I have a super simple question about using superposition to solve this circuit and it looks like I am running into a misconception about why my analysis is wrong.

I have attached a simple circuit with two voltage sources of equal magnitude in parallel and a simple resistor for a load in parallel as well. If we do a simple nodal analysis, it's clear that the current through the resistor is V/R and the current drawn from each source is V/2R.

Now consider we use superposition to solve this problem. First, we would turn each voltage source off individually by substituting it with a short circuit and only leave one voltage source on and find the effect due to the source by itself. In the end, we would sum all of the results up to get the total result. So if we turn the left source off, it will create a short circuit to the ground and we would get an infinite current to GND.

Now if we turn the right source off, the same thing happens and we get infinite current shorted to GND. In both cases, no current flows through the resistor and if we sum the results up at the end we will get the current through the load is 0A.

What is happening here? How come we are not able to use superposition to solve this simple circuit? It seems to be both linear and bilateral.

enter image description here

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    \$\begingroup\$ It violates real circuits which must have a small ESR and V's must be matched closely. \$\endgroup\$ Commented Dec 2, 2021 at 5:10
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    \$\begingroup\$ This is why ideal components can't exist. For every ideal thing there's a situation you can put it into where things don't make sense and break down. It is like you've taken two ideals: unstoppable force and immovable object and asked what happens when they meet each other. \$\endgroup\$
    – DKNguyen
    Commented Dec 2, 2021 at 5:11
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    \$\begingroup\$ Two voltage sources in parallel is an undefined situation. An ideal voltage source is a mathematical construct that says that the potential difference between the two nodes is a certain value. This is the equivalent of saying that x = V1, and x = V2. It only makes mathematical sense if V1 = V2 with an infinite resolution, which is physically not possible. If you did this in real life you would have the voltage "fight" to try to set the node to their voltage. \$\endgroup\$ Commented Dec 2, 2021 at 5:13
  • \$\begingroup\$ While applying superposition, you will short the other source, and what will the rest of the circuit look like? \$\endgroup\$
    – Syed
    Commented Dec 2, 2021 at 6:39
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    \$\begingroup\$ You don't need this circuit for paradox scenario, take an ideal voltage source, short it and apply KVL. Boom! \$\endgroup\$
    – Mitu Raj
    Commented Dec 3, 2021 at 5:38

5 Answers 5


The circuit isn't as simple as it looks, as it contains an impossible situation--two voltage sources in parallel.

The two voltage sources are both trying to assert a voltage between the same two points, but the voltage between two points must be unique--and here, you're asserting that it has two values at the same time! If you were to build this in reality, you'd have a very large current circulating through the two voltage sources, until something blew a fuse.

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    \$\begingroup\$ I don't see anything in the question that says that the two voltage sources have different values. On the contrary, I think they are expected to have the same value, based on other statements made by the OP. Placing to ideal voltages sources in parallel is fine if they have the same value, but of course they should usually be combined to a single source. \$\endgroup\$ Commented Dec 2, 2021 at 14:58
  • \$\begingroup\$ @ElliotAlderson When superposition is performed, they must necessarily have different values during the solution process even if equal in the initial formulation of the problem - only one (or a subset) gets turned on at a time, and then the results superimposed. That's the crux of the issue, which is hinted at by this answer, but not acknowledged explicitly. \$\endgroup\$
    – nanofarad
    Commented Dec 2, 2021 at 17:06
  • \$\begingroup\$ But that's not what you said. You said that "two voltage sources in parallel" was "an impossible situation". It's not impossible, but they must be combined before performing superposition. \$\endgroup\$ Commented Dec 2, 2021 at 17:09
  • \$\begingroup\$ @ElliotAlderson Okay, I did neglect to say that the problem is different if you have two equal-magnitude sources in parallel, but there is still a problem: you end up with a degree of freedom in your solution, because the current flowing in the loop that contains only the two voltage sources could be any magnitude at all. I don't have time right now to edit that into the answer, but I'll come back to this later. \$\endgroup\$
    – Hearth
    Commented Dec 2, 2021 at 20:28

Other people gave explanations on why your model is flawed and hence produces weird results.

I'll give you a mathematical explanation, instead.

TL;DR: Simply put, you applied superimposition in a subtly wrong way!

Superimposition theorem cannot be applied in your case.

In the following I'll explain what's the catch.

The problem is that many textbooks don't actually teach you the exact, mathematically rigorous, formulation of the theorem (which is a mathematical theorem of circuit theory). They simply state that superimposition can be applied to any linear circuit, possibly hand-waving away any subtle corner case.

I couldn't find a ready reference online (hep!), so I had to resort to my trusty copy of the seminal book of Desoer and Kuh "Basic Circuit Theory" (1969). Alas it's a 1991 Italian reprint, so I can't quote the exact theorem in English as the authors wrote it.

Suffice it to say that the theorem has a very important hypothesis which most textbooks neglect: the circuit must have a single zero-state solution, whatever the waveform of ALL the independent sources might be.

Since your circuit has no state (being purely resistive, without any energy-storage elements), it's behavior can be determined using just algebraic equations (that is, no differential equations). Therefore that requirement about the zero-state solution just boils down to the circuit having just a single solution for every possible waveform of the two (equal) generators.

Maybe you might be surprised that a linear circuit might have multiple solutions, but that often happens with "pathological" circuits like yours (usually they are extremely idealized models of real circuits).

Anyway, your circuit fails to comply with that requirement.

In fact, let's call the quantities in the circuit like this:

  • V: voltage across the three elements (polarity upward);
  • Is1: current through 1st generator (direction upward);
  • Is2: current through 2nd generator (direction upward);
  • Ir: current through resistor (direction downward).

KVL is trivial here, so it gives us no useful equation. KCL gives us the equation \$ I_r = I_{s1} + I_{s2} \$. Together with Ohm's law, \$ V = R \cdot I_r \$, those are all the equations of the circuit.

That set of equations has no single solution, since any pair of \$ I_{s1}, I_{s2} \$ values whose sum equals \$ I_r = \frac V R \$ will satisfy the system!

You might be tempted to say that, for symmetry Is1 must be equal to Is2, but that is just a physical consideration which has nothing to do with math (perfectly valid for practical circuits, but useless when proving math theorems). Is1 and Is2 are completely independent from one another, as far as circuit theory goes. Otherwise either one or the other would be a dependent source (and those cannot not be switched off during superimposition application)!

Hence you cannot apply the superimposition theorem to this circuit, because it has no single solution!

  • \$\begingroup\$ I see what you are saying here but what if I add a resistor at the positive output of each voltage source now? I still have a state with Is1 + Is2 = Ir but this circuit CAN be solved using superposition. Is there a single zero-state solution if I add the resistors in? Does adding the resistors force my Is1 and Is2 to be an exact value and hence superposition works? \$\endgroup\$
    – maxonezhou
    Commented Dec 3, 2021 at 1:47
  • \$\begingroup\$ @maxonezhou I think that this circuit "has" a "single" solution. The problem is just between the 2 sources ( "infinite" internal current which is physically not realistic, but mathematically acceptable, just use internal resistance as "low" as you want. If the internal resistors are equal, the voltage across R is always (E1+E2)/2 ). This is why I used the "limit" concept. \$\endgroup\$
    – Antonio51
    Commented Dec 3, 2021 at 8:30
  • \$\begingroup\$ I see, in the beginning .... that for "node analysis" ... "-- as long as no independent voltage sources form a loop .... ". it is the same limitations that occur with a simulator! We can't make a schematic where we find a loop of two voltage sources, etc ... This is a message error from the simulators ... and this is gone when "it" says that it add "resistor" at the litigious point. \$\endgroup\$
    – Antonio51
    Commented Dec 3, 2021 at 8:49
  • \$\begingroup\$ @maxonezhou Yes, adding a series resistor makes the circuit solvable by superimposition. The problem is not the equation Is1+Is2=Ir, that's KCL. The problem in the original circuit is that there is no other equation that constraints Is1 and Is2. The overall equation system has no single solution. Adding those resistors creates two meshes in the circuit, so now you have two more equations from KVL, so you have a single solution (two independent mesh currents, Is1 and Is2 and two independent equations from KVL). \$\endgroup\$ Commented Dec 3, 2021 at 12:03

Update : Hypothesis : for "node analysis" ...

"-- as long as no independent voltage sources form a loop .... ".

It is the same limitation that occurs with a simulator!

We can't analyse a schematic where we find a loop of two voltage sources, etc ...

This is a message error from the simulators ... and this is gone when "it" says that it add "resistor" at the litigious point, or one add resistors ...

But ...

One can "solve" this circuit. Add a resistor (internal impedance for each source).

Apply "superposition", then, take the "limit" when these resistors go to zero ... unless I am wrong ...

enter image description here

If a "mathematical problem" appears, then it is obvious that something (internal impedance) is really missing. NB: Do you remember that some "functions" (statistical or others) can be defined at "one point"? Example: Dirac pulse?

This "operating mode" is the same applied when you discharge a "charged" capacitor into another "discharged" capacitor (capacitors have equal value capacitance). When making the "balance" of energy stored into the two capacitors (at the end) and the energy stored in the first one initially ... there is a "loss" of half the initial energy, which is "disappeared" ... in the wire, whatever resistance is ...


I have attached a super simple circuit with two voltage sources of magnitude in parallel and a simple resistor for a load in parallel as well. If we do a simple nodal analysis, it's clear that the current through the resistor is V/R and the current drawn from each source is V/2R.

I think there is a key word missing in your statement: -

two voltage sources of \$\color{red}{\boxed{{\text{equal}}}}\$magnitude in parallel

I mean, why would you say "two voltage sources of magnitude in parallel"

So, if they have equal magnitude, then they morph into one voltage source. I say this because it is foolish to start trying to apply a circuit theorem without taking 10 seconds to look for simplifications. This means that your circuit becomes this: -

enter image description here

And clearly, Ohm's law is the most appropriate circuit theory to use.


Perhaps another way to consider this circuit: Since the 2 voltage sources must be equal to one another, they are not independent of each other. So this circuit has only one independent voltage source, say V1 = V on the left, and one dependent voltage source, say V2 = mV on the right where m = 1.

Therefore V2 cannot be turned on/off independently of V1. Superposition is inapplicable as there is only one independent source.

  • \$\begingroup\$ That is not a "dependent voltage source" \$\endgroup\$ Commented Dec 16, 2021 at 23:24
  • \$\begingroup\$ The circuit as depicted would not make sense unless the 2 ideal voltage sources are the equal, which means they cannot be independent of each other. Hence one must be a dependent voltage source. In other words, the circuit symbol for one of the two voltage sources is inconsistent and should have been depicted as a dependent source that takes the same value as the other, independent voltage source. Then the circuit diagram will be consistent with the problem and make sense. This is just one simple way to resolve the apparent inconsistency or contradiction in the original analysis. \$\endgroup\$
    – Lim
    Commented Dec 17, 2021 at 0:32
  • \$\begingroup\$ _ one must be a dependent voltage source_. Or, more likely, neither one is, and sparks will fly. \$\endgroup\$ Commented Dec 21, 2021 at 19:10

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