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--- 184_notes:examples:week7_wire_dimensions [2017/10/04 13:11] – [Example: Changing the Dimensions of a Wire] tallpaul
+++ 184_notes:examples:week7_wire_dimensions [2021/06/14 23:40] (current) – schram45
@@ Line 1: / Line 1: @@
+[[184_notes:r_energy|Return to Energy in Circuits]]
 =====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 $\Delta V_1 = 1 \text{ V}$. In all three segments of the wire, determine the magnitude of the electric field inside and the power transmitted.
@@ Line 6: / Line 8: @@
   * 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: $\Delta V_1 = 1 \text{ V}$.
+  * Voltage: $\Delta V_1 = 1 \text{ V}$, $\Delta V_{battery} = 12 \text{ V}$.
 ===Lacking===
@@ Line 12: / Line 14: @@
 ===Approximations & Assumptions===
-  * The circuit is in a steady state.
+  * The circuit is in a steady state: This allows the current should be the same in all three sections.
-  * Approximating the battery as a mechanical battery.
+  * Approximating the battery as a mechanical battery: Batteries normally dont keep their energy forever and slowly die over time. Using a mechanical battery means our battery produces a steady source of energy in this problem.
-  * The wire has a circular cross-section.
+  * No outside influence on the circuit: This simplifies the model and isolates our circuit from any outside sources of charge or energy that could effect our calculations.
-  * No outside influence on the circuit.
-  * The wire is made of the same material throughout.
 ===Representations===
-  * We represent the situation with diagram below. We number the segments for simplicity of representing the quantities we are interested in.
+  * 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").
-{{ 184_notes:7_wire_dim.png?400 |Circuit Diagram}}
+[{{ 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
+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 (notice we divide diameter by 2 in order to get the radius of the circular cross-section): $$E_2=\frac{A_1}{A_2}E_1 = \frac{\pi (d_1/2)^2}{\pi (d_2/2)^2}E_1=9E_1=4.5\text{ V/m}$$
-$$I_{B\rightarrow D} = I_{out}-I_3 = I_{in}-I_3 = I_{A\rightarrow B}-I_3 = 1 \text{ A}$$
-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
+<WRAP TIP>
-$$I_{C\rightarrow D} = I_{out} = I_{in} = I_2+I_3 = 7 \text{ A}$$
+===Assumptions====
+In order to do this calculation there are two important assumptions that must be made
+  * The wires have a circular cross section: This allows us to use the formula for the area of a circle to come up with the correct proportion.
+  * The wires are made of the same material throughout: There are two terms in the electron current equation that are material propeties and these will cancel out for each segment of wire if they are made of the same material. This allows the electric field to only vary with cross sectional area.
+</WRAP>
-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
+A simple application of the [[184_notes:current#Current_in_Different_Parts_of_the_Wire|Current Node Rule]] tells us that $I_2=I_1$. The voltage is easily found from the constant electric field: $\Delta V_2=E_2 L_2 = 9 \text{ V}$. The power dissipated through the segment is then $$P_2=I_2 \Delta V_2 = 45 \text{ W}$$
-$$I_{D\rightarrow battery} = I_{out} = I_{in} = I_{B\rightarrow D}+I_{B\rightarrow D} = 8 \text{ A}$$
-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.
+For segment 3, we can reason based on the thicknesses of the segments that $E_3=E_1$. This yields $$E_3 = 0.5 \text{ V/m}$$
+We can use the same reasoning as before to say that $I_3=I_2=I_1$. We can also use the same equation to find voltage: $\Delta V_3=E_3 L_3 = 2 \text{ V}$. The power is calculated as before. $$P_3=I_3 \Delta V_3 = 10 \text{ W}$$
-{{ 184_notes:6_nodes_with_arrows.png?300 |Circuit with Nodes}}
+Notice that we could have also used [[184_notes:r_energy#Energy_around_the_Circuit|Kirchoff's Loop Rule]] to find the voltage of different segments. For now, it will serve as a nice check on our math. If we travel along the direction of conventional current (counterclockwise in our representation), voltage decreases, so $\Delta V_1$, $\Delta V_2$, $\Delta V_3<0$, whereas we have $\Delta V_{battery}>0$. These four potential differences form a loop, so by Kirchoff's Loop Rule they should add to 0:
+$$\Delta V_{battery}+\Delta V_1+\Delta V_2+\Delta V_3 = 12 \text{ V} - 1 \text{ V} - 9 \text{ V} - 2 \text{ V} = 0$$
+Sometimes we will not have as much information as we did here, and using the Loop Rule will be required. For now, it serves as a nice check.