Confused about choosing current direction in nodal analysis

AI Thread Summary
The discussion centers on confusion regarding current direction in nodal analysis for a circuit problem. The participant initially struggles with the consistency of current direction choices, leading to incorrect voltage calculations. They clarify that all currents entering or leaving a node must be treated consistently in equations, which is crucial for accurate results. After revising their equations, they successfully find the correct voltage value of 1.73 V. The conversation concludes with a question about the number of equations needed for supernodes, confirming that only n-1 equations are necessary when dealing with multiple supernodes.
mgmt113
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Homework Statement


The problem states: Determine the voltage labeled v in the following circuit: https://ibb.co/hz0Q37
upload_2018-2-16_19-21-39.png


The variable is the voltage across the 2 ohm resistor. The ground I chose for this example is located in the center of the circuit.

Homework Equations


V=IR I guess.

The Attempt at a Solution


I have a huge confusion regarding current directions. How come the textbook says that you can choose any current direction you want as long as you're consistent, but when I choose them to be a certain way trying to be consistent, I get the wrong answer? When I check solved problems, they assume current directions and I would have not gotten the same answers if I chose my directions. So, can you really choose the directions you want, and am I missing something important?

These are the equations I got. I got all the terms on one side to avoid confusion. If the current is entering the node, then I assign it a negative sign, if it's leaving the node, it stays positive.

For Node 1 (the one in the uppermost part of the circuit)
1 + (V3-V1)/10 + V1/2 = 0

For Supernode 2-4

-1 + (V2-V4)/12 - V3/20 + 5 -V1/2 = 0

For Supernode 3-4

V3/20 + (V1-V3)/10 -5 + (V4-V2)/12 = 0

The answer for V1 that I get is -0.5 V, however the solution manual says it's 1.7 V.

Are the directions I posted on the image right? If not, why? And is there an strategy to avoid getting confused at directions?

Thank you very much for your help. I really appreciate it.

 

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If you are consistent then it does not matter what direction you choose. If you get a negative answer, that just means you chose the "wrong" direction and the current is really flowing in the opposite direction from what you chose.
 
mgmt113 said:
1 + (V3-V1)/10 + V1/2 = 0
This is not consistent.
The basic equation is that the sum of flows into (or out of, if you prefer) a node is zero.
With the arrows drawn as you have them, you would write (flow 3 to 1)=(flow 1 to 2) + (flow 1 to 0).
That leads to
(V3-V1)/10 = 1 + V1/2.
 
haruspex said:
This is not consistent.
The basic equation is that the sum of flows into (or out of, if you prefer) a node is zero.
With the arrows drawn as you have them, you would write (flow 3 to 1)=(flow 1 to 2) + (flow 1 to 0).
That leads to
(V3-V1)/10 = 1 + V1/2.
Forgive me if I'm wrong. I thought that if the current was leaving the node, I could write V1-V3, but because it is entering the node, then it's -(V1-V3) and that's why I have V3-V1. Thanks!
 
mgmt113 said:
Forgive me if I'm wrong. I thought that if the current was leaving the node, I could write V1-V3, but because it is entering the node, then it's -(V1-V3) and that's why I have V3-V1. Thanks!
Yes, but that is not where you went wrong.
In your equation
mgmt113 said:
1 + (V3-V1)/10 + V1/2 = 0
you have added a mixture of currents entering the node and currents leaving the node. That doesn't work. For the sum to be zero, they must all be entering or all leaving.
 
mgmt113 said:

Homework Statement



For Node 1 (the one in the uppermost part of the circuit)
1 + (V3-V1)/10 + V1/2 = 0

For Supernode 2-4

-1 + (V2-V4)/12 - V3/20 + 5 -V1/2 = 0

For Supernode 3-4

V3/20 + (V1-V3)/10 -5 + (V4-V2)/12 = 0
You have another problem here. Your three equations are not linearly independent. You can see (after you get the signs correct) that adding the first two equations results in the third equation. So you have two linearly independent equations and four unknowns. You need to come up with two more equations from your diagram. There are very simple ones you can use. They are not current balance equations.
 
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Thank you so much for your help! I've been understanding it wrong the whole time.

These are the equations that I came up with:

For Node 1:
-0.6 V1 + 0.1 V3 = 1

For Supernode 3-4:
0.1V1 + 0.083 V2 - 0.15 V3 - 0.083 V4 = -5

For the sources inside the supernodes:

V4-V3=10
V2=5

The answer for v is 1.73 V. Yay!

One more question: When dealing with n supernodes, do I have to only take into account n-1 equations for solving it, as they will repeat themselves?

Thank you so much again!
 
mgmt113 said:
Thank you so much for your help! I've been understanding it wrong the whole time.

These are the equations that I came up with:

For Node 1:
-0.6 V1 + 0.1 V3 = 1

For Supernode 3-4:
0.1V1 + 0.083 V2 - 0.15 V3 - 0.083 V4 = -5

For the sources inside the supernodes:

V4-V3=10
V2=5

The answer for v is 1.73 V. Yay!

One more question: When dealing with n supernodes, do I have to only take into account n-1 equations for solving it, as they will repeat themselves?

Thank you so much again!
If you have k nodes and n current sources, and can get one potential difference equation for each source, then the number of current balance equations you need is going to be k-n-1.
 
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