
#1
Jan1607, 05:09 AM

P: 492

1. The problem statement, all variables and given/known data
For the circuit below find [itex]I_1[/itex] and [itex]I_2[/itex]: 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 14, I get a crazy answer like 1.35 mA for [itex]I_1[/itex]. Does that seem right? Any suggestion on what to do about the onemore variable, onemore equation problem? I tried a supernode between nodes 2 and 3. Didn't help though. 



#2
Jan1607, 05:27 AM

Engineering
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P: 6,385

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. 



#3
Jan1707, 04:47 AM

P: 492

Ok, so I get these for the node voltages(4,3,2,2,1):
[tex]V_4\,\,0[/tex] [tex]V_3\,+\,5V[/tex] [tex]V_2\,\,0[/tex] [tex]V_2\,\,2\,\,V_4\,=\,0[/tex] [tex]V_2\,=\,V_4\,+\,2[/tex] [tex]V_1\,\,0[/tex] 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 [itex]i\,=\,\frac{V}{R}[/itex]. I am seriously stuck now though!!! Can someone walk me through the most logical way to proceed from here, I am really confused. Thanks 



#4
Jan1707, 06:35 AM

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P: 6,385

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



#5
Jan1807, 12:04 AM

P: 161

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. 



#6
Jan1807, 12:08 AM

P: 492

OK, I have added 3 currents to the diagram (in green).
[tex]V_4\,=\,V_a\,=\,V_3[/tex] [tex]V_2\,=\,V_4\,\,2\,=\,V_b[/tex] [tex]V_b\,=\,V_a\,\,2[/tex] [tex]V_1\,=\,V_\,2\,\,2\,V_a[/tex] Just what you said above. Now, I use [itex]i\,=\,\frac{V}{R}[/itex] to get the new currents in green. KCL: [tex]I_3\,+\,I_4\,+\,I_5\,+\,\frac{V_b}{4K\Omega}\,=\,0[/tex] [tex]I_3\,=\,\frac{(5\,V)}{4000\Omega}\,=\,0.00125\,A\,=\,1.25\,mA[/tex] [tex]I_4\,=\,\frac{V_2\,\,V_3}{2000\Omega}[/tex] [tex]I_5\,=\,\frac{V_1\,\,V_3}{4000\Omega}[/tex] Now if I combine those four equations above: [tex]\frac{V_2\,\,V_3}{2000\Omega}\,+\,\frac{V_1\,\,V_3}{4000\Omega}\,+\,\frac{V_4\,\,2}{4000\Omega}\,=\,1.25\,A[/tex] How do you proceed? 



#7
Jan1807, 12:23 AM

P: 492

Ok, lets try NODE 1 first:
[tex]\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[/tex] Is that correct? 



#8
Jan1807, 12:30 AM

P: 161

And yes, combine these terms into an equation. Note that Vb = V2. Also, at the moment, let's forget about V2 = V42. Now write down again the nodal equation for node 3. 



#9
Jan1807, 12:32 AM

P: 161





#10
Jan1807, 12:41 AM

P: 492

Cool! NODE 2 now:
[tex]I_1\,+\,\left(\frac{V_2}{2000\Omega}\right)\,+\,\left(\frac{V_2\,\,V_3}{2000\Omega}\right)\,=\,I_2[/tex] Is that right? Is [itex]V_b[/itex] still equal to [itex]V_2[/itex]? 



#11
Jan1807, 12:52 AM

P: 161





#12
Jan1807, 01:27 AM

P: 492

The mistake fixed?
[tex]I_1\,+\,\left(\frac{V_2}{2000\Omega}\right)\,+\,\left(\frac{V_3\,\,V_2}{2000\Omega}\right)\,=\,I_2[/tex] 



#13
Jan1807, 01:29 AM

P: 161

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




#14
Jan1807, 01:36 AM

P: 492

OK, for NODE 3:
[tex]\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)[/tex] And for NODE 4: [tex]\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)[/tex] Are those right? Or did I mess up the path with the 5V independent voltage source and 4Kohm resistor? 



#15
Jan1807, 01:57 AM

P: 161

Very good. Now, I would like to have these equations simplified a bit. As an example, for node 4, you wrote:
[tex]\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)[/tex] I want it simplified to become: [tex]\frac{1}{3}V_1  \frac{1}{4}V_2  \frac{7}{12}V_4 + I_2 = 0[/tex] 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. 



#16
Jan1807, 02:21 AM

P: 492

NODE3: [tex]V_1\,+\,3\,V_2\,\,4\,V_3\,=\,5[/tex]
NODE2: [tex]2000\,I_1\,\,2000\,I_2\,\,2\,V_2\,+\,V_3\,=\,0[/tex] NODE1: [tex]4000\,I_1\,\,\frac{19}{3}\,V_1\,+\,V_3\,+\,\frac{4}{3}\,V_4\,=\,0[/tex] Right? 



#17
Jan1807, 02:48 AM

P: 161

Okay, that's close enough. Let me edit a bit...
Node 1: [tex]\frac{19}{12}V_1 + \frac{1}{4}V_3 + \frac{1}{3}V_4  I_1 = 0[/tex] Node 2: [tex]V_2 + \frac{1}{2}V_3 + I_1  I_2 = 0[/tex] Node 3: [tex]V_1 + 3V_2  4V_3 = 5[/tex] Node 4: [tex]\frac{1}{3}V_1  \frac{1}{4}V_2  \frac{7}{12}V_4 + I_2 = 0[/tex] 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. 



#18
Jan1807, 03:14 AM

P: 492

OK, I did N1 + N2 + N4.
[tex]\frac{5}{4}\,V_1\,\,\frac{5}{4}\,V_2\,+\,\frac{3}{4}\,V_3\,\,\frac{1}{4}\,V_4\,=\,0[/tex] Is this what the supernode would look like? 


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