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184_notes:examples:week7_resistance_wire [2017/10/04 15:56] – [Resistance of a Wire] tallpaul | 184_notes:examples:week7_resistance_wire [2018/06/19 14:54] (current) – curdemma | ||
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+ | [[184_notes: | ||
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=====Resistance of a Wire===== | =====Resistance of a Wire===== | ||
- | Suppose you have a wire whose resistance | + | Suppose you have a wire whose resistance |
===Facts=== | ===Facts=== | ||
- | * The original wire has $L = 2 \text{ cm}$, $A = 1 \text{ mm}^2$, and $R = 50 \text{ m}Omega$. | + | * The original wire has $L = 2 \text{ cm}$, $A = 1 \text{ mm}^2$, and $R = 60 \text{ m}\Omega$. |
- | * The length could be increased to $L_{new} = 4 \text{ cm}$. | + | * The length could be increased to $L_{new} = 6 \text{ cm}$. |
* The cross-sectional area could be increased to $A_{new} = 3 \text{ mm}^2$. | * The cross-sectional area could be increased to $A_{new} = 3 \text{ mm}^2$. | ||
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===Representations=== | ===Representations=== | ||
- | * We represent the resistance of a simple wire such as this with: $$R = \frac{L}{\sigma A}$$ | + | * We represent the resistance of a simple wire with: $$R = \frac{L}{\sigma A}$$ |
====Solution==== | ====Solution==== | ||
- | 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 | + | All we need here is our representation for the resistance of the wire. In the first change |
- | $$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ | + | |
- | + | ||
- | We do a similar analysis for node $B$. Incoming current | + | |
- | $$I_{B\rightarrow D} = I_{out}-I_3 = I_{in}-I_3 = I_{A\rightarrow B}-I_3 = 1 \text{ A}$$ | + | |
- | + | ||
- | For node $C$, incoming current is $I_2$ and $I_3$. There is no outgoing current defined yet! $I_{C\rightarrow D}$ must be outgoing to balance. To satisfy the Node Rule, we set | + | |
- | $$I_{C\rightarrow D} = I_{out} | + | |
- | + | ||
- | 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 | + | If instead, we made the other change, we would have tripled the cross-sectional area ($1 \text{ mm}^2 \rightarrow |
- | {{ 184_notes: | + | These answers should make sense physically. The longer the wire, the more material there is for the electrons to push through, so the resistance is higher. The bigger the area, the more electrons are able to flow through the wire in at a given time so the resistance should be smaller. |