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--- 184_notes:examples:week7_wire_dimensions [2017/10/04 00:49] – [Example: Application of Node Rule] tallpaul
+++ 184_notes:examples:week7_wire_dimensions [2017/10/04 13:51] – [Solution] tallpaul
@@ Line 1: / Line 1: @@
 =====Example: Changing the Dimensions of a Wire=====
-Suppose you have a simple circuit whose wire changes in thickness. The wire is 8 meters long. The first 2 meters of the wire are 3 mm thick. The next 2 meters are 1 mm thick. The last 4 meters are 3 mm thick. The wire is connected to a 12-Volt battery and current is allowed to flow. You use an ammeter and a voltmeter to find that the current through the first 2 meters of wire is $I_1 = 5 \text{ A}$, and the voltage across the first two meters is $\Delta V_1 = 1 \text{ V}$.
+Suppose you have a simple circuit whose wire changes in thickness. The wire is 8 meters long. The first 2 meters of the wire are 3 mm thick. The next 2 meters are 1 mm thick. The last 4 meters are 3 mm thick. The wire is connected to a 12-Volt battery and current is allowed to flow. You use an ammeter and a voltmeter to find that the current through the first 2 meters of wire is $I_1 = 5 \text{ A}$, and the voltage across the first two meters is $\Delta V_1 = 1 \text{ V}$. In all three segments of the wire, determine the magnitude of the electric field inside and the power transmitted.
-{{ 184_notes:6_nodes.png?300 |Circuit with Nodes}}
 ===Facts===
-  * $I_1=8 \text{ A}$, $I_2=3 \text{ A}$, and $I_3=4 \text{ A}$.
+  * Segment lengths: $L_1=2 \text{ m}$, $L_2=2 \text{ m}$, and $L_3=4 \text{ m}$.
-  * $I_1$, $I_2$, and $I_3$ are directed as pictured.
+  * Segment diameters: $d_1=3 \text{ mm}$, $d_2=1 \text{ mm}$, and $d_3=3 \text{ mm}$.
+  * Current: $I_1 = 5 \text{ A}$.
+  * Voltage: $\Delta V_1 = 1 \text{ V}$, $\Delta V_{battery} = 12 \text{ V}$.
 ===Lacking===
-  * All other currents (including their directions).
+  * Power and electric field in all segments
 ===Approximations & Assumptions===
-  * The current is not changing (circuit is in steady state).
+  * The circuit is in a steady state.
-  * All current in the circuit arises from other currents in the circuit.
+  * Approximating the battery as a mechanical battery.
-  * No resistance in the battery (approximating the battery as a mechanical battery)
+  * The wire has a circular cross-section.
+  * No outside influence on the circuit.
+  * The wire is made of the same material throughout.
 ===Representations===
-  * We represent the situation with diagram given.
+  * We represent the situation with diagram below. We number the segments for simplicity of representing the quantities we are interested in (see above in "Facts").
-  * We represent the Node Rule as $I_{in}=I_{out}$.
+{{ 184_notes:7_wire_dim.png?400 |Circuit Diagram}}
 ====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
+Let's start with segment 1. The electric field is constant since the wire is uniform with respect to the rest of the segment, so we get $$E_1 = \frac{\Delta V_1}{L_1} = 0.5 \text{ V/m}$$
-$$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$$
+The power dissipated through the segment is just $$P_1=I_1 \Delta V_1 = 5 \text{ W}$$
-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, 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}}
+Now, for segment 2. We can use [[184_notes:r_energy#Conservation_of_Charge_in_Circuits|what we know]] about charge in steady state circuits to determine the electric field: $$E_2=\frac{A_1}{A_2}E_1 = \frac{\pi d_1^2}{\pi d_2^2}E_1=9E_1=4.5\text{ V/m}$$