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184_notes:examples:week6_node_rule [2017/09/27 13:58] – [Solution] tallpaul | 184_notes:examples:week6_node_rule [2018/02/03 22:29] – [Example: Application of Node Rule] tallpaul | ||
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=====Example: | =====Example: | ||
- | Suppose you have the circuit below. | + | Suppose you have the circuit below. |
- | + | {{ 184_notes:6_nodeless.png?300 |Circuit}} | |
- | {{ 184_notes:6_nodes.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 $I_{in}=I_{out}$, | ||
- | ===Lacking=== | + | ===Goal=== |
- | * All other currents (including their directions). | + | * Find all the currents in the circuit |
- | + | ||
- | ===Approximations & Assumptions=== | + | |
- | * The current is not changing. | + | |
- | * All current in the circuit arises from other currents in the circuit. | + | |
===Representations=== | ===Representations=== | ||
- | * We represent | + | For simplicity of discussion, we label the nodes in an updated representation: |
- | * We represent the Node Rule as $I_{in}=I_{out}$. | + | {{ 184_notes: |
====Solution==== | ====Solution==== | ||
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$$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$$ | $$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$$ | ||
+ | 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, | ||
+ | {{ 184_notes: |