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Example: Application of Node Rule
Suppose you have the circuit below. You are given a few values: I1=8 A, I2=3 A, and I3=4 A. Determine all other currents in the circuit, using the Current Node Rule. Draw the direction of the current as well.
Facts
- I1=8 A, I2=3 A, and I3=4 A.
- I1, I2, and I3 are directed as pictured.
- The Node Rule is Iin=Iout, for any point along the current.
Goal
- Find all the currents in the circuit and their directions.
Representations
Solution
Let's start with node A. Incoming current is I1, and outgoing current is I2. How do we decide if IA→B is incoming or outgoing? We need to bring it back to the Node Rule: Iin=Iout. Since I1=8 A and I2=3 A, we need IA→B to be outgoing to balance. To satisfy the Node Rule, we set IA→B=Iout−I2=Iin−I2=I1−I2=5 A
We do a similar analysis for node B. Incoming current is IA→B, and outgoing current is I3. Since IA→B=5 A and I3=4 A, we need IB→D to be outgoing to balance. To satisfy the Node Rule, we set IB→D=Iout−I3=Iin−I3=IA→B−I3=1 A
For node C, incoming current is I2 and I3. There is no outgoing current defined yet! IC→D must be outgoing to balance. To satisfy the Node Rule, we set IC→D=Iout=Iin=I2+I3=7 A
Lastly, we look at node D. Incoming current is IB→D and IC→D. Since there is no outgoing current defined yet, ID→battery must be outgoing to balance. To satisfy the Node Rule, we set ID→battery=Iout=Iin=IB→D+IB→D=8 A
Notice that ID→battery=I1. 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 diagram with directions is shown below.