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  #1  
Old 11-27-2008, 03:31 PM
adashiu adashiu is offline
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Default Nodal analysis with 2 voltage sources

Can someone solve this circuit?


Find v and i in the circuit.

I know that we have to make somethng with supernode, but where the supernode is here? How to develop this problem? Please help me

Btw it is nice forum, today i have found that. I hope I will be able to help you in other topics.

My attempts:

Im not sure, but I think that reference node will be that under 3 ohm resistor ( ? ) I have no idea how to cope with a problem of 2 supernodes in this circuits ( If I am right, that there are 2 supernodes). Actually Im green....

Greetings, Adam
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  #2  
Old 11-27-2008, 06:09 PM
guitarguy12387 guitarguy12387 is offline
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Hey Adam,

If it were me, i'd call the bottom node ground and make my super node around the 3V source. That way, you'll only need one supernode. Generally, my approach to supernodes is to call a pair of nodes a supernode when you only know the relative voltage drop between the two and not the absolute voltage with respect to ground. This is generally when a voltage source is connected between two non-reference nodes.

Hope that helps!
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  #3  
Old 11-27-2008, 06:13 PM
blazedaces's Avatar
blazedaces blazedaces is offline
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Quote:
Originally Posted by adashiu View Post
Can someone solve this circuit?


Find v and i in the circuit.

I know that we have to make somethng with supernode, but where the supernode is here? How to develop this problem? Please help me

Btw it is nice forum, today i have found that. I hope I will be able to help you in other topics.

My attempts:

Im not sure, but I think that reference node will be that under 3 ohm resistor ( ? ) I have no idea how to cope with a problem of 2 supernodes in this circuits ( If I am right, that there are 2 supernodes). Actually Im green....

Greetings, Adam
First of all, the ground (or reference node) is indeed under the 3 ohm resistor, but the node includes all connections to it. I'm not sure that you understood that... See, there's actually only 1 supernode in this problem.

A supernode occurs when you have a voltage source not touching the ground (or reference node). The voltage source on the left is between the node in the upper left and the reference node. Only the 3V voltage source is between two nodes that are not ground.

Do you understand what you need to do in case of a single supernode? Because if you had two there would be no difference, the procedure would be the same...

Perhaps take a look at THIS PDF for more of a tutorial on nodal analysis and supernodes. It has examples too...

-blazed
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  #4  
Old 11-27-2008, 08:32 PM
adashiu adashiu is offline
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Thanks for replying. Anyway if You could tell me how do we know the direction of a current in a branch?? Im going to have a coursework and Im damn idiot, out of any knowledge....
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  #5  
Old 11-27-2008, 08:48 PM
adashiu adashiu is offline
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My solution :

I made supernode with a 3V source between. After that I made a KCL for supernode :

i1=i2+i3+i4

7V/4Ω=V1/3Ω+V2/2Ω+V2/6Ω
21V=4V1+8V2

KVL for loop with a supernode(containg 3V source, 3 Ohm and 2 Ohm):

-3V+V2-3V1=0
V1=V2-3V

After solving system of equations we have

V2=33/12
V1=-1/4

Unfortuately the answer is -0,2V Where is a boob? :/
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  #6  
Old 11-27-2008, 10:21 PM
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blazedaces blazedaces is offline
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Look, I really would like to help you, but to be honest with you I was always bad with nodal analysis. I preferred mesh analysis so I don't know nodal analysis too well.

As a result, I'm probably not the best person to give you a good final answer. I'm sure there's someone else on the site who could though.

But... I'll give it a try anyways... to begin with, the voltage through the 4 ohm resistor should be based on both the voltage at node 1 and 2, since it's between the two nodes.

i1 = (V1 - V2)/4Ohms, assuming your current is going from left to right...

You should have a node between the 4 ohm resistor and the 7V voltage source. Therefore, your supernode equation probably looks more like this:

i1 = i2 + i3 + i4

(V1 - V2)/4Ohms = V2/3Ohms + V3/2Ohms + V3/6Ohms

And your constraint equation for the 3 V source should be:

3Volts = V3 - V2

Lastly, node 1 is not part of the supernode and has a simple enough equation:

V1 - V0 (ground so it's equal to 0) = 7V
=> V1 - 0 = 7V
=> V1 = 7V

Plug that back into your system of equations and solve for all three nodal voltages, which by the way you should kind of try to understand that there's no such thing as voltage at a node, or a point. Voltage means potential difference, or the difference of potential between two points. When we say nodal voltages, we really mean potentials at that node... but anyway.

Good luck. I hope you arrive at the correct answer,

-blazed
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  #7  
Old 11-27-2008, 10:44 PM
guitarguy12387 guitarguy12387 is offline
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Quote:
7V/4Ω=V1/3Ω+V2/2Ω+V2/6Ω
I think this is where you may have messed up.

Blazedaces' answer looks pretty good to me. I haven't plugged the numbers, but looks like a good approach.

But back to the problem. I think its important that you understand (as blazed said) that when applying KCL, you have to make sure that the 'voltage' is the voltage drop across the circuit element. For example, when trying to find the current through a resistor, you need to know the voltage drop across it.

More specifically, the first thing i'd do is apply KCL at the supernode. Out of convention, i generally reference currents going out of the node.

So these are your equations:

(V - 7)/4ohm + (V - 0)/3ohm + (Vo - 0)/2ohm + (Vo - 0)/6ohm = 0
Vo - V = 3volts.

Notice that i wrote in 0volts in the first equation in an effort to emphasize that voltages are potential difference between two points.

Quote:
Anyway if You could tell me how do we know the direction of a current in a branch??
Basically what you want to do is 'guess' which direction the current is going through a node when solving the circuit. If you end up with a negative current, then you know you guessed 'wrong.'

Hopefully that helps!
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  #8  
Old 11-28-2008, 04:42 AM
Ratch Ratch is offline
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adashiu,

Quote:
Can someone solve this circuit?
Yes.

Quote:
Find v and i in the circuit.
OK, one equation with one unknown. No worrying about which direction the charge flows. Find the voltage to the left of the 3v source and everything can be figured out from that. So we have for the nodal method:

(v-7)/4 + v/3 + (v+3)/2 + (v+3)/6 = 0 which is easily solved as v = -0.2v, i = (v+3)/2 = 1.4a

Ratch
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  #9  
Old 11-28-2008, 08:47 AM
The Electrician The Electrician is offline
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Assuming you designate the 3 nodes left to right along the top edge as v1, v2 and v3, with the bottom edge being your reference, proceed like this.

Your first equation is just a constraint equation, which simply says that v1 is 7 volts..

Your second equation is dealing with the supernode.

Your third equation is another constraint equation, which says that v3-v2=3 volts.

The three equations will be:

1*v1 + 0*v2 + 0*v3 = 7

-1/4*V1 + (1/4+1/3)*v2 + (1/2+1/6)*v3 = 0

0*v1 + (-1)*v2 + 1*v3 = 3

Setting up to solve these equations let's write them in matrix form:

Code:
  [    1           0            0      ]      [ 7 ]
  [  -1/4       1/4+1/3      1/2+1/6   ]   =  [ 0 ]
  [    0          -1            1      ]      [ 3 ]
Select All
Solving this with a matrix solver, either a fancy calculator or perhaps Matlab, you get:

[ 7 ]
[ -.2 ]
[ 2.8 ]

These are the 3 node voltages. We already knew that v1=7, and we now have v2 = -.2 volts and v3 =2.8 volts. Knowing v3, we can easily calculate i as 2.8/2 = 1.4 amps.

This solution is more involved than Ratch's, but it illustrates the formal method, which you may have to use in more complicated circuits.
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