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--- 184_notes:examples:week7_wire_dimensions [2017/10/04 13:41] – [Solution] tallpaul
+++ 184_notes:examples:week7_wire_dimensions [2018/06/19 14:18] – [Example: Changing the Dimensions of a Wire] curdemma
@@ Line 12: / Line 12: @@
 ===Approximations & Assumptions===
-  * The circuit is in a steady state.
+  * The circuit is in a steady state - which means the current should be the same in all three sections.
   * Approximating the battery as a mechanical battery.
   * The wire has a circular cross-section.
@@ Line 21: / Line 21: @@
   * 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 segment 1. The power dissipated through the segment is just $$P_1=I_1 \Delta V_1 = 5 \text{ W}$$
+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 $$E_1 = \frac{\Delta V_1}{L_1} = 0.5 \text{ V/m}$$
-The electric field is constant since the wire is uniform with respect to the rest of the segment, so we get $$E_1 = \frac{\Delta V_1}{L_1} = 0.5 \text{ V/m}$$
+The power dissipated through the segment is just $$P_1=I_1 \Delta V_1 = 5 \text{ W}$$
+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): $$E_2=\frac{A_1}{A_2}E_1 = \frac{\pi (d_1/2)^2}{\pi (d_2/2)^2}E_1=9E_1=4.5\text{ V/m}$$
+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 $$P_2=I_2 \Delta V_2 = 45 \text{ W}$$
+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. $$P_3=I_3 \Delta V_3 = 10 \text{ W}$$
+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.