Solving a weird circuit using nodal analysis?

[x86]

Homework Statement

Homework Equations

V=IR
Kirchoffs Current Law

The Attempt at a Solution

I picked the middle node (the one surrounded by 3 resistors) as my reference. V1 and V2 are my other two nodes.

I'm also assuming that V1>V2, V1<reference

Using Kirchoffs Current Law:

(Node 1)
-(0-V1)/6 +6 + (V1-V2)/9 -4 -(0-V1)/2=0

(Node 2)
4 + V2/4 - 6 -(V1-V2)/9 = 0

(I_0 current)
I_0 = (0-V1)/6

Of course, I keep getting the wrong answer and just can't solve this problem

 

[gneill]

Your node equations look okay to me. Can you show more of your work?

 

[x86]

Your node equations look okay to me. Can you show more of your work?

I think I figured out that one of my assumptions was incorrect. For instance, I both assumed V1>V2 and V1<V2 for current flow. (a sneaky error indeed)

I am confused about one thing though, if my reference wasn't at the center point (and I am analyzing the node between the three resistors), how would I model current flowing in from between the two 6 mA current sources?

 

[gneill]

I am confused about one thing though, if my reference wasn't at the center point (and I am analyzing the node between the three resistors), how would I model current flowing in from between the two 6 mA current sources?

6 mA in and 6 mA out, for a net of zero contribution. In the node equation you'd write something like 6 - 6 + ...

 

[x86]

6 mA in and 6 mA out, for a net of zero contribution. In the node equation you'd write something like 6 - 6 + ...

Right, but for the bottom current source, aren't there two paths for the current? One path would be right, into the node. And another pathwould be up? So surely not all of this 6 mA flows into the node?

 

[gneill]

Right, but for the bottom current source, aren't there two paths for the current? One path would be right, into the node. And another pathwould be up? So surely not all of this 6 mA flows into the node?

It all flows into the node, and the other 6 mA source all flows out of the node. You don't have to worry about the trajectories of individual charge carriers, only the net results.

 

[x86]

It all flows into the node, and the other 6 mA source all flows out of the node. You don't have to worry about the trajectories of individual charge carriers, only the net results.

I'm still a little confused, why does all the current flow into the node? there are two paths the bottom current source can take (right and up). That means that 6 = (current flowing into node + current flowing past node)

 

[gneill]

All the current flows into the node because the source is connected to the node. The fact that another source of the same size is also connected to the node and carries away the same current is incidental.

If you wanted to you could swap the positions of the 4k resistor and the upper 6 mA source and not change a thing about the circuit or the resulting equations. The components would still be connected to the same nodes.

 

[x86]

All the current flows into the node because the source is connected to the node. The fact that another source of the same size is also connected to the node and carries away the same current is incidental.

If you wanted to you could swap the positions of the 4k resistor and the upper 6 mA source and not change a thing about the circuit or the resulting equations. The components would still be connected to the same nodes.

So when you talk about node, I'm guessing you mean this entire red line?

If that is the case, then I could see why we say there is 6 mA flowing into the node, and 6 mA flowing out of the node, because of the current sources

 

[gneill]

Yes, that's right.

 

[Simon Bridge]

One of the disciplines in this sort of analysis is to draw current arrows on the diagram around each node.
That basically means you assume currents and not voltages, just make sure each node has at least one outflowing and one inflowing arrow.
You can finess the process by making informed guesses about which way the arrows "should" point - but it doesn't matter because, if you guess wrong, the value of the current will some out negative. A negative current just means that the associated arrow should be the other way around.
When you do the sum currents = 0, try instead currents in = currents out. Some people find the second one less confusing.

Then draw potential-difference arrows on the resistors (later: impedences). The PD arrow across a resistor just points against the direction that the current flows through the resistor.

I'll do a bit of a walkthough for your problem, using this approach:

In your diagram - numbering nodes left-to-right and top-to bottom from 1 to 7.
Looking ahead - you can see n3 has 6mA into in along one wire and 6mA out along another wire. The third wire, therefore, carries no current and can be removed from the diagram. The resulting circuit, if you redraw it, should look nicer to analyze.
It is possible to simplify even further by pairing the other nodes that have just a wire between them - you can see the 6k and 2k resistors are in parallel ferinstance.

But we don't have to do that, and I want to demonstrate how it all comes out automatically even if we are not clever enough to notice all that stuff.

I'll choose notation so that I2, I4, I6, and I9 are the current's through the 2k, 4k, 6k, and 9k, resistors respectively. I hope you can see why that can be useful. Comparig with the problem statement, this means we should comment that I0=I6.

The I9 current is chosen to go from bottom to top while the rest flows away from their common node (n4). All currents in milliAmps.
Drawing the arrows on the diagram is key to this approach - stops you from getting mixed up.

n1. I9 in from the left, 6mA in from below, I1 out to the right. i.e. $I_9+6=I_1$
n2. I1 in from the left, 4mA out to the right, I4 in from below: $I_1+I_4=4$
n3. 6mA in from below, 6ma out upwards, I3 out to the right: $6+I_3=6$
n4. I3 in from the left, I4 out up, I2 out to the right, I6 out down: $I_2+I_4+I_6=I_3$
n5. 4mA and I2 are flowing in, I7 must flow out and down: $4+I_2=I_7$
n6. I9 out to the left, 6 out up, I5 in from the right: $I_9+6=I_5$
n7. I5 and I6 flow in, I7 flows out to the right: $I_5+I_6=I_7$
additionally: $I_0=|I_6|$ (from the problem statement.)

The above list can be constructed almost mindlessly and contains all the donkeywork for the problem.
We can write out the voltage equations too ... the notation has been chosen so that, for instance, $V_n=I_nR_n$
Step around loops that contain more than one resistor, listing, in a row, the PD's you step over.
When you step in the same direction as a PD arrow, that voltage has a + sign in front of it otherwise it has a negative sign.
Then you write "= 0" at the end of the list and it turns into an equation.

i.e. $9I_9\; -4I_4\; +6I_6\; = 0$ ... I fiddled the spacing so you can see the way the list turns into an equation.

check units: 1kOhm x 1milliAmp = Ohms x Amps = Volts.
Note: there is no PD across current sources, the PD arrow for a voltage source goes from negative to positive.

I think you can see how this works.
Above, there are seven simultaneous equations and seven unknowns, so the problem can be done without doing the voltages so we won't.
Why do more maths than you have to?

Next step is to switch the brain back on to do some algebra ... you basically have to solve the system of equations for I6.
You don't need much cleverness for this if you don't want to - you can solve it by row-echelon reduction. All the physics was done at the node-by-node, and you did that by drawing arrows on the diagram: the rest is just taking care when applying an algorithm. I'll just point out a few features based on the questions above.

For n3, notice that this simplifies to $I_3=0$. So there is no current between n3 and n4 (as anticipated: it's nice t be right!). This means you could actually just remove that wire without changing the circuit.

Using this result in n5 tells you $I_2+I_4+I_6=0$ ... in English, this means that the assumed current direction through at least one of those resistors is wrong. It's not too surprising because I chose the direction for being easy to describe rather than by any physical arguments. We are not worried, it just means one or two of the values will come out negative. This is why I put the absolute-value signs into the equation for $I_0$ - just in case.

Getting the node equations right does not, therefore, involve making assumptions about voltages - the voltages are kept internally consistent just by asking sure the arrows are consistent - which is usually easier to see when you make a mistake.

All this is about strategies to minimize mistakes during long multi-step calculations. Not much actual physics.
The exercise has education value like that - it's mostly about discipline and focus and how you develop strategies.
There is some outlook/attitude training there too, since node analysis is all about conservation laws in complicated situations.
Teachers can be sneaky like that.

 

[donpacino]

a few things.

1. simon: The goal of the homework assignment is about nodal analysis, which involves a process that does not define each and every current in the system. The beauty of nodal analysis is you simply take the voltage differentials over the resistances (ohms law) as the currents. You can "blindly" write equations at every unknown node and arrive at the answer.

2. x86: like was mentioned before your equations look correct. One thing that i found might make things easier. in some instances you have equations that look like this
-(0-V1)...

in this case it is often simpler to write (V1-0). It standardizes things and makes it easier. I've found that it is easiest to always do the current flowing out of a node, or vise versa.

3. As was previously mentioned, you do not need to make assumptions about what the voltages are. All that matters is that your equations satisfy KCL and ohms law. If you are actually getting the answer wrong, you are most likely solving your systems of equations incorrectly.