Since I began studying, I was taught that when analysing a circuit using multiple sources, independent sources could be turned off, meaning that I could transform a voltage source into a short, and a current source into an open circuit and analyze the circuit by how each individual source acts on it.

However, regarding dependent sources, every electric circuits book I've read says the same thing:

"Dependent sources should never turn off when using the superposition principle analysis, they should be left intact since they are controlled by variables somewhere in the circuit".

However recently I've noticed that I can get the same results if I treat the dependent source as a regular independent source. Ive simulated this, compared to the end-of-chapter problem solutions and it all turns out the same. Take the following problem from Alexander/Sadiku's Book as an example, which states:

"Find Vx using the superposition principle"


simulate this circuit – Schematic created using CircuitLab

Normally I would leave the dependent source on, turn off the 4A current source and find Vx1, then I would turn off the 6A current source and find Vx2, add them up and find Vx.

However, if I treat the dependent current source as an independent current source, I get the same results, Here's the procedure I followed:

Turning off both current sources:


simulate this circuit

Finding Vx using Ohm's law results in:

$$Vx1 = -(4Ix / 10) * 8$$ $$Vx1 = -3.2Ix$$

Now I shut down all sources (including the dependent voltage source like if it were an independent source) and leave the 6A source on:


simulate this circuit

Using the current divider formula and multiplying by 8 I get Vx2


$$Vx2 = (2 / 10) * 6 * 8 = 9.6V$$

Finally, I turn off all sources but the 4A current source:


simulate this circuit

Using the same procedure as above, by current divider I find Vx3:

$$Vx3 = (2 / 10) * 4 * 8 = 6.4V$$

Finally, the total voltage Vx is the sum of each individual voltage Vx (Vx1, Vx2, Vx3):

$$Vx = Vx1 + Vx2 + Vx3 = -3.2Ix + 9.6 + 6.4$$

I can find Ix by Kirchhoff's Current Law, the currents in the top node of the circuit are:

$$Ix + Vx/8 = 6 + 4$$

Solving for Ix

$$Ix = 6 + 4 - Vx/8$$

Using the above equation to eliminate Ix from Vx:

$$Vx = -3.2(6 + 4 - Vx / 8) + 9.6 + 6.4$$

Solving for Vx:

$$Vx = -26.667V$$

And this is the simulation I made on Multisim:

Simulation of the circuit

So my question is: What is going on? Why is it "prohibited" to treat dependent sources this way when the results are the same?

  • 2
    \$\begingroup\$ If you turn off both current sources then you have the anomalous situation of Ix equalling 4Ix. This can onlynean Ix is zero. Just a thought . \$\endgroup\$
    – Andy aka
    Commented Apr 21, 2014 at 22:39
  • \$\begingroup\$ Thats once you substitute directly but if you wait until the end to do the substitution then Ix is not zero. Again, I think that by doing this, we find out what the contribution of the dependent source to the circuit will be once everything is turned on, if we just stay with what the contribution to the circuit is when everything is off, it will of course be zero. P.S. I just read the article Alfred Centauri posted. ITS AMAZING, it really opens up a world of possibility using superposition just like I did in this example. That article should be mandatory for every EE circuit theory class. \$\endgroup\$
    – S.s.
    Commented Apr 22, 2014 at 4:16
  • \$\begingroup\$ Not a thought, Andy. Simulation give 0. \$\endgroup\$
    – Antonio51
    Commented Jul 4, 2021 at 6:34

3 Answers 3


Superposition of dependent sources isn't prohibited: Superposition of Dependent Sources is Valid in Circuit Analysis.

The author has investigated the presentation of superposition in circuits texts by surveying twenty introductory books on circuit analysis. Fourteen explicitly state that if a dependent source is present, it is never deactivated and must remain active (unaltered) during the superposition process. The remaining six specifically refer to the sources as being independent in stating the principle of superposition. Three of these present an example circuit containing a dependent source which is never deactivated. The other three do not present an example in which dependent sources are present. From this limited survey, it is clear that circuits texts either state or imply that superposition of dependent sources is not allowed. The author contends that this is a misconception.

As a simple example using superposition of a dependent source consider the following circuit:


simulate this circuit – Schematic created using CircuitLab

By superposition, we can write the equation for \$V_x\$ by inspection:

$$V_x = V_s\frac{R_2}{R_1 + R_2} + 5i_x R_1|| R_2 $$

We also have, by inspection

$$i_x = \frac{V_s - V_x}{R_1} $$


$$V_x = V_s\frac{R_2}{R_1 + R_2} + 5 \frac{V_s - V_x}{R_1}R_1|| R_2$$

It's just algebra from here. No need for node equations or mesh equations.

The key to successfully using superposition with dependent sources is the following:

Do not attempt to solve for a numeric answer until the superposition sum has been written.

  • \$\begingroup\$ "To apply superposition to dependent sources, the controlling variables must not be set to zero when a source is deactivated." I think that is cheating. Superimposition states that you must turn off all the generators, so if a controlling variable becomes zero, well the controlled source is zero, end of story. The method provided in the paper is valid but it uses a different approach in my opinion... I just can't properly explain that in English, I'll think of it. \$\endgroup\$ Commented Apr 21, 2014 at 23:04
  • 2
    \$\begingroup\$ @VladimirCravero, the difference is this: The superposition principle says that we can add the numeric results from each individual source to arrive at the correct answer. However, one can generalize the technique to dependent sources if one postpones solving for numeric values until after the superposition sum. This is crucial to the success of the technique. Whether this is 'cheating' is irrelevant - the generalized technique works and, as I've found from experience, allows one to often write the solution by inspection. \$\endgroup\$ Commented Apr 21, 2014 at 23:12
  • \$\begingroup\$ Yeah I am convinced it works, I just need a little to let it settle and understand it. \$\endgroup\$ Commented Apr 21, 2014 at 23:13
  • \$\begingroup\$ YES!! the second comment is exactly what I thougt (read my previous comment to Vladimir's response) I believe that by leaving the numeric results "as a variable" and solve at the end, we are finding out what would the contribution of the dependent source would be to the circuit once we consider all the rest of the sources. . I dont think its cheating, in my opinion, mathematically speaking theres nothing in this case that tells you when you can solve for a numeric answer. \$\endgroup\$
    – S.s.
    Commented Apr 22, 2014 at 3:30
  • 1
    \$\begingroup\$ @JoeM, also, see this paper: eprints.soton.ac.uk/271202/1/superposition.pdf In particular: It should also be apparent that Leach’s admonition “provided the controlling variable is not set to zero when the source is deactivated” is not quite complete; the error is to set the controlling variable to anything other than its value in the full, original circuit. \$\endgroup\$ Commented Apr 23, 2014 at 3:08

The solution lies to the answer of the question, " why we want superposition theorem". the answer is to do the analysis easily. Now if you want to turn off the dependent source turn it off, no problem, and if you find it easy to leave it on, then that is also fine, since , the linearity is followed by the elements and voltage and current function.


The answer is simple: you are making a mistake.

When only the dependent source is on you get \$V_x=0\$. Let's add the voltages on the loop, starting on the low side of \$R_1\$ and going counter clockwise: $$4i_x-R_2i_x-R_1i_x=0 \rightarrow i_x(4-8-2)=-6i_x=0 \rightarrow i_x=0 \rightarrow V_x=-i_xR_2=0$$

You can actually find networks where you can treat dependent sources as independent, imagine a net where a dependent source is a controlled voltage source with only one terminal connected, it's clear that it will not affect the circuit in any way so you can treat it pretty much how you want.

All in all the fact that such examples exist does not mean that such a method is general.

Think of a simple common source amplifier:


simulate this circuit – Schematic created using CircuitLab

If you have to find \$v_d\$ as a function of \$v_{in}\$ and use your method you can easily see that you'll find out that \$v_d=0\$ no matter what \$v_{gs}\$ is, that is wrong.

  • \$\begingroup\$ I dont believe Im making a mistake, a mistake would imply wrong answers, and all answers Ive done doing this method are correct, I understand that if only a dependent source is on the voltage would be zero since a dependent source produces no voltage. However the way I see it is that by doing this approach we consider what the contribution of the dependent source to the circuit will be once its in its "on" state, of course if we stayed only with the dependent source the voltage would be zero. I could state more examples along with the simulation proving that the results are correct. \$\endgroup\$
    – S.s.
    Commented Apr 22, 2014 at 3:21
  • \$\begingroup\$ You are making a mistake because \$i_x=0\$, as Andy said too. \$\endgroup\$ Commented Apr 22, 2014 at 6:05
  • \$\begingroup\$ Yet the result is correct, every time, so again, taking into account what I read, this is not a mistake, and if you still consider it a mistake, then its a positive one. \$\endgroup\$
    – S.s.
    Commented May 22, 2014 at 13:55

Not the answer you're looking for? Browse other questions tagged or ask your own question.