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--- 184_notes:examples:week6_node_rule [2017/09/27 13:44] – [Example: Application of Node Rule] tallpaul
+++ 184_notes:examples:week6_node_rule [2017/09/27 14:09] – [Solution] tallpaul
@@ Line 5: / Line 5: @@
 ===Facts===
-  * The dipole charges are $q=5 \mu\text{C}$, $-q=-5 \mu\text{C}$.
+  * $I_1=8 \text{ A}$, $I_2=3 \text{ A}$, and $I_3=4 \text{ A}$.
-  * The dipole distance is $1 \text{ m}$.
+  * $I_1$, $I_2$, and $I_3$ are directed as pictured.
-  * The cylinder has radius $4 \text{ m}$ and length $16 \text{ m}$.
 ===Lacking===
-  * $\Phi_e$ through the cylinder
+  * All other currents (including their directions).
 ===Approximations & Assumptions===
-  * The axis of the cylinder is aligned with the dipole.
+  * The current is not changing.
-  * The dipole and cylinder are centered with respect to each other.
+  * All current in the circuit arises from other currents in the circuit.
-  * The electric flux through the cylinder is due only to the dipole (i.e., no other charges exist).
-  * The charges are point charges, which indeed means we can model them as a dipole.
 ===Representations===
-  * We represent the situation with the following diagram.
+  * We represent the situation with diagram given.
+  * We represent the Node Rule as $I_{in}=I_{out}$.
 ====Solution====
-First, notice that we probably do not want to do any calculations here, since the it will not be fun to take a dot-product of the dipole's electric field and the area-vector, and it will get very messy very quickly when we start integrating over the surface of the cylinder. Instead, we evaluate the situation more qualitatively. Consider the electric field vectors of the dipole near the surface of the cylinder:
+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
-{{ 184_notes:5_dipole_field_lines.png?300 |Dipole Electric Field Lines}}
+$$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$$
-Notice that the vectors near the positive charge are leaving the cylinder, and the vectors near the negative charge are entering. Not only this, but they are mirror images of each other. Wherever an electric field vector points out of the cylinder on the right side, there is another electric field vector on the left that is pointing into the cylinder at the same angle. These mirror image vectors also have the same magnitude, though it is a little tougher to visualize.
+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
+$$I_{B\rightarrow D} = I_{out}-I_3 = I_{in}-I_3 = I_{A\rightarrow B}-I_3 = 1 \text{ A}$$
-We could write this as a comparison between the left and right side of the cylinder. The $\text{d}\vec{A}$-vectors are mirrored for left vs. right, and the $\vec{E}$-vectors are mirrored, but opposite $\left(\vec{E}_{left}=-\vec{E}_{right}\right)$. We tentatively write the equality,
+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
-$$\Phi_{left}=-\Phi_{right}$$
+$$I_{C\rightarrow D} = I_{out} = I_{in} = I_2+I_3 = 7 \text{ A}$$
-Putting it together, we tentatively write:
+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
-$$\Phi_{\text{cylinder}}=\Phi_{left}+\Phi_{right}=0$$
+$$I_{D\rightarrow battery} = I_{out} = I_{in} = I_{B\rightarrow D}+I_{B\rightarrow D} = 8 \text{ A}$$
-We gain more confidence when we read the [[184_notes:gauss_ex|next section of notes]], where we define "Gauss' Law". This law states that the total flux through a close surface is the amount of charge enclosed divided by $\epsilon_0$, the [[https://en.wikipedia.org/wiki/Vacuum_permittivity|permittivity of free space]].
-$$\Phi_{\text{total}}=\int \vec{E} \cdot \text{d}\vec{A}=\frac{Q_{\text{enclosed}}}{\epsilon_0}$$
+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, no resistance in the battery). 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.
-Since the total charge of the dipole is $0$, then indeed the charge enclosed is $0$, and we were correct with our reasoning about the electric field and flux above.
+{{ 184_notes:6_nodes_with_arrows.png?300 |Circuit with Nodes}}