184_notes:examples:week6_node_rule

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184_notes:examples:week6_node_rule [2017/09/27 14:08] – [Solution] tallpaul184_notes:examples:week6_node_rule [2021/06/08 00:51] (current) schram45
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-=====Example: Application of Node Rule===== +[[184_notes:current|Return to current in wires]]
-Suppose you have the circuit below. Nodes are labeled for simplicity of discussion. 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 [[184_notes:current#Current_in_Different_Parts_of_the_Wire|Current Node Rule]]. Draw the direction of the current as well.+
  
-{{ 184_notes:6_nodes.png?300 |Circuit with Nodes}}+=====Example: Application of Node Rule===== 
 +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 [[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}}]
  
 ===Facts=== ===Facts===
   * $I_1=8 \text{ A}$, $I_2=3 \text{ A}$, and $I_3=4 \text{ A}$.   * $I_1=8 \text{ A}$, $I_2=3 \text{ A}$, and $I_3=4 \text{ A}$.
   * $I_1$, $I_2$, and $I_3$ are directed as pictured.   * $I_1$, $I_2$, and $I_3$ are directed as pictured.
 +  * The Node Rule is $I_{in}=I_{out}$, for any point along the current.
  
-===Lacking=== +===Goal=== 
-  * All other currents (including their directions). +  * Find all the currents in the circuit and their directions.
- +
-===Approximations & Assumptions=== +
-  * The current is not changing. +
-  * All current in the circuit arises from other currents in the circuit.+
  
 ===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}}] 
 + 
 +<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>
  
 ====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>
 +
 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 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
 $$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$$ $$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$$
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 $$I_{C\rightarrow D} = I_{out} = I_{in} = I_2+I_3 = 7 \text{ A}$$ $$I_{C\rightarrow D} = I_{out} = I_{in} = I_2+I_3 = 7 \text{ A}$$
  
-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+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
 $$I_{D\rightarrow battery} = I_{out} = I_{in} = I_{B\rightarrow D}+I_{B\rightarrow D} = 8 \text{ A}$$ $$I_{D\rightarrow battery} = I_{out} = I_{in} = I_{B\rightarrow D}+I_{B\rightarrow D} = 8 \text{ A}$$
  
-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, no resistance in the battery). 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. +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.
  
 +[{{ 184_notes:6_nodes_with_arrows.png?300 |Circuit with Nodes}}]
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  • Last modified: 2017/09/27 14:08
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