Would it be correct to use use source transformation on a dependent source. For example, would it be legal to turn a voltage dependent voltage source in series with a resistor, into a voltage dependent current source, in parallel with the resistor? Does the regular application of Ohm's law apply.

I've looked around the internet and can't seem to find any definite answers.

  • \$\begingroup\$ Have you tried doing it in one of your exercises? \$\endgroup\$
    – Nogurenn
    Oct 9, 2013 at 7:00
  • \$\begingroup\$ just because it might work for a given problem, doesn't mean it will work in general, so I wanted to be sure. \$\endgroup\$
    – golvok
    Oct 11, 2013 at 0:06

3 Answers 3


Since this is quite within reach I will try to go through the whole story.

Let's take a generic network N and spot one generator (either current or voltage), for now we will not consider it as dependant.

I will name quantities with capital letters, i.e. V, I, meaning DC values, AC phasors, Laplace's variables but also -pushing a little- time domain quantities or whatever can be useful provided it can be handled by corresponding calculation rules.

Noted prerequisite already known from circuit analisys:

  • any voltage generator needs to have a series impedance to be transformable into a current one and dually any current generator needs a parallel impedance to be transformable into a voltage one.

  • "remaining" network N' does not have to linear (even if in this case computations will soon get hard).

enter image description here

we can assert:

  • the two networks N are equivalent provided \$V_\text{k}=Z_\text{k}I_\text{k}\$

  • which means: all quantities in "remanining network" N' are unaffected. This
    includes any voltage (like \$V_\text{rs}\$), any current (like \$I_\text{uv}\$) throughout N' and hence voltage and current at n-m port (\$V_\text{mn}\$ and \$I_\text{nm}\$) too.

Switching to controlled generators and stripping off fancy N' graphics we get to this two circuits

enter image description here

Now we can try to run through all possible control variable:

x is internal to N': no problem at all.
Any quantity inside N' up to the border's ones \$V_\text{mn}\$ and \$I_\text{nm}\$ is not affected by external generator being current or voltage. From inside we cannot tell which is connected.
So in this case controlled generator can be transformed freely.

External control variable: we just have six possibilities -depicted in blue below-, so we can easily run through them all.

enter image description here

What we need to do is relate those six external quantities to internal ones, so we can recover those lost in converting from current to voltage generator.

Current generator case:
we could possibly have \$V_\text{k}\$, \$I_\text{k}\$ and \$I_\text{Zk}\$ as control variable but these can easily be found as $$ \left\{ \begin{align} V_\text{k} &=V_\text{mn} \\ I_\text{Zk} &=-\frac{V_\text{mn}}{Z_\text{k}} \\ I_\text{ik} &=I_\text{nm}-I_\text{Zk}=I_\text{nm}+\frac{V_\text{mn}}{Z_\text{k}} \end{align} \right. $$

So having expressed all possible control variables as function of -unchanged- internal ones we are allowed to transform from current to voltage generator.

Voltage generator case:
we just have to do the same with the three possible quantities across/thru the external parts (\$I_\text{k}\$, \$V_\text{Zk}\$ and \$V_\text{vk}\$ )

$$ \left\{ \begin{align} I_\text{k} &=I_\text{nm} \\ V_\text{Zk} &=-Z_\text{k}I_\text{nm} \\ V_\text{vk} &=V_\text{mn}-V_\text{Zk}=V_\text{mn}+Z_\text{k}I_\text{nm} \end{align} \right. $$

so we have now proved the other way round transformation, from voltage to current generator, is always possible too.


Voltage and current controlled sources can be transformed each other. To do so it might be necessary to do some math on control variable, but this is going to be always possible.


The answer is yes. A dependent source is simply one whose current or voltage is not a constant, but a function of some input parameter \$x\$. So in the source transformation algebra, you simply regard the quantities \$V\$ and \$I\$ not as constants but as \$V(x)\$ and \$I(x)\$; in other words \$V(x) = I(x)R\$ and \$I(x) = \frac{V(x)}{R}\$. Now this \$x\$ is of course derived from some current or voltage elsewhere in the circuit, but that is irrelevant.

  • \$\begingroup\$ Alright. Cool, me and my roommates were having a discussion about this, and I wanted a final word on the matter. I was thinking that it wouldn't behave the same as an independent source, given that it might depend on some other parts of the circuit, and the source transformation would affect those parts. However, that doesn't make much sense, and is obviously not the case. \$\endgroup\$
    – golvok
    Oct 11, 2013 at 0:15

The answer is yes, source transformation is valid for dependant sources. And it does not affect the other parts of the circuit. Those who are looking for an official source can refer to this: C.K. Alexander and M.N.O. Sadiku, Fundamentals of electric Circuits, 4th ed. , McGraw Hill, 2011, ch. 4, sec. 4.4, pp. 136


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