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--- 184_notes:examples:week6_node_rule [2018/02/03 22:29] – tallpaul
+++ 184_notes:examples:week6_node_rule [2018/02/03 22:33] – [Solution] tallpaul
@@ Line 12: / Line 12: @@
 ===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}}
 ====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
 $$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$$