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184_notes:examples:week6_node_rule [2017/09/27 13:58] – [Solution] tallpaul | 184_notes:examples:week6_node_rule [2021/06/08 00:51] (current) – schram45 |
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| [[184_notes:current|Return to current in wires]] |
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=====Example: Application of Node Rule===== | =====Example: Application of Node Rule===== |
Suppose you have the circuit below. Nodes are labeled for simplicity of discussion. 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 [[184_notes:current#Current_in_Different_Parts_of_the_Wire|Current Node Rule]]. Draw the direction of the current as well. | 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 [[184_notes:current#Current_in_Different_Parts_of_the_Wire|Current Node Rule]]. Draw the direction of the current as well. |
| [{{ 184_notes:6_nodeless.png?300 |Circuit}}] |
{{ 184_notes:6_nodes.png?300 |Circuit with Nodes}} | |
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===Facts=== | ===Facts=== |
* I1=8 A, I2=3 A, and I3=4 A. | * I1=8 A, I2=3 A, and I3=4 A. |
* I1, I2, and I3 are directed as pictured. | * I1, I2, and I3 are directed as pictured. |
| * The Node Rule is Iin=Iout, for any point along the current. |
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===Lacking=== | ===Goal=== |
* All other currents (including their directions). | * Find all the currents in the circuit and their directions. |
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===Approximations & Assumptions=== | |
* The current is not changing. | |
* All current in the circuit arises from other currents in the circuit. | |
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===Representations=== | ===Representations=== |
* We represent the situation with diagram given. | For simplicity of discussion, we label the nodes in an updated representation: |
* We represent the Node Rule as $I_{in}=I_{out}$. | [{{ 184_notes:6_nodes.png?300 |Circuit with Nodes}}] |
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| <WRAP TIP> |
| ===Assumption=== |
| We will assume we have a perfect battery to supply a steady current to the circuit and will not die over time. |
| </WRAP> |
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====Solution==== | ====Solution==== |
| Okay, there is a lot going on with all these nodes. Let's make a plan to organize our approach. |
| <WRAP TIP> |
| === 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. |
| </WRAP> |
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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 | 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
| IA→B=Iout−I2=Iin−I2=I1−I2=5 A
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| 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
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| 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
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| 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
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| 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 representation with directions is shown below. |
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| [{{ 184_notes:6_nodes_with_arrows.png?300 |Circuit with Nodes}}] |