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--- 184_notes:examples:week7_resistance_wire [2017/10/03 23:46] – created tallpaul
+++ 184_notes:examples:week7_resistance_wire [2018/06/19 14:54] (current) – curdemma
@@ Line 1: / Line 1: @@
-=====Example: Application of Node Rule=====
+[[184_notes:resistivity|Return to resistors and conductivity]]
-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.
-{{ 184_notes:6_nodes.png?300 |Circuit with Nodes}}
+=====Resistance of a Wire=====
+Suppose you have a wire whose resistance is 60 m$\Omega$. The wire has a length of 2 cm, and has a cross-sectional area of 1 mm$^2$. What would be the resistance if you increase the length of the wire to 6 cm (keeping the area the same)? What would be the resistance if you increase the cross-sectional area to 3 mm$^2$ (keeping the original length of wire)?
 ===Facts===
-  * $I_1=8 \text{ A}$, $I_2=3 \text{ A}$, and $I_3=4 \text{ A}$.
+  * The original wire has $L = 2 \text{ cm}$, $A = 1 \text{ mm}^2$, and $R = 60 \text{ m}\Omega$.
-  * $I_1$, $I_2$, and $I_3$ are directed as pictured.
+  * The length could be increased to $L_{new} = 6 \text{ cm}$.
+  * The cross-sectional area could be increased to $A_{new} = 3 \text{ mm}^2$.
 ===Lacking===
-  * All other currents (including their directions).
+  * Resistances of new wires.
 ===Approximations & Assumptions===
-  * The current is not changing (circuit is in steady state).
+  * The conductivity of the wire does not change.
-  * All current in the circuit arises from other currents in the circuit.
+  * The wire's material is uniform.
-  * No resistance in the battery (approximating the battery as a mechanical battery)
 ===Representations===
-  * We represent the situation with diagram given.
+  * We represent the resistance of a simple wire with: $$R = \frac{L}{\sigma A}$$
-  * We represent the Node Rule as $I_{in}=I_{out}$.
 ====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 the Node Rule, we set
+All we need here is our representation for the resistance of the wire. In the first change to the wire, we triple it's length ($2 \text{ cm} \rightarrow 6 \text{ cm}$). Our new resistance then is found by $$R_{new} = \frac{L_{new}}{\sigma A} = \frac{3L}{\sigma A} = 3R = 180 \text{ m}\Omega$$
-$$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$$
-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
+If instead, we made the other change, we would have tripled the cross-sectional area ($1 \text{ mm}^2 \rightarrow 3 \text{ mm}^2$). Our new resistance would then be $$R_{new} = \frac{L}{\sigma A_{new}} = \frac{L}{\sigma 3 A} = \frac{1}{3}R = 20 \text{ m}\Omega$$
-$$I_{B\rightarrow D} = I_{out}-I_3 = I_{in}-I_3 = I_{A\rightarrow B}-I_3 = 1 \text{ A}$$
-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
+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.
-$$I_{C\rightarrow D} = I_{out} = I_{in} = I_2+I_3 = 7 \text{ A}$$
-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 diagram with directions is shown below.
-{{ 184_notes:6_nodes_with_arrows.png?300 |Circuit with Nodes}}