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--- 184_notes:examples:week6_node_rule [2017/09/27 13:53] – [Solution] 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. 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.
+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_nodes.png?300 |Circuit with Nodes}}
 ===Facts===
   *  $I_1=8 \text{ A}$ ,  $I_2=3 \text{ A}$ , and  $I_3=4 \text{ A}$ .
   *  $I_1$ ,  $I_2$ , and  $I_3$  are directed as pictured.
+  * 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===
-  * 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====
-Let's start with node  $A$ . Incoming current is  $I_1$ , and outgoing current is $
+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
+ $I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$
+We do a similar analysis for node  $B$ . Incoming current is  $I_{A\rightarrow B}$ , and outgoing current is  $I_3$ . Since  $I_{A\rightarrow B}=5 \text{ A}$  and  $I_3=4 \text{ A}$ , we need  $I_{B\rightarrow D}$  to be outgoing to balance. To satisfy the Node Rule, we set
+ $I_{B\rightarrow D} = I_{out}-I_3 = I_{in}-I_3 = I_{A\rightarrow B}-I_3 = 1 \text{ A}$
+For node  $C$ , incoming current is  $I_2$  and  $I_3$ . There is no outgoing current defined yet!  $I_{C\rightarrow D}$  must be outgoing to balance. To satisfy the Node Rule, we set
+ $I_{C\rightarrow D} = I_{out} = I_{in} = I_2+I_3 = 7 \text{ A}$
+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
+ $I_{D\rightarrow battery} = I_{out} = I_{in} = I_{B\rightarrow D}+I_{B\rightarrow D} = 8 \text{ A}$
+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 representation with directions is shown below.
+[{{ 184_notes:6_nodes_with_arrows.png?300 |Circuit with Nodes}}]