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Find Io in the network nodal analysis

by Color_of_Cyan
Tags: analysis, nodal
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Color_of_Cyan
#1
Jan22-13, 02:39 PM
P: 306
1. The problem statement, all variables and given/known data





Find Io in the network

2. Relevant equations
V = IR

current division, voltage division,

KCL, KVL

Nodal Analysis (or mesh)


3. The attempt at a solution

So I think I need to use nodal analysis for this problem but not sure where to start with that.


I think it goes something like

24V - 12V = (2Ω)(I2), but not sure at all.


Do you use KVL / KCL at all? If yes, then how can I apply it here?

Tips for more places to start would also be helpful.


Thank you
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gneill
#2
Jan22-13, 04:52 PM
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P: 11,813
No need to invoke nodal analysis here. Just work backwards from the output and fill in voltages and currents as you go. For example, with 24V at the output, what's the current through the 4Ω resistor? What then is the current through the 2Ω resistor? What's the potential change across the 2Ω resistor? So then the potential at the top of the 18Ω resistor is... continue in that vein. Employ KVL, KCL, Ohm's Law, as required.
Color_of_Cyan
#3
Jan23-13, 04:02 AM
P: 306
so redrawing it like this:




I can't really tell which direction goes which and if I add or subtract that good with KVL but would this be on the correct track? :


24V - (4Ω)(I1) = 0

(4Ω)(I1) - (2Ω)(I2) - (18Ω)(I3) = 0

(18Ω)(I3) - 12V = 0

(2Ω)(I4) + (4Ω)(I5) = 0

Io = 2A + I5 + I3 + I1 ?

Where would I go from here or are there mistakes?


(and this was meant to go in the Engineering forum not physics forums by the way, because I was tired, heh)

haruspex
#4
Jan23-13, 04:17 AM
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Find Io in the network nodal analysis

Quote Quote by Color_of_Cyan View Post
(4Ω)(I1) - (2Ω)(I2) - (18Ω)(I3) = 0
That doesn't look right in relation to the arrows shown for the currents. Try again.
Color_of_Cyan
#5
Jan23-13, 05:09 AM
P: 306
I can't really tell which would be negative / positive in the equation though, any tips?

How about (2Ω)(I2) - (18Ω)(I3) + (4Ω)(I1) = 0, for the one you said?
haruspex
#6
Jan23-13, 05:32 AM
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Labelling the junctions T1 to T4 in the top row and B1 to B4 in the bottom row, the diagram shows two routes from T3 to B4. The voltage drop must be the same in both routes.
Color_of_Cyan
#7
Jan23-13, 03:13 PM
P: 306
So, you can say (4Ω)(I5) = (2Ω)(I4) = 12 V then ?

Would that make I4 = 6A and I5 = 3A?


Not sure what to do with 2 voltage sources in regards to finding current through a resistor.. (ie how would I find the current through the 2Ω resistor ? )


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