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-   -   CIRCUIT ANALYSIS: 7 resistors, 2 Indep. Volt Source, V.C.C.S, V.C.V.S. - find I (http://www.physicsforums.com/showthread.php?t=151625)

 VinnyCee Jan16-07 05:09 AM

CIRCUIT ANALYSIS: 7 resistors, 2 Indep. Volt Source, V.C.C.S, V.C.V.S. - find I

1. The problem statement, all variables and given/known data

For the circuit below find $I_1$ and $I_2$:

http://img155.imageshack.us/img155/8...oblem84ro6.jpg

2. Relevant equations

KVL
KCL
Ohm's Law

3. The attempt at a solution

I tried the problem many times, but I always get crazy answers. It seems that every time I need one new equation to have a system of solveable equations, I have to add a new variable and hence I need another equation. It's a vicious cycle that when I get up to 13 variables for all of the V's at the resistors and different I's at nodes 1-4, I get a crazy answer like -1.35 mA for $I_1$. Does that seem right?

Any suggestion on what to do about the one-more variable, one-more equation problem? I tried a super-node between nodes 2 and 3. Didn't help though.

 AlephZero Jan16-07 05:27 AM

Start by finding the voltages. You can write down the voltage at nodes 4, 3, 2 (two different ways) and 1 (in that order) without knowing any currents.

The two ways of getting the voltage at node 2 gives you a relation between Va and Vb, so eliminate one of them.

Then start finding the currents in terms of the voltages.

You don't need to add any more nodes.

 VinnyCee Jan17-07 04:47 AM

Ok, so I get these for the node voltages(4,3,2,2,1):

$$V_4\,-\,0$$

$$V_3\,+\,5V$$

$$V_2\,-\,0$$

$$V_2\,-\,2\,-\,V_4\,=\,0$$
$$V_2\,=\,V_4\,+\,2$$

$$V_1\,-\,0$$

Are these right? If not, how am I supposed to make these voltage equations?

Also, I am stuck again, I don't know where to go from here (even if the voltage EQs are correct)!

Should I use KVL or KCL? I tried to add 2 current variables at Node 2. I used KCL there and I made some EQs for the currents there (that I had an R for) using $i\,=\,\frac{V}{R}$. I am seriously stuck now though!!!

Can someone walk me through the most logical way to proceed from here, I am really confused. Thanks

 AlephZero Jan17-07 06:35 AM

I don't understand exactly what those expresssions are.

The way I would do this is by looking at the circuit and thinking about what you know, not trying to apply the K. laws in a mechanical way.

It's "obvious" from the circuit diagram that

V4 = VA

V3 = V4 = VA (the current source has no internal resistance so no voltage across it)

V2 = V4 - 2 = VA - 2 (from the 2V voltage source)
and also V2 = VB so VB = VA - 2

V1 = V2 - 2VA = -2 - VA

Now try and find a node which doesn't have the unknown currents I1 and I2 flowing into it: there is one, node 3. Find all the currents flowing into node 3 by Ohms law. By KCL they add up to zero. That will give you an equation for VA.

Now you know all the voltages, you can use Ohms law and KCL to find the other currents.

 doodle Jan18-07 12:04 AM

Okay Vinny, we'll use nodal analysis or KCL for the problem if that is fine with you. And for the moment, let's avoid supernodes but stick with the conventional nodes.

To get you started, do this for me: Note down the 4 equations corresponding to the 4 nodes using KCL, i.e., the sum of currents entering/leaving the node equals zero. And do that with only the variables V1, V2, V3, V4, I1, I2 and no other variables.

Quote:
 Quote by AlephZero (Post 1215488) ... V3 = V4 = VA (the current source has no internal resistance so no voltage across it)
Not true. The current source could present a voltage drop/gain without any internal resistance.

 VinnyCee Jan18-07 12:08 AM

OK, I have added 3 currents to the diagram (in green).

http://img444.imageshack.us/img444/1...84part2fc9.jpg

$$V_4\,=\,V_a\,=\,V_3$$

$$V_2\,=\,V_4\,-\,2\,=\,V_b$$

$$V_b\,=\,V_a\,-\,2$$

$$V_1\,=\,V_\,2\,-\,2\,V_a$$

Just what you said above. Now, I use $i\,=\,\frac{V}{R}$ to get the new currents in green.

KCL: $$I_3\,+\,I_4\,+\,I_5\,+\,\frac{V_b}{4K\Omega}\,=\,0$$

$$I_3\,=\,\frac{(5\,V)}{4000\Omega}\,=\,0.00125\,A\,=\,1.25\,mA$$

$$I_4\,=\,\frac{V_2\,-\,V_3}{2000\Omega}$$

$$I_5\,=\,\frac{V_1\,-\,V_3}{4000\Omega}$$

Now if I combine those four equations above:

$$\frac{V_2\,-\,V_3}{2000\Omega}\,+\,\frac{V_1\,-\,V_3}{4000\Omega}\,+\,\frac{V_4\,-\,2}{4000\Omega}\,=\,-1.25\,A$$

How do you proceed?

 VinnyCee Jan18-07 12:23 AM

Ok, lets try NODE 1 first:

$$\left(\frac{-V_1}{1000\Omega}\right)\,+\,\left[\left(\frac{V_3\,-\,V_1}{4000\Omega}\right)\,+\,\left(\frac{V_4\,-\,V_1}{3000\Omega}\right)\right]\,=\,I_1$$

Is that correct?

 doodle Jan18-07 12:30 AM

Quote:
 Quote by VinnyCee (Post 1216276) $$V_4\,=\,V_a\,=\,V_3$$
As I have said earlier, to claim that V3 = V4 is incorrect. Read my earlier post.

Quote:
 Quote by VinnyCee (Post 1216276) Just what you said above. Now, I use $i\,=\,\frac{V}{R}$ to get the new currents in green. KCL: $$I_3\,+\,I_4\,+\,I_5\,+\,\frac{V_b}{4K\Omega}\,=\,0$$
Well yes, that is right. But you do know that the above expression gives only 1 nodal equation, that is, the nodal equation for node 3. You would have to do the same thing for the other 3 nodes.

Quote:
 Quote by VinnyCee (Post 1216276) $$I_3\,=\,\frac{(5\,V)}{4000\Omega}\,=\,0.00125\,A\,=\,1.25\,mA$$
This is not right- you're forgetting V3.

Quote:
 Quote by VinnyCee (Post 1216276) $$I_4\,=\,\frac{V_2\,-\,V_3}{2000\Omega}$$ $$I_5\,=\,\frac{V_1\,-\,V_3}{4000\Omega}$$
These are correct.

And yes, combine these terms into an equation. Note that Vb = V2. Also, at the moment, let's forget about V2 = V4-2. Now write down again the nodal equation for node 3.

 doodle Jan18-07 12:32 AM

Quote:
 Quote by VinnyCee (Post 1216287) Ok, lets try NODE 1 first: $$\left(\frac{-V_1}{1000\Omega}\right)\,+\,\left[\left(\frac{V_3\,-\,V_1}{4000\Omega}\right)\,+\,\left(\frac{V_4\,-\,V_1}{3000\Omega}\right)\right]\,=\,I_1$$ Is that correct?
That's right. :)

 VinnyCee Jan18-07 12:41 AM

Cool! NODE 2 now:

$$I_1\,+\,\left(\frac{-V_2}{2000\Omega}\right)\,+\,\left(\frac{V_2\,-\,V_3}{2000\Omega}\right)\,=\,I_2$$

Is that right?

Is $V_b$ still equal to $V_2$?

 doodle Jan18-07 12:52 AM

Quote:
 Quote by VinnyCee (Post 1216315) Cool! NODE 2 now: $$I_1\,+\,\left(\frac{-V_2}{2000\Omega}\right)\,+\,\left(\frac{V_2\,-\,V_3}{2000\Omega}\right)\,=\,I_2$$ Is that right? Is $V_b$ still equal to $V_2$?
A small mistake here, check the equation again. Yes, Vb = V2. Two more nodal equations to go (nodes 3 and 4). Keep it up!

 VinnyCee Jan18-07 01:27 AM

The mistake fixed?

$$I_1\,+\,\left(\frac{-V_2}{2000\Omega}\right)\,+\,\left(\frac{V_3\,-\,V_2}{2000\Omega}\right)\,=\,I_2$$

 doodle Jan18-07 01:29 AM

That's right, now move on to the other 2 equations.

 VinnyCee Jan18-07 01:36 AM

OK, for NODE 3:

$$\left(\frac{V_1\,-\,V_3}{4000\Omega}\right)\,+\,\left(\frac{V_2\,-\,V_3}{2000\Omega}\right)\,+\,\left(\frac{5\,-\,V_3}{4000\Omega}\right)\,=\,\left(\frac{-V_2}{4000\Omega}\right)$$

And for NODE 4:

$$\left(\frac{-V_4}{4000\Omega}\right)\,+\,\left(\frac{V_1\,-\,V_4}{3000\Omega}\,+\,I_2\right)\,=\,\left(\frac{V_2}{4000\Omega}\righ t)$$

Are those right? Or did I mess up the path with the 5V independent voltage source and 4Kohm resistor?

 doodle Jan18-07 01:57 AM

Very good. Now, I would like to have these equations simplified a bit. As an example, for node 4, you wrote:

$$\left(\frac{-V_4}{4000\Omega}\right)\,+\,\left(\frac{V_1\,-\,V_4}{3000\Omega}\right)\,+\,I_2\,=\,\left(\frac{V_2}{4000\Omega}\righ t)$$

I want it simplified to become:

$$\frac{1}{3}V_1 - \frac{1}{4}V_2 - \frac{7}{12}V_4 + I_2 = 0$$

Specifically, I have ignored the '000 in the R's (the V's are still as before but the I's are now in milliamperes) and arranged the equations such that on the left side are the unknowns, ordered V1, V2, V3, V4, I1, I2 and on the right side, the constants. There's a reason for doing all these of course. :)

Do the same for the other 3 equations and we will proceed from there.

 VinnyCee Jan18-07 02:21 AM

NODE3: $$V_1\,+\,3\,V_2\,-\,4\,V_3\,=\,-5$$

NODE2: $$2000\,I_1\,-\,2000\,I_2\,-\,2\,V_2\,+\,V_3\,=\,0$$

NODE1: $$-4000\,I_1\,-\,\frac{19}{3}\,V_1\,+\,V_3\,+\,\frac{4}{3}\,V_4\,=\,0$$

Right?

 doodle Jan18-07 02:48 AM

Okay, that's close enough. Let me edit a bit...

Node 1: $$-\frac{19}{12}V_1 + \frac{1}{4}V_3 + \frac{1}{3}V_4 - I_1 = 0$$

Node 2: $$-V_2 + \frac{1}{2}V_3 + I_1 - I_2 = 0$$

Node 3: $$V_1 + 3V_2 - 4V_3 = -5$$

Node 4: $$\frac{1}{3}V_1 - \frac{1}{4}V_2 - \frac{7}{12}V_4 + I_2 = 0$$

I prefer to have the equations in the manner above, with the I's in milliamperes. Now, note that there's a voltage source across nodes 1 and 2, similarly a voltage source between nodes 2 and 4. Due to that, we ought to form a supernode, that is, to combine nodes 1, 2 and 4 into a single supernode. The result is the elimination of the unknown variables I1 and I2. To do that, try combining equations 1, 2 and 4 above into a single equation such that the unknowns I1 and I2 disappear.

 VinnyCee Jan18-07 03:14 AM

OK, I did N1 + N2 + N4.

$$-\frac{5}{4}\,V_1\,-\,\frac{5}{4}\,V_2\,+\,\frac{3}{4}\,V_3\,-\,\frac{1}{4}\,V_4\,=\,0$$

Is this what the supernode would look like?

http://img412.imageshack.us/img412/9...84part3lo3.jpg

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