Differences

This shows you the differences between two versions of the page.

--- 184_notes:examples:week6_node_rule [2017/09/27 13:44] – [Example: Application of Node Rule] tallpaul
+++ 184_notes:examples:week6_node_rule [2021/06/08 00:51] (current) – schram45
@@ Line 1: / Line 1: @@
+[[184_notes:current|Return to current in wires]]
 =====Example: Application of Node Rule=====
-Suppose you have the circuit below. Nodes are labeled for simplicity of discussion. you are given a few values: $I_1=8 \text{ A}$, $I_2=3 \text{ A}$, and $I_3=4 \text{ A}$. Determine all other currents in the circuit, using the [[184_notes:current#Current_in_Different_Parts_of_the_Wire|Current Node Rule]]. Draw the direction of the current as well.
+Suppose you have the circuit below. You are given a few values: $I_1=8 \text{ A}$, $I_2=3 \text{ A}$, and $I_3=4 \text{ A}$. Determine all other currents in the circuit, using the [[184_notes:current#Current_in_Different_Parts_of_the_Wire|Current Node Rule]]. Draw the direction of the current as well.
+[{{ 184_notes:6_nodeless.png?300 |Circuit}}]
-{{ 184_notes:6_nodes.png?300 |Circuit with Nodes}}
 ===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 $16 \text{ m}$.
+  * The Node Rule is $I_{in}=I_{out}$, for any point along the current.
-===Lacking===
+===Goal===
-  * $\Phi_e$ through the cylinder
+  * Find all the currents in the circuit and their directions.
-===Approximations & Assumptions===
-  * The axis of the cylinder is aligned with the dipole.
-  * The dipole and cylinder are centered with respect to each other.
-  * 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.
+For simplicity of discussion, we label the nodes in an updated representation:
+[{{ 184_notes:6_nodes.png?300 |Circuit with Nodes}}]
+<WRAP TIP>
+===Assumption===
+We will assume we have a perfect battery to supply a steady current to the circuit and will not die over time.
+</WRAP>
 ====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:
+Okay, there is a lot going on with all these nodes. Let's make a plan to organize our approach.
-{{ 184_notes:5_dipole_field_lines.png?300 |Dipole Electric Field Lines}}
+<WRAP TIP>
+=== Plan ===
+Take the nodes one at a time. Here's the plan in steps:
+  * Look at all the known currents attached to a node.
+  * Assign variables to the unknown currents attached to a node.
+  * Set up an equation using the Node Rule. If not sure about whether a current is going in or coming out of the node, guess.
+  * Solve for the unknown currents.
+  * If any of them are negative, then we guessed wrong two steps ago. We can just flip the sign now.
+  * Repeat the above steps for all the nodes.
+</WRAP>
+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
+$$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$$
+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
+$$I_{C\rightarrow D} = I_{out} = I_{in} = I_2+I_3 = 7 \text{ A}$$
-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.
+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
+$$I_{D\rightarrow battery} = I_{out} = I_{in} = I_{B\rightarrow D}+I_{B\rightarrow D} = 8 \text{ A}$$
-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,
+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 representation with directions is shown below.
-$$\Phi_{left}=-\Phi_{right}$$
-Putting it together, we tentatively write:
+[{{ 184_notes:6_nodes_with_arrows.png?300 |Circuit with Nodes}}]
-$$\Phi_{\text{cylinder}}=\Phi_{left}+\Phi_{right}=0$$
-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]].
-$$\Phi_{\text{total}}=\int \vec{E} \cdot \text{d}\vec{A}=\frac{Q_{\text{enclosed}}}{\epsilon_0}$$
-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.