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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 22:34] – [Solution] tallpaul | ||
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===Representations=== | ===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: | {{ 184_notes: | ||
====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. | ||
+ | </ | ||
+ | |||
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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$$I_{D\rightarrow battery} = I_{out} = I_{in} = I_{B\rightarrow D}+I_{B\rightarrow D} = 8 \text{ A}$$ | $$I_{D\rightarrow battery} = I_{out} = I_{in} = I_{B\rightarrow D}+I_{B\rightarrow D} = 8 \text{ A}$$ | ||
- | 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, | + | 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: | {{ 184_notes: |