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--- 184_notes:examples:week6_node_rule [2018/02/03 22:29] – tallpaul
+++ 184_notes:examples:week6_node_rule [2021/06/08 00:51] (current) – schram45
@@ Line 1: / Line 1: @@
+[[184_notes:current|Return to current in wires]]
 =====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}}
+[{{ 184_notes:6_nodeless.png?300 |Circuit}}]
 ===Facts===
@@ Line 12: / Line 14: @@
 ===Representations===
-  * For simplicity of discussion, we label the nodes in an updated representation:
+For simplicity of discussion, we label the nodes in an updated representation:
-{{ 184_notes:6_nodes.png?300 |Circuit with Nodes}}
+[{{ 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====
+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
  $I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$
@@ Line 28: / Line 47: @@
  $I_{D\rightarrow battery} = I_{out} = I_{in} = I_{B\rightarrow D}+I_{B\rightarrow D} = 8 \text{ A}$
-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 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}}
+[{{ 184_notes:6_nodes_with_arrows.png?300 |Circuit with Nodes}}]