Example: Application of Node Rule

Return to current in wires

Suppose you have the circuit below. You are given a few values: $I_{1} = 8\text{~A}$, $I_{2} = 3\text{~A}$, and $I_{3} = 4\text{~A}$. Determine all other currents in the circuit, using the Current Node Rule. Draw the direction of the current as well.

Circuit

Facts

$I_{1} = 8\text{~A}$, $I_{2} = 3\text{~A}$, and $I_{3} = 4\text{~A}$.
$I_{1}$, $I_{2}$, and $I_{3}$ are directed as pictured.
The Node Rule is $I_{in} = I_{out}$, for any point along the current.

Goal

Find all the currents in the circuit and their directions.

Representations

For simplicity of discussion, we label the nodes in an updated representation:

Circuit with Nodes

Assumption

We will assume we have a perfect battery to supply a steady current to the circuit and will not die over time.

Solution

Okay, there is a lot going on with all these nodes. Let’s make a plan to organize our approach.

Plan

Take the nodes one at a time. Here’s the plan in steps:

Look at all the known currents attached to a node.
Assign variables to the unknown currents attached to a node.
Set up an equation using the Node Rule. If not sure about whether a current is going in or coming out of the node, guess.
Solve for the unknown currents.
If any of them are negative, then we guessed wrong two steps ago. We can just flip the sign now.
Repeat the above steps for all the nodes.

Let’s start with node $A$. Incoming current is $I_{1}$, and outgoing current is $I_{2}$. How do we decide if $I_{A \rightarrow B}$ is incoming or outgoing? We need to bring it back to the Node Rule: $I_{in} = I_{out}$. Since $I_{1} = 8\text{~A}$ and $I_{2} = 3\text{~A}$, we need $I_{A \rightarrow B}$ to be outgoing to balance. To satisfy the Node Rule, we set

$$ I_{A \rightarrow B} = I_{out} - I_{2} = I_{in} - I_{2} = I_{1} - I_{2} = 5\text{~A} $$

We do a similar analysis for node $B$. Incoming current is $I_{A \rightarrow B}$, and outgoing current is $I_{3}$. Since $I_{A \rightarrow B} = 5\text{~A}$ and $I_{3} = 4\text{~A}$, we need $I_{B \rightarrow D}$ to be outgoing to balance. To satisfy the Node Rule, we set

$$ I_{B \rightarrow D} = I_{out} - I_{3} = I_{in} - I_{3} = I_{A \rightarrow B} - I_{3} = 1\text{~A} $$

For node $C$, incoming current is $I_{2}$ and $I_{3}$. There is no outgoing current defined yet! $I_{C \rightarrow D}$ must be outgoing to balance. To satisfy the Node Rule, we set

$$ I_{C \rightarrow D} = I_{out} = I_{in} = I_{2} + I_{3} = 7\text{~A} $$

Lastly, we look at node $D$. Incoming current is $I_{B \rightarrow D}$ and $I_{C \rightarrow D}$. Since there is no outgoing current defined yet, $I_{D \rightarrow battery}$ must be outgoing to balance. To satisfy the Node Rule, we set

$$ I_{D \rightarrow battery} = I_{out} = I_{in} = I_{B \rightarrow D} + I_{B \rightarrow D} = 8\text{~A} $$

Notice that $I_{D \rightarrow battery} = I_{1}$. This will always be the case for currents going in and out of the battery (approximating a few things that are usually safe to approximate, such as a steady current). In fact, we could have treated the battery as another node in this example. Notice also that if you incorrectly reason about the direction of a current (incoming or outgoing), the calculation will give a negative number for the current. The Node Rule is self-correcting. A final representation with directions is shown below.

Circuit with Nodes