# Nodal Analysis

Discussion in 'Homework Help' started by tim_3491, Apr 1, 2008.

1. ### tim_3491 Thread Starter New Member

Apr 1, 2008
7
0
Hi

I have been doing this problem with mesh analysis and using matrices to solve the currents etc. However i get a different answer everytime. I understand nodal analysis but this question has an ammeter in it so now i don't no what to do.

I have the following equations for the attached document
A: I8=I2+I1
B: I8=I3
C: I2+0.5A=I4
D: I3=I6+I7
E:
F:

And then for the currents

I1=(Va-Ve)/10
I2=(Va-Vc)/20
I3=(Vb-Vd)/30
I4=(Vc-Ve)/100
I5=(Vf-Ve)/100
I6=(Vd-Vf)/100 => Vd/100
I7=0.5A
I8=Vb-Va+5V

But i don't know what to do with the equations and whether or not i can leave 0.5A in there and use it.

Any help would be appreciated

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2. ### RmACK Active Member

Nov 23, 2007
54
0
Sounds really familiar. I went through the same thing 3 years ago...

Still, here's my two cents.
I8=I3 is wrong. I8=-I3 as the arrows point opposite ways.
E: I5=I1+I4
F: I6=I5
Also your statement I8=Vb-Va+5 is wrong, where is the resistance to relate voltages Vb & Va to current I8?? Just use the I3=-I8 here.

Hope this helps a little, I really need to go to sleep or I'd help you more.

3. ### RmACK Active Member

Nov 23, 2007
54
0
Here is what I did:
I know that all the currents out of a node sum to zero. I also will be calling the voltages at points on the circuit A B C etc instead of Va, Vb, etc for clarity.
Remember F=0

Point E
E/100+(E-C)/100+(E-A)/10=0
E+E-C+10E-10A=0
-10A-C+12E=0 (1)

Point C
(C-E)/100+(C-A)/20-0.5=0
C-E+5C-5A-50=0
-5A+6C-E-50=0 (2)

Point A&B lumped together as a supernode
(A-E)/100+(A-E)/20+(B-d)/30=0
6A-6E+3A-3C-2B-2D=0
9A+2B-3C-2D-6E=0 (3)

POINT D
(D-B)/30+D/100+0.5=0
10D-10B+3D+150=0
-10B+13D+150=0 (4)

Oh and indidentally, A=B+5 so B=A-5 (5)

POINT F
F-E/100+F-D/100=0 etc so F=0 and you get
D=-E or E=-D (6)

Now I use (5) to convert all B's to A's and (6) to convert all E's to D's by substitution getting the following set of equations:

From (1)
-10A-C-12D=0 (7)
From (2)
-5A+6C+D-50=0 (8)
FROM (3)
11A-3C+4D-10=0 (9)
From (4)
-10A+13D+200=0
D=(10A-200)/13 (10)
NOW WE TAKE (7)-2(8):
-10A-C-12D+10A-12C-2D+100=0
-13C-14D+100=0
C=(100-14D)/13 (11)

Now I substitute (11) and into (7) to convert all C's into D's
-10A-C-12D=0
-10A-((100-14D)/13)-12D=0
NOW SUBSTITUTE IN (10) TO CONVERT ALL D'S TO A'S)
-10A-(100/13)+((14/13)-12)((10A-200)/13)=0
Multiply by 13
-130A-100+((14/13)-12)(10A-200)=0
Multiply by 13 again
-1690A-1300+(-142)(10A-200)=0
-1690A-1300-1420A+28400=0
3110A=27100
A=27100/3110=8.714V YAY!
THEN B=A-5=3.714V
D=(10A-200)/13=-8.682V
etc etc you can work out voltages and currents from there.
Attached is a pspice simulation showing that my answers are correct. Hope this helps you.

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