inadequate to describe any real physical situation. Because in reality there is no such thing as an ideal current source or an ideal open circuit (although it is generally the current source that is "less ideal" than the open circuit).


In comments you asked,

In the ideal state, where internal resistance of the current source is infinite, does Ohm's law still apply?

Ohm's law applies to resistors, and defines what it means when we model something with a resistor.

I think you're forgetting something very basic, like that the internal resistance of a current source is in parallel with the ideal element, not in series. If you have a non ideal current source and a resistive load, the model looks like this:

strong text

Here, I1 is the "ideal part" of the current source, and R1 is the source's internal resistance. R2 is the load resistor.

Using the current divider rule, you should see right away that the closer R1 gets to infinity, the more of the current from I1 goes through R2. That is, the closer this is to an ideal current source, delivering exactly the same current to the load regardless of the load resistance.

If you take the limit as R1 goes to infinity, and R2 goes to zero (you make the current source ideal, and the load into a short circuit), then the current through R2 goes to I, the source current. And, by Ohm's law, the voltage between nodes a and b goes to zero.

Whether you put the voltmeter measuring the voltage between a and b on the left side of your schematic or the right is immaterial. Whether you have one ideal voltmeter or two or three hundred is immaterial as they are ideal voltmeters that don't affect any of the voltages or currents in the circuit.

The point is that making the current source internal resistance higher makes the output voltage closer to $I_{source}\times R_{load}$ (what you would expect from Ohm's law) rather than farther from it.

  • $\begingroup$ First, I really appreciate your taking the time to answer my question. I am not entirely sure about your answer. In the first case, regarding Eq. 1, I think an ideal current source is supposed to have infinite internal resistance (see my comment above). Therein lies the issue. The voltmeter over the short wire we agree: V = 0 because R = 0 and I is fixed. But the voltage over the current source is less clear. Here I think R is approaching infinity, but I am not sure. $\endgroup$ – cccube Jun 19 '20 at 18:27
  • $\begingroup$ @cccube, the current source has infinite internal resistance. But there is another device (the shorting wire) connected in parallel with it. So when you measure voltage across it, you measure the effect of both the source, and the other device. Your two voltmeters are connected to the same two points in the circuit, so they are measuring the exact same thing. $\endgroup$ – The Photon Jun 19 '20 at 18:41
  • $\begingroup$ If you want to consider the case where you measure the voltage across the current source before you connect the shorting wire, and then measure again after connecting the shorting wire, then see my last paragraph. $\endgroup$ – The Photon Jun 19 '20 at 18:42
  • $\begingroup$ Well... I was thinking of both cases, simultaneous connection of the two voltmeters and sequential, but wanted to start with the sequential to make sure I understood what you were saying. So let me first start with the sequential again, and go to your last paragraph. I am still not sure what the voltmeter Va alone would read over the current source (internally). I know physically this would not exist but this is a thought experiment, so thank you for bearing with me. Would Va read "undefined"? $\endgroup$ – cccube Jun 19 '20 at 19:01
  • $\begingroup$ @cccube, if I wrote the mathematical equation "5=0", what would be the solution? Asking what voltage an ideal current source produces into an open circuit is equally nonsensical/impossible. $\endgroup$ – The Photon Jun 19 '20 at 19:34

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