# Is there a paradox in the superposition principle?

I came across this problem for the first time when I started thinking about the clever circuit of Howland current source (see What is the brilliant idea behind the Howland current source?). Let me explain what this is all about with a simple example.

Imagine a simple current source assembled by a voltage source V1 and resistor R. At best, it would be short connected, and the current would be simply I = V1/R.

Unfortunately, it is loaded by another voltage source V2 with lower voltage... and the current decreases. We can explain why in two ways - in terms of voltages and in terms of currents.

First we can say that, according to KVL, the effective “current-creating” voltage VR = V1 - V2 across the resistor R decreases… аnd this decreases the current I = (V1 - V2)/R. That is, the current decreases since another opposing voltage is subtracted from the initial voltage.

But with the same success we can say that, according to the superposition principle, the current decreases since another opposing current I2 = V2/R is subtracted from the initial current I1 = V1/R.

If we replace the voltage source V2 with a resistor R2, we can continue to think the same way… but in the second case we will not be able to make physical sense of what is happening.

That is right, because the superposition principle assumes at least two sources… and here there is only one. The load voltage V2 cannot create current I2 = V2/R1 since it is zero when V1 is zero. That is, this statement makes no physical sense.

At the same time, the principle of superposition is formally valid… and this is the paradox…

I still have an idea of ​​how to see physical meaning. If we measure V2 and replace the resistor R2 with a voltage source with the same voltage, nothing will change… but it will make some physical sense.

• Superposition with one source Is wasting time and of course, with one source, I2 cannot exist. Commented Oct 26, 2019 at 14:31
• Sorry, I don't see any paradox. I don't think you understand how to use superposition. If there are $N$ independent sources you construct and analyze $N$ circuits, where each circuit has just one of the independent sources while the other sources are deactivated. If there is just one source you have a trivial case: you analyze one circuit with that source remaining active. Commented Oct 26, 2019 at 14:49
• Exactly. If you have V1 and V2 you have 2 sources and superpostion applies. If you have V1 and R2 you have 1 source and superposition is useless (but does still apply) Commented Oct 26, 2019 at 14:54
• I agree with all of you that in the latter case (resistor load) the principle of superposition is useless for calculation purposes. In my opinion, this is because the principle requires voltage sources be independent and with a specified voltage value. Even if we replace the voltage drop of VR2 with equivalent voltage of another source (I remembered that there is such "substitution theorem"), there would still be little use for the purposes of calculations because we will need to calculate this drop beforehand… Commented Oct 28, 2019 at 16:26
• ... The point is that I am interested in the principle of superposition not so much for the purposes of calculations but rather for understanding and explaining circuits. And in this case (Howland current source) I wanted to find out what exactly this V/R1 is in the expression I = V1/R1 - V2/R1. It gave me the impression that, even in this case, the principle was valid ... and that seemed puzzling to me. That's why I asked this question. I'm sorry if it seems too naive and elementary to you... but I put up with the fact that thinking at the lowest level in circuitry is a thankless job... Commented Oct 28, 2019 at 16:27

Sorry, I don't see any paradox. I don't think you understand how to use superposition. If there are $$\N\$$ independent sources you construct and analyze $$\N\$$ circuits, where each circuit has just one of the independent sources while the other sources are deactivated. If there is just one source you have a trivial case: you analyze one circuit with that source remaining active.