Question

Node-voltage method: If you have two voltage sources in the circuit would you have v1 and v2? I wrote v1 in the diagram I sketched. I just don't know when there's another v that should be indicated in the diagram. File attached

Answer 1

You are off to a good start. The two nodes (green dots) at the top and the reference node the ground symbol (lower green dot) are very good choices. I would for clarity, label the upper left green dot (node) as Va and the upper right green dot (node) as Vb. Then using Kirchoffs Current Law (KCL) at each node (Va, and Vb) develop two equations, one for each node. This will result in two equations with two unknows (Va, and Vb) which you can solve using algebra (substitution, elimination or a matrix). I used substitution, solving first for Vb and then Va.

Answer 2

I used your labels for the individual node currents, I did however change the direction for Ie in order to keep the results positive. If you don't you will get a negative value for Ie which would indicate that your direction (Ie flowing out of Vb is) incorrect.

Answer 3

If you should need further guidance, I will develop the equation for the Va node where the sum of Ia, Ib, and Ic are zero (KCL) for you, and then you can use that as an example to solve the Vb currents.

Answer 4

Here are the equations for the currents.
Ia =(128 - Va)/5
Ib = Va/60
Ic = (Va - Vb)/4
Id = Vb/80
Ie =(320-Vb)/10 (note that I changed the direction of Ie.

These are the currents associated with the two nodes Va and Vb.

Answer 5

Kirchoff's Law states that the sum of the currents at node equal zero.
At node Va I am assuming the Ia is entering the node and Ib and Ic are leaving the node so the equation at Node Va would look like this:
Ia - Ib - Ic = 0 We now substitute the values in the above post, giving us:
(128 - Va)/5 - Va/60 - (Va - Vb)/4 = 0
Simplify and then work out for Node Vb.

Answer 6

Note, you will have to do Node Vb, before solving for Node Va. You will then have two equations with two unknowns. Good luck with your studies.

Answer 7

@radar Thank you so much. Sorry I'm getting back to you this late. I like how you explained what to do better than the explanation given in my own course and in the textbook. Very straightforward and very helpful. I might have more circuit questions in the long-run so I'll be needing your help!

Answer 8

Will be glad to help when I can and you are certainly welcome. Circuit analysis can be confusing in the way they are depicted. If you have read my profile, you will find that it has been years since I have worked with them.