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Here's a circuit: circuit

As you see there's a dependant source, 0.01Vx, depending on Vx, the voltage at the resistor of 150 ohm.

I don't exactly know how to handle this circuit properly. The presence of the dependant voltage source bugs me. I wan't to find Ia. Vs and Vab are undefined - I want only to expressions in terms of relations to Vs and Vab. I also want to know how to apply nodal analysis correctly.

I don't get how to apply properly KCL at the node 2.

First, I apply nodal analysis. I define node 3 as the 0 voltage node (ground). Voltage at node 1 is Vx.

I end up with these equations:

\begin{equation} \left[ \frac{1}{25} + \frac{1}{150} + \frac{1}{50} \right] \cdot Vx = \frac{Vs}{25} - \frac{0.01Vx}{50} + \frac{Vab}{50} \end{equation}

First, is that correct?

Second, applying KCL bugs me. As said, the dependant source confuses me. I basically must find the current between node1 and node2 and apply KCL. But... The dependant voltage source... does it add another current? I would maybe try: \begin{equation} I_{a} = -\frac{0.01\cdot Vx}{50} + \frac{Vab}{50} \end{equation}

But I'm not sure if Vab must be accounted in the KCL or not. I get headaches thinking about it.

Could you help?

Thank you!

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If you wish to solve the circuit using node voltage analysis, you would not bother to write a KCL equation at node 2.

Remember, when doing node voltage analysis, one is solving for the node voltages.

But, the voltage at node 2 is given: \$V_2 = V_{ab}\$

So, you might think that you must write a KCL equation for node 1 but, in fact, you don't because there is a voltage source connected there too.

Simply use KVL to write:

$$V_x + 0.01V_x = V_{ab} \rightarrow V_x = \dfrac{V_{ab}}{1.01}$$

Now, you know the node voltages so you can find the resistor currents. Can you take it from here to find \$I_a\$?


Finally, about Ia. I am also confused by the presence of 0.01Vx. Would applying KCL only means finding current between node 1 and 2 or do we have to involve 0.01Vx too?

Since you know the node voltages, you know the currents through the resistors connected to node 1. Thus, if you write a KCL equation there, the only unknown is the current through the dependent source so use this KCL equation to solve for the dependent source current.

Now that you've found the dependent source current, KCL at node 2 involves only one unknown current, the current \$I_a\$.


The reason why I applied node analysis is because I am studying it these days, and wanted to apply it correctly. Did I?

To correctly apply node voltage analysis, you must enclose the dependent voltage source and parallel resistor inside a supernode. The KCL equation for the supernode is:

$$\dfrac{V_x - V_s}{25} + \dfrac{V_x}{150} = I_a $$

There are two unknowns so you need another equation which is the KVL equation I wrote above.

Note that the 50 ohm resistor is not a factor in the equation. This is due to the fact that it is in parallel with a voltage source which means that the only circuit variable the 50 ohm resistor affects is the current through the dependent source.

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  • \$\begingroup\$ Thanks! Though. The reason why I applied node analysis is because I am studying it these days, and wanted to apply it correctly. Did I? Also, You say voltage at node 2, V2, is V2 = Vab. What I am confused with is that... There is also the 0.01Vx source which has an influence on node 2. So would it be more appropriate to say V2 is a combination of 0.01Vx and Vab, like for example Vab + 0.01Vx or am I wrong? Finally, about Ia. I am also confused by the presence of 0.01Vx. Would applying KCL only means finding current between node 1 and 2 or do we have to involve 0.01Vx too? \$\endgroup\$ – Yannick Dec 3 '13 at 15:06
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    \$\begingroup\$ No, the voltage between node 2 and node 3 is Vab period. Since node 3 is the reference node, the voltage at node 2 is Vab and is not a combination of anything else in the circuit. Likewise, the voltage between node 2 and node 1 is 0.01Vx period. This follows from the definition of a voltage source. See the update to my answer. \$\endgroup\$ – Alfred Centauri Dec 3 '13 at 15:45
  • \$\begingroup\$ THANK YOU! Understood perfectly. Chosen answer of course! \$\endgroup\$ – Yannick Dec 3 '13 at 17:04

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