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Node method again... |
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| Jul5-06, 09:10 AM | #1 |
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Node method again...
I'm trying to solve an exercise on the node method but I'm not quite sure if the equations for the node tensions are right,so I was hoping if someone could give me a hand...Here are the node equations:
Node 1: v1/50 + (v1-50)/80 + (v1-v0)/40=0 Node 2: v0/200 + (v0-v1)/40 + (v0-50)/800 - 0,75=0 The link for the exercise is: http://i75.photobucket.com/albums/i2...nodemethod.jpg Thanks in advance for the help! |
| Jul5-06, 01:37 PM | #2 |
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Not quite right. Call the voltage to ground at node 2 "V2" and try again. Vo as labelled is not V2. Vo is the voltage across the top resistor, not to ground.
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| Jul5-06, 11:56 PM | #3 |
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Hummm....I guess I'm not understanding quite well what you are saying...Could you or someone help me correct my equations?
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| Jul6-06, 08:29 AM | #4 |
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Node method again...
So...V0 would be something like: v0=(v2-v1)/800 ?
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| Jul6-06, 11:57 AM | #5 |
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Just go ahead and re-write the equations one more time using V2 as a term. Don't worry about Vo for now. In the end you will have V2, and that's enough to solve for Vo. |
| Jul6-06, 02:55 PM | #6 |
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Thanks!I wasn't attending to the fact that it was the voltage what we wanted to know,I thoughtthe current instead...I think I'm getting it now...
So,the equations should be something like: Eq. 1:(v1-50)/80 + v1/50 + (v1-v2)/40=0 Eq. 2: (v2-v1)/40 + v2/200 + v2/800 -0,75=0 Is this right? |
| Jul6-06, 02:59 PM | #7 |
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Almost, but in Equation 2, you need to also account for the current flowing through the 800 Ohm resistor up on top. Add that current out of node 2 into Equation 2, and then you can solve for V1 and V2, which gives you Vo.
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| Jul6-06, 03:25 PM | #8 |
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But,Isn't that current flowing through the 800 ohm resistor given by v2/800?
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| Jul6-06, 04:58 PM | #9 |
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No. What is the voltage on the left side of the 800 Ohm resistor? It's not zero. So the current isn't (V2-0)/800.
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| Jul7-06, 07:38 AM | #10 |
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I got it...The current is (V2-50)/800.Thanks for the help!
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