# Nodal Analysis

Discussion in 'Homework Help' started by Gatesy, Nov 11, 2009.

1. ### Gatesy Thread Starter New Member

Nov 11, 2009
5
0
hi ive been asked to complete nodal and mesh analysis on the above circiut in order to find the voltage across R6. i have managed to complete the mesh analysis and also i have used SPICE so i know i am looking for 4v across R6 im at a complete loss now however with nodal. i think i have formulea correct for N1. N2 & N3 however i am stuck at this point

N1= (1/ 2k)N1 - (1/1k8)N2 = 1v/2k = 5mA
N2 = -(1/2k)N1 + (1/2k + 1/8k)N2 + (1/4k8)N3 = 16/2k = 2mA
N3 = 0*N1 + (-1)N2 +(1)N3 = 14.4/4k8 = 3mA

i think the above is correct however i am well & truly stuck now.

any help would be much appriciated.

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2. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,724
496
In what way are you stuck? You have 3 equations in 3 unknowns. Why don't you just solve them?

3. ### Gatesy Thread Starter New Member

Nov 11, 2009
5
0
i am unsure how to find the final node voltage. VL1. i was given a hint when i got the problem that N3 & N4 will be equal. but to find VL1 i dont know what step to take next.

4. ### Gatesy Thread Starter New Member

Nov 11, 2009
5
0
those calculations above were wrong anyway i have the node voltages at N1= 1v
N2 = 8.4V N3= 10.8V & NL= 4V i found these using SPICE so when i sub teh numbers into my equations above im nowhere near. back to the drawing board

5. ### Firestorm Senior Member

Jan 24, 2005
353
0
Try using a supernode around N2 N3 and N4. Start from the left, and work your way Node to Node.
http://img511.imageshack.us/img511/9836/nodalproblem.png

If you have any questions let me know. I usually not big on working out a whole problem, but it looks like you put forth the effort before asking questions.

Chet

6. ### Firestorm Senior Member

Jan 24, 2005
353
0
Disregard the current in the top right of the diagram. I was doing 2 different ways, so the current direction and the equation conflict one another. It should be Vn4-Vl1/6.8k given that current direction. It is shown in the equations worked below on that page.