Sign convention to use in Nodal Analysis

Question

^{simulate this circuit – Schematic created using CircuitLab}

We know that current flows from higher potential to lower potential in a resistor. In this question, we make use of nodal analysis and name the nodal voltages V_x and V_Th.

When I write the equations how do I assume the direction of current, whether the current is flowing from V_x to V_Th or the other way? My equations are :

In the first node, 1A current is entering the node V_x and V_x/6 current is leaving as it is connected to ground. But what about the other part of the equation? Will it be incoming i.e. 1+(V_x+14-V_Th)/14 or outgoing i.e. 1-(V_x+14-V_Th)/14.

Answer 1

In short, there isn't a convention. The result of your calculations is a dependent on the assumption you make when forming the equation.

For example, for V_x, the 1A source will only allow current to flow into the node (by definition), and you have also placed the 14v branch on the same side of the equation using a '+'. Therefore you have assumed the same direction of current flow for both of these branches (same side of the equation) and since you have written the 1A as a positive value, you have assumed a convention of positive value meaning current flowing into the node.

What is confusing the matter, is that you have placed the resistor branch on the opposite side of the equation. This means that you have assumed that the resistor branch has current flowing out of the node, and therefore a positive value for this side of the branch is signifying a different direction to a positive value on the other side.

Ultimately, what you do depends on how you want to keep track of your logic. If you want to make implicit assumptions of current flow directions, and then keep track of the direction you are assuming, so that you can then interpret your positive values depending on which side of the equation they are, that's fine.

What I would suggest is easier (especially as circuits get larger and harder to logically track through), is to assume a global direction for all nodes. I usually take an assumption of all branches pushing current into the node, and that current flow into a node is positive. This means that my equations take the form of: $$X + Y + Z = 0$$ Where now if a X, Y, or Z come out positive, I know that its current flowing into the node, and if any of them come out negative, its flowing out of the node. The nice thing about this method is that the maths keeps track of all your directions for you, and you only need to remember one rule in order to interpret your answers, instead of also keeping track of the specific direction assumption you made for each branch.

The only thing you need to bear in mind is that the direction and sign needs to be conserved across the circuit, so for example the node V_x will have an equation of: $$1+\frac{{V_{Th} - 14 - V_{x}}}{14}+\frac{-V_x}{6} = 0$$ But for V_Th it will be: $$-\frac{{V_{Th} - 14 - V_{x}}}{14} - 3 + \frac{-V_{Th}}{5} = 0$$ As we have already assumed the direction of the 14v branch to be into V_x, it is written as a negative on this node to conserve the positive-into-node convention. Similarly the current source of 3A is pulling current out of the node, so it is written in as a negative.

Answer 2

First, I will present a method that uses Mathematica to solve this problem. I know that this approach is not 'smart' but this method will work all the time, even when the circuit is (way) more complicated than this one. Also, this method will check your work.

Well, we are trying the analyze the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \begin{alignat*}{1} \text{I}_\text{a}&=\text{I}_1+\text{I}_2\\ \\ \text{I}_2&=\text{I}_\text{b}+\text{I}_5\\ \\ \text{I}_5&=\text{I}_3+\text{I}_4\\ \\ \text{I}_6&=\text{I}_3+\text{I}_4\\ \\ 0&=\text{I}_\text{b}+\text{I}_0+\text{I}_6\\ \\ \text{I}_1&=\text{I}_\text{a}+\text{I}_0 \end{alignat*} \end{cases}\tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \begin{alignat*}{1} \text{I}_1&=\frac{\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2&=\frac{\text{V}_2-\text{V}_3}{\text{R}_2}\\ \\ \text{I}_3&=\frac{\text{V}_3}{\text{R}_3}\\ \\ \text{I}_4&=\frac{\text{V}_4}{\text{R}_4} \end{alignat*} \end{cases}\tag2 $$

We also know that \$\text{V}_2-\text{V}_1=\text{V}_0\$.

Using \$(2)\$ we can rewrite \$(1)\$ as follows:

$$ \begin{cases} \begin{alignat*}{1} \text{I}_\text{a}&=\frac{\text{V}_1}{\text{R}_1}+\frac{\text{V}_2-\text{V}_3}{\text{R}_2}\\ \\ \frac{\text{V}_2-\text{V}_3}{\text{R}_2}&=\text{I}_\text{b}+\text{I}_5\\ \\ \text{I}_5&=\frac{\text{V}_3}{\text{R}_3}+\frac{\text{V}_3}{\text{R}_4}\\ \\ \text{I}_6&=\frac{\text{V}_3}{\text{R}_3}+\frac{\text{V}_3}{\text{R}_4}\\ \\ 0&=\text{I}_\text{b}+\text{I}_0+\text{I}_6\\ \\ \frac{\text{V}_1}{\text{R}_1}&=\text{I}_\text{a}+\text{I}_0 \end{alignat*} \end{cases}\tag3 $$

Now, we can set up a Mathematica code to solve for all the voltages and currents:

In[1]:=Clear["Global`*"];
FullSimplify[
 Solve[{Ia == I1 + I2, I2 == Ib + I5, I5 == I3 + I4, I6 == I3 + I4, 
   0 == Ib + I0 + I6, I1 == Ia + I0, I1 == V1/R1, I2 == (V2 - V3)/R2, 
   I3 == V3/R3, I4 == V3/R4, V2 - V1 == V0}, {I0, I1, I2, I3, I4, I5, 
   I6, V1, V2, V3}]]

Out[1]={{I0 -> -((
    Ib R3 R4 + 
     Ia R1 (R3 + R4) + (R3 + R4) V0)/((R1 + R2) R3 + (R1 + R2 + 
        R3) R4)), 
  I1 -> (Ia R2 R3 - Ib R3 R4 + 
    Ia (R2 + R3) R4 - (R3 + R4) V0)/((R1 + R2) R3 + (R1 + R2 + 
       R3) R4), 
  I2 -> (Ib R3 R4 + 
    Ia R1 (R3 + R4) + (R3 + R4) V0)/((R1 + R2) R3 + (R1 + R2 + 
       R3) R4), 
  I3 -> (R4 (Ia R1 - Ib (R1 + R2) + 
      V0))/((R1 + R2) R3 + (R1 + R2 + R3) R4), 
  I4 -> (R3 (Ia R1 - Ib (R1 + R2) + 
      V0))/((R1 + R2) R3 + (R1 + R2 + R3) R4), 
  I5 -> ((R3 + R4) (Ia R1 - Ib (R1 + R2) + 
      V0))/((R1 + R2) R3 + (R1 + R2 + R3) R4), 
  I6 -> ((R3 + R4) (Ia R1 - Ib (R1 + R2) + 
      V0))/((R1 + R2) R3 + (R1 + R2 + R3) R4), 
  V1 -> (R1 (Ia R2 R3 - Ib R3 R4 + 
      Ia (R2 + R3) R4 - (R3 + R4) V0))/((R1 + R2) R3 + (R1 + R2 + 
       R3) R4), 
  V2 -> (-Ib R1 R3 R4 + Ia R1 (R3 R4 + R2 (R3 + R4)) + 
    R2 R3 V0 + (R2 + R3) R4 V0)/((R1 + R2) R3 + (R1 + R2 + R3) R4), 
  V3 -> (R3 R4 (Ia R1 - Ib (R1 + R2) + 
      V0))/((R1 + R2) R3 + (R1 + R2 + R3) R4)}}

Now, we can find:

• \$\text{V}_\text{th}\$ we get by finding \$\text{V}_3\$ and letting \$\text{R}_4\to\infty\$: $$\text{V}_\text{th}=\frac{\text{R}_3\left(\text{V}_0+\text{I}_\text{a}\text{R}_1-\text{I}_\text{b}\left(\text{R}_1+\text{R}_2\right)\right)}{\text{R}_1+\text{R}_2+\text{R}_3}\tag4$$
• \$\text{I}_\text{th}\$ we get by finding \$\text{I}_4\$ and letting \$\text{R}_4\to0\$: $$\text{I}_\text{th}=\frac{\text{V}_0+\text{I}_\text{a}\text{R}_1-\text{I}_\text{b}\left(\text{R}_1+\text{R}_2\right)}{\text{R}_1+\text{R}_2}\tag5$$
• \$\text{R}_\text{th}\$ we get by finding: $$\text{R}_\text{th}=\frac{\text{V}_\text{th}}{\text{I}_\text{th}}=\frac{\text{R}_3\left(\text{R}_1+\text{R}_2\right)}{\text{R}_1+\text{R}_2+\text{R}_3}\tag6$$

Where I used the following Mathematica codes:

In[2]:=FullSimplify[
 Limit[(R3 R4 (Ia R1 - Ib (R1 + R2) + 
     V0))/((R1 + R2) R3 + (R1 + R2 + R3) R4), R4 -> Infinity]]

Out[2]=(R3 (Ia R1 - Ib (R1 + R2) + V0))/(R1 + R2 + R3)

In[3]:=FullSimplify[
 Limit[(R3 (Ia R1 - Ib (R1 + R2) + 
     V0))/((R1 + R2) R3 + (R1 + R2 + R3) R4), R4 -> 0]]

Out[3]=(Ia R1 - Ib (R1 + R2) + V0)/(R1 + R2)

In[4]:=FullSimplify[%2/%3]

Out[4]=((R1 + R2) R3)/(R1 + R2 + R3)

Using your values we get:

• $$\text{V}_\text{th}=-8\space\text{V}\tag7$$
• $$\text{I}_\text{th}=-2\space\text{A}\tag8$$
• $$\text{R}_\text{th}=4\space\Omega\tag9$$