# Why are dependent sources not disabled with the superposition principle?

I've been wondering this recently. My intuition says that a voltage drop across a dependent source is no different than any other circuit element whose response depends on the inputs of one of the independent sources. But I don't feel like that reasoning is rigorous enough. How exactly would this principle regarding treatment of dependent sources be derived?

• We use both for static and transient conditions or swept frequency response. Dependant sources are common for Ac analysis and DC analysis and superposition works in both cases as long as it is linear. For example clipping causes analytical errors unless you have specific boundary conditions – Tony Stewart EE75 Jun 3 '17 at 16:44

• If you disabled dependent sources when modeling the contribution of a dependent source to the circuit, they'd have no effect on the circuit. If you did another round of superposition to get the contribution from the dependent source, there'd be no output because there wouldn't be any independent source present to drive the dependent source's input. So you might as well have never put them in your circuit model to begin with. Which obviously defeats whatever purpose you had in including the dependent source in your model.

• Dependent sources are no different from other elements in that they respond to stimulus from the independent sources. For example, you could model a resistor as a CCVS whose input and output ports happen to be connected in series. So any argument you have for removing dependent sources during superposition solutions also applies to resistors, capacitors, and inductors. And if you removed those from your circuit, you'd have no circuit left. So obviously that's not what you should do.

• These are two ways to think about it but not the only two ways. You might want to hold off accepting an answer until users around the world have had a chance to answer. – The Photon Jun 3 '17 at 16:52

But we can use the Superposition with the dependent sources. There is only one proviso - the controlling parameter "source" cannot be removed in any of the individual cases where any one active source is under consideration.

See the example:

And we can use the superposition add find $I_1$ first:

Case A: Independent voltage source acting alone

$$I_1 = \frac{20V}{6 \Omega}$$

Case B: Dependent source acting alone

$$I_1= \frac{2Ω•I_1}{12\Omega}$$

The superposition of the 2 cases to give $I_1$ is then:

$$I_1 = \frac{20V}{12Ω} + \frac{2Ω•I_1}{12Ω}$$

$$I_1 = \frac{5}{3} + \frac{2Ω•I_1}{12Ω}$$

$$I_1 = \frac{5}{3} + \frac{I_1}{6}$$

$$I_1 = \frac{I_1 + 10}{6}$$

$$6•I_1 = I_1 + 10$$

$$5•I_1 = 10$$

$$I_1 = 10/5 = 2A$$

And finally $U_{TH} = 2A•6Ω = 12V$

And here you have another example

Case a : Current source acting alone By the current divider rule, the current flowing to create voltage$Vr_a$ is

$$I_{2 \Omega}=-2•\frac{5}{5+2}=-\frac{10}{7}\;A$$

which will produce a voltage $Vr_a$ across the 2 ohm resistor

$$V_{r\_a}=-\frac{20}{7} \;V$$

Case b: Independent voltage source acting alone

Using the voltage divider rule with the 4V shared across the 5 ohm and 2 ohm resistors.

$$V_{r\_b}=-4• \frac{2}{5+2}=-\frac{8}{7} \; V$$

Case c: Dependent source of value 3*Vr acting alone Again by the voltage divider rule, the voltage 3*Vr is shared across the 5 ohm & 2 ohm resistors.

$$V_{r\_c}=+3V_r • \frac{2}{5+2} = +\frac{6}{7}V_r$$

The superposition of the 3 cases to give Vr is then:

$$V_r=V_{r\_a}+V_{r\_b}+V_{r\_c}=-\frac{20}{7}-\frac{8}{7} +\frac{6}{7}V_r$$

Multiplying through by 7 and collecting terms in Vr gives

$$7 • V_r-6•V_r=-20-8$$

or

$$V_r=-28V$$

And additional info can be found here: https://leachlegacy.ece.gatech.edu/papers/superpos.pdf