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--- 184_notes:examples:week7_wire_dimensions [2017/10/04 13:14] – [Example: Changing the Dimensions of a Wire] tallpaul
+++ 184_notes:examples:week7_wire_dimensions [2017/10/04 15:37] – [Solution] tallpaul
@@ Line 24: / Line 24: @@
 ====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 use the equation for the area of the circular cross-section): $$E_2=\frac{A_1}{A_2}E_1 = \frac{\pi (d_1/2)^2}{\pi (d_2/2)^2}E_1=9E_1=4.5\text{ V/m}$$
-$$I_{B\rightarrow D} = I_{out}-I_3 = I_{in}-I_3 = I_{A\rightarrow B}-I_3 = 1 \text{ A}$$
+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 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
+For segment 3, we can reason based on the thicknesses of the segments that $E_3=E_1$. This yields $$E_3 = 0.5 \text{ V/m}$$
-$$I_{C\rightarrow D} = I_{out} = I_{in} = I_2+I_3 = 7 \text{ A}$$
+We can use the same reasoning as before to say that $I_3=I_2=I_1$. We can also find voltage pretty easily: $\Delta V_3=E_3 L_3 = 2 \text{ V}$. The power is calculated as before. $$P_3=I_3 \Delta V_3 = 10 \text{ W}$$
-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
+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, 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:
-$$I_{D\rightarrow battery} = I_{out} = I_{in} = I_{B\rightarrow D}+I_{B\rightarrow D} = 8 \text{ A}$$
+$$\Delta V_{battery}+\Delta V_1+\Delta V_2+\Delta V_3 = 12 \text{ V} - 1 \text{ V} - 9 \text{ V} - 2 \text{ V} = 0$$
+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.
-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}}