Node Voltage Help

Discussion in 'Homework Help' started by Ryuk, Mar 6, 2013.

1. Ryuk Thread Starter New Member

Oct 9, 2012
18
0 Assume ground reference is at bottom node.

So in this circuit diagram, some of the currents are shown their directions with an arrow. As I understand NVM, I need to pick whether currents going into a node are positive or negative and vice versa for going out when writing equations equal to zero. However, how do I know which way the current is flowing when there's no arrow specified?

For example, at node V1, I don't know which direction the current is flowing through the middle 8Ω resistor. I know that with regard to the equation, I write (V1-0)/8Ω for the node V1 equation that's equal to zero. However, since I don't know the direction of the current, how do I know if I'm supposed to make that + or -?

Another thing, how can current flow away from ground? Ground is the lowest potential so how is that possible? I see op amp circuits all the time that show current flowing away from ground. It confuses me.

2. WBahn Moderator

Mar 31, 2012
24,555
7,691
You can pick either the polarity of the voltage across a component or the direction of the current through a component arbitrarily. But once you pick one of them (or if it is already indicated), then you need to assign the other one so that it is consistent. For loads, such as resistors, positive current flows through the device from the positive terminal to the negative terminal (of the voltage across that device). For a source, the opposite is true.

If you are using the equation (V1-0)/8Ω to calculate the current (and, BTW, it's not 8Ω, but rather it's 8kΩ), then that means that the current is flowing from the V1 side to the 0 (ground) side. So you assign the current direction in that resistor as flowing from top to bottom (from Node 1 to GND). If it turns out that V1 < 0V, then the voltage across the resistor is negative and the current is also negative, meaning that it is actually flowing in the opposite direction of the assigned arrow.

Ground is not necessarily the lowest potential. It is nothing more than a node that we have chosen to assign a value of 0V to in order to use it as a reference for the voltages on all of the other nodes. If a node has a voltage lower than the node we chose as our reference, then it has a negative voltage relative to that reference. Since, by definition, a voltage is a potential difference between two nodes, we can only talk about the voltage of a node in reference to another node. We get to pick which node is our reference and we can any node we want.

In your circuit, the voltage at the bottom of the 2mA current source is going to be below ground because it will generate whatever voltage is needed to force 2mA to flow away from the ground node through a 4kΩ resistor. That will result in the voltage on that node being 0V-2mA*4kΩ=-8V.

3. Ryuk Thread Starter New Member

Oct 9, 2012
18
0
I'm still confused. For instance, at node V1, we have the 1 mA current flowing away and the current across the left resistor also flows away from the node since the resistor has terminals from + to -. However, I can randomly assign whichever way current flows for the other two resistors? Well...at least one of them has to flow in otherwise the circuit wouldn't make sense?

Ok, have a look at this circuit (ground is at bottom node again) at node B. I know that the correct equation is (Vb - Va)/1Ω + (Vb - 0)/2Ω + (Vb - 2)/2Ω = 0. That equation is stating however that all currents are flowing away from the node. That's not possible though. I know one of the currents is going to come out negative but I still can't understand why the set up of the equation is done this way when the equation is stating all currents are flowing away. Last edited: Mar 7, 2013
4. WBahn Moderator

Mar 31, 2012
24,555
7,691
Keep in mind that just because we label a voltage or a current with a certain polarity does not mean that we labeled it correctly. You crank the analysis and if the end result for a certain voltage or current is negative then all that means is that the actual voltage or current is of the opposite polarity.

You can freely declare that all of the currents are going into (or out of) a particular node. All that means is that you know that at least one of them is going to end up being negative.

5. DiodeMan New Member

Feb 3, 2013
13
0
I do not claim to be an expert, but in my experience questions similar to this would be solved by using superposition theorem. However, I'm no electrical engineer, so I'm unsure if that's how this question would properly be solved.

6. WBahn Moderator

Mar 31, 2012
24,555
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The issues the OP is trying to get a handle on are independent of analysis method. They are fundamental understanding of voltages and currents and the difference between symbolic values and actual values.

7. Ryuk Thread Starter New Member

Oct 9, 2012
18
0
So if I assume going into the node V1 is positive, the equation can be -1 - (V1 - 4)/8kΩ + (V1 - 0)8kΩ + (V1 - V2)/4kΩ = 0. Correct? I want to assume the unknown currents are going in.

Maybe so, but I'm taking a circuit analysis course where we're required to solve circuits with all methods we learn...including mesh current, source transformations, and superposition. I don't have much of a problem with the other methods but node voltage is just not clicking with me.

8. WBahn Moderator

Mar 31, 2012
24,555
7,691
Okay, you need to use units in your work. (V1-4) is meaningless because V1 is a voltage and 4 is a number. If it is really 4V, then you need to say 4V. The exception is a value of exactly 0 because 0 can have any units whatsoever and it doesn't change anything. So that's the one area where I will cut you some slack (and I will often leave the units off of 0 myself, though generally not in an answer).

If you want to assume that all of the currents are going into Node 1, then you need to write the equation with each term constructed for a current going into Node 1.

Let's write the four currents separately as I_N, I_S, I_E, and I_W (north, south, east, and west for ease of reference) with the directions chosen so that the currents are into Node 1 if the current is positive.

We know that the sum of the currents into the node has to be equal to the sum of the currents out of the node, so

I_N + I_S + I_E + I_W = 0

For the three nodes connected to resistors, the expression for the current going into Node 1 is (Voltage_on_other_node - V1)/Resistance.

So:

I_W = (4V - V1)/8kΩ

I_S = (0V - V1)/8kΩ

I_E = (V2 - V1)/4k

For the current source, since it is 1mA flowing away from Node 1, we have

I_N = -1mA

Our equation is thus

I_N + I_S + I_E + I_W = 0

-1mA + (0V - V1)/8kΩ + (V2 - V1)/4k + (4V - V1)/8kΩ = 0

While this is similar to your equation, it doesn't match completely.

Looking at your:

[-1] - [(V1 - 4)/8kΩ] + [(V1 - 0)8kΩ] + [(V1 - V2)/4kΩ] = 0

Your first term in square brackets is a current going into Node 1.

Your remaining terms in brackets are all currents going out of Node 1.

The second term is changed to a current going into Node 1 because of the minus sign in front of it, but the others are still currents going out of Node 1.