
#1
Sep1910, 06:13 PM

P: 66

1. The problem statement, all variables and given/known data
I'm confused about how to use KCL when doing Nodal Analysis. Please consider the diagram below: 2. Relevant equations My first point of confusion is that the diagram shows the current "4Ix" going up, but I thought current always went from higher voltage to lower voltage? (which would be down, in this case). I also am very confused on how to setup the KCL equation, as you'll see in my attempt below.. 3. The attempt at a solution To use KCL, we assume that all of the currents are either entering or leaving the node and then add them up, right? Since one of the currents is already shown as entering, would it be logical to write my equation with them all entering? I tried to write the KCL equation like this: The first term is the current entering from the left, the second is term the current going up  it's negative because it's entering through the negative sign (ohms law)  and the third term is the current entering on the right, also negative for the same reason as before. I know this is wrong, but I've spent hours trying to figure this out and am still totally confused. Please help me understand! 



#2
Sep1910, 10:20 PM

P: 393

you can define the direction of current any way you want. when you solve the unknown, you may find the current is negative, indicating the actual current is flowing in the opposite direction as your guess.
as for your nodal equations, they are almost right except for when you seemingly randomly throw in negatives with the odd reason of "ohms law" [tex]I_{in} = I_{out}[/tex] this is because charge must be conserved. What flows into a node must flow out of the node. It is a common standard to define current to leave toward any branch with a resistor in it and to define the current of a branch with a current source as the same direction as that source (after all, why not?) so the nodal equation you seek to write out would have the foundation as such (in words, not numbers): current entering the node from the current source = current leaving the node toward the 1 ohm resistor + current leaving the node toward the 5 ohm resistor don't add any silly negatives! just stick with the math. Current approaches the positive terminal of a resistor. As an example, "current leaving the node toward the 1 ohm resistor" would be like this mathematically: [tex]\frac{V_o  V_1}{1}[/tex] It's worth nothing, also, you could also define all the currents as entering the node for the branches with resistors. You will have V1  V0 in the numerator for the same example resistor instead of v0  v1. That would be the equivalency of subtracting the once labeled "in" current onto the "out" side. And when you distribute the subtraction through (v0  v1) you get (v1  v0), so it is mathematically the same. It all goes back to my original point: you can define the directions any way you want. 


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