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I'm back again with a nodal analysis question. I'm getting better, but I've come to a problem again.

Consider the following circuit with the bottom node as ground: A circuit with voltage sources in series with resistors in its branches

I specifically don't know how to deal with the two lower left branches. When I apply KVL I get equations of the form v1 = v1 and v2 = v2, which is not helpful. One of them is also a dependent source, so I can't just use source transformations as far as I know.

I know that this problem ought to be a lot easier with mesh current analysis. Unfortunately, I'm doing this problem for educational purposes and we're expected to be able to solve this problem with nodal analysis too.

Thanks in advance again, Joshua

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    \$\begingroup\$ How are you using KVL? Can you give a direct example of how you're getting these "V1 = V1" equations? Why do you say that the supernode spans the entire circuit? \$\endgroup\$ – Greg d'Eon Apr 7 '15 at 11:02
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You don't need to do anything weird to handle the lower left branches. Treat the dependent voltage source the same as you would an independent source. If \$V_1\$ is the node between the 10 and 20 ohm resistors, \$V_2\$ is the node between the 40 ohm resistor and the 120 V source, and \$V_3\$ is the node between the dependent current source and the 80 ohm resistor:

$$\frac{V_1 - 100\ \mathrm V}{10\ \Omega} + \frac{V_1 - 4V_O}{20\ \Omega} + \frac{V_1 - V_2}{40\ \Omega} = 0$$

You've introduced a new variable (\$V_O\$), so you need another equation to solve the circuit. This equation should define \$V_O\$ in terms of the node voltages. In this case, \$V_O\$ is simply equal to one of the node voltages:

$$V_O = V_3$$

You can use a similar technique for the dependent current source.

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Let X be node voltage at junction of 10/20/40 resistors, then

(100-X)/10 + (4Vo-X)/20 = [X-(Vo-120)]/40 = Io

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