Nodal analysis question

  • #1
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While performing nodal analysis problems, I am always unsure of which node gets subtracted from during KCL. For example, if I have (V1-V2)/2k, how do I know that it shouldn't be (V2-V1)/2k?
 

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  • #2
berkeman
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While performing nodal analysis problems, I am always unsure of which node gets subtracted from during KCL. For example, if I have (V1-V2)/2k, how do I know that it shouldn't be (V2-V1)/2k?
Welcome to the PF.

I use the convention that the sum of all currents *out* of each node is zero. That gives me the direction for each voltage subtraction. Makes sense?

EDIT -- to be a bit more clear. Since I'm summing the currents out of a particular node, the node's voltage is the first one in the subtraction equations for that node.
 
  • #3
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Welcome to the PF.

I use the convention that the sum of all currents *out* of each node is zero. That gives me the direction for each voltage subtraction. Makes sense?
Thanks, that makes sense. Say I have I1+I2-I3=0. How do you determine if I1=(V1-V2)/12k compared to I1=(V2-V1)/12k. Does it have something do to based on the reference node?
 
  • #4
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Welcome to the PF.

I use the convention that the sum of all currents *out* of each node is zero. That gives me the direction for each voltage subtraction. Makes sense?

EDIT -- to be a bit more clear. Since I'm summing the currents out of a particular node, the node's voltage is the first one in the subtraction equations for that node.
OH I think I understand it better now. So when I have the sum of currents equal to zero (KCL), the direction of the current determines which one is subtracted?
 
  • #5
berkeman
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OH I think I understand it better now. So when I have the sum of currents equal to zero (KCL), the direction of the current determines which one is subtracted?
Yes. Can you post an example circuit and show your reasoning now? :smile:
 
  • #6
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Yes. Can you post an example circuit and show your reasoning now? :smile:
I can't figure out how to post a picture from my phone, but the way I'm doing it now is when the current goes through the resistor, I'm taking the node on the negative end and subtracting it from the node on the positive end of the resistor
 
  • #7
berkeman
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I can't figure out how to post a picture from my phone, but the way I'm doing it now is when the current goes through the resistor, I'm taking the node on the negative end and subtracting it from the node on the positive end of the resistor
When summing the currents *out* of a node, subtract the far voltage from the near voltage (the near voltage is at your node). Don't worry what the values of the actual voltages are at this step. So ignore the current directions shown in the schematic below, and just write the two node equations for the sum of the currents out equals zero for each...

https://www.ibiblio.org/kuphaldt/electricCircuits/DC/00221.png
00221.png
 
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  • #8
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So @ V1: I2+I3-I1=0
-> (V1-0)/R2 + (V1-V2)/R3 - (B1-V1)/R1

Would this be right?
 
  • #9
berkeman
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(V1-0)/R2 + (V1-V2)/R3 - (B1-V1)/R1
Not quite. I would write it like this:

(V1-0)/R2 + (V1-V2)/R3 + (V1-B1)/R1 = 0

Remember to keep it in the form of the sum of all currents out of the node. When you start changing signs so it's not a sum anymore, it can be easy to get confused and make errors. :smile:
 
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  • #10
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Not quite. I would write it like this:

(V1-0)/R2 + (V1-V2)/R3 + (V1-B1)/R1 = 0

Remember to keep it in the forum of the sum of all currents out of the node. When you start changing signs so it's not a sum anymore, it can be easy to get confused and make errors. :smile:
I see what you mean, it's easier to think of it like that and have them all as a sum and go from there. Thank you for the help you just saved my test grade tomorrow!
 
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