Need help regarding node analysis and supersposition problems

[palui123]

I've been doing this question for like 2 days and what I find is wrong answer all over again. This is my last resort and I hope my hope is bright. Please if anyone knows how to do it. Include the steps and explanation. I need to understand the concept X_X.!

http://s1230.photobucket.com/albums/ee495/palui123/?action=view&current=Q2.jpg

28A and 2A what the hell is that?

http://s1230.photobucket.com/albums/ee495/palui123/?action=view&current=Q3.jpg

Is not that I don't know to this Superposition but what's with the 6V 0.5ohm , and 12V 1ohm . . What does it means? I don't see any resistor symbol in those source voltages.

 

[palui123]

No one helping? or it is just the image that I post is illegal...?

 

[gneill]

I've been doing this question for like 2 days and what I find is wrong answer all over again. This is my last resort and I hope my hope is bright. Please if anyone knows how to do it. Include the steps and explanation. I need to understand the concept X_X.!

We can help you to solve the problems, but we can't do the work for you. That's forum policy.

<snip picture>

28A and 2A what the hell is that?

Those are fixed current sources.

How about taking one problem at a time. The first problem asks you to use nodal analysis to find the currents in all the resistors in the network. Presumably you have covered nodal analysis in class so you have some idea how it goes? I've attached a schematic of the first problem ("Question 2" it was called in the picture), and labeled the obvious nodes with voltage labels V1, V2, and V3. Can you start by writing nodal equations corresponding to the node with V1?

 

[palui123]

Node 1
I=I1+I2+I3

28=v1/2+(v1-v2)/2+(v1-v3)/10

(x2) 56=V1+V1-V2+(v1-v3)/5

(x5) 280=5V1+5V1+5V2+V1-V3

280=11V1+5V2-V3

Am I correct?? I did it many times. If not correct me. .

 

[gneill]

It would appear that the sign for the V2 term is incorrect; You changed the sign of the V2 term when you multiplied through by 5. I get the equation:

280 = 11V1 - 5V2 - V3

Try the other two nodes and let's see how you do.

 

[palui123]

My bad. . .

Node 2

I4+I3=I6+2A

(V2-V3)/1+(V1-V3)/10=(V3/4)+2

(x10) 10V2-10V3+V1-V3=(10V3/4)+20

(x4) 40V2-40V3+4V1-4V3=10V3+80

4V1+40V2-54V3=80

Node 3

I2=I4+I5

(V1-V2)/2 = V2/5 + (V2-V3)/1

(x5) (5V1-5V2)/2 = V2+5V2-5V3

(x2) 5V1-5V2=2V2+10V2-10V3

5V1-17V2-10V3=0


Correct me when I'm wrong . After those steps what must I do?

 

[gneill]

Your Node 2 equation looks fine (although you could have divided through by 2).

For your Node 3 equation, the you forgot to change the sign of the V3 term when you moved it from the RHS to the LHS.

Once you've straightened out that sign, you will have three equations involving the node voltages V1, V2, and V3. Solve them. With the node voltages you will be in a position to determine all the resistor currents in the circuit in the same fashion that you wrote their symbolic values into your node equations -- find the potential across the resistor and divide by the resistor value.

 

[palui123]

Owhowh owh. . . okok after this I should use matrix solving to solve those 3 equation right? btw Do you have any other recommended method other than using Matlab and Matrix??

 

[gneill]

Owhowh owh. . . okok after this I should use matrix solving to solve those 3 equation right? btw Do you have any other recommended method other than using Matlab and Matrix??

Right.

For a 3 x 3 system you might just use straightforward substitution/elimination. It shouldn't get too messy. Even Cramer's Rule is not difficult here (3 x 3 determinants aren't hard to do by hand). Otherwise, use your favorite math package.

 

[palui123]

I summarize up the equation I have.

Node 1

280=11V1-5V2-V3

Node 2

40=2V1+20V2-27V3

Node 3

0=5V1-17V2+10V3


Correct?
after this what method should I use. If I use Matrix method . I find out it will become a big number.

 

[palui123]

I didn't see ur reply. . . testing now. . .

 

[palui123]

OHOHOHOHO! THANNK U VERY MUCH . IVe DOne it. . .HMM it was a silly mistake that took me so long. . . How about the "Question 3" what does it mean by "6v 0.5ohm" and "12V 1ohm" . Never been through it before. . .

 

[gneill]

OHOHOHOHO! THANNK U VERY MUCH . IVe DOne it. . .HMM it was a silly mistake that took me so long. . . How about the "Question 3" what does it mean by "6v 0.5ohm" and "12V 1ohm" . Never been through it before. . .

Glad to hear that you succeeded.

When a voltage supply is specified in the form "6.0V 0.5Ω" it means that it is not an ideal voltage supply, but a model of a real supply that has some internal resistance associated with it. The 0.5Ω is a "hidden" resistor, if you'd like to think of it that way, that "lives" inside the battery.

For this problem simply add the battery internal resistances to the existing external resistors that are in series with the batteries and proceed as usual.

 

[palui123]

Roger that. . .