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184_notes:examples:week7_ohms_law [2017/10/04 16:06] – [Example: Application of Node Rule] tallpaul | 184_notes:examples:week7_ohms_law [2018/06/19 14:54] (current) – curdemma | ||
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=====Example: | =====Example: | ||
- | Suppose you have the circuit | + | Suppose you have a simple |
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- | {{ 184_notes: | + | |
===Facts=== | ===Facts=== | ||
- | * $I_1=8 \text{ A}$, $I_2=3 \text{ | + | * $\Delta V = 9\text{ |
- | * $I_1$, $I_2$, and $I_3$ are directed as pictured. | + | * $R = 120 \Omega$ |
===Lacking=== | ===Lacking=== | ||
- | * All other currents (including their directions). | + | * Current |
===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
- | * The current is not changing (circuit is in steady state). | + | * The wire has very very small resistance when compared to the 120 $\Omega$ resistor. |
- | * All current in the circuit | + | * The circuit |
- | * No resistance in the battery (approximating | + | * Approximating |
+ | * There is no outside influence on the circuit. | ||
===Representations=== | ===Representations=== | ||
- | * We represent | + | * We represent |
- | * We represent the Node Rule as $I_{in}=I_{out}$. | + | * We represent the situation with following circuit diagram. |
+ | |||
+ | [{{ 184_notes: | ||
====Solution==== | ====Solution==== | ||
- | Let's start with node $A$. Incoming current is $I_1$, | + | We have assumed that the battery |
- | $$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$$ | + | |
- | + | ||
- | We do a similar analysis for node $B$. Incoming current | + | |
- | $$I_{B\rightarrow D} = I_{out}-I_3 = I_{in}-I_3 = I_{A\rightarrow B}-I_3 = 1 \text{ A}$$ | + | |
- | + | ||
- | For node $C$, incoming current | + | |
- | $$I_{C\rightarrow D} = I_{out} = I_{in} = I_2+I_3 | + | |
- | + | ||
- | Lastly, we look at node $D$. Incoming current is $I_{B\rightarrow D}$ and $I_{C\rightarrow D}$. Since there is no outgoing current defined yet, $I_{D\rightarrow battery}$ must be outgoing | + | |
- | $$I_{D\rightarrow battery} = I_{out} = I_{in} = I_{B\rightarrow D}+I_{B\rightarrow D} = 8 \text{ | + | |
- | + | ||
- | 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, | + | |
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- | {{ 184_notes: | + |