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184_notes:examples:week6_node_rule [2017/09/27 13:48] – [Example: Application of Node Rule] tallpaul | 184_notes:examples:week6_node_rule [2017/09/27 14:10] – [Solution] tallpaul | ||
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* We represent the situation with diagram given. | * We represent the situation with diagram given. | ||
* We represent the Node Rule as $I_{in}=I_{out}$. | * We represent the Node Rule as $I_{in}=I_{out}$. | ||
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====Solution==== | ====Solution==== | ||
- | First, notice that we probably | + | 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 |
- | {{ 184_notes: | + | $$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$$ |
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+ | 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 | ||
+ | $$I_{B\rightarrow D} = I_{out}-I_3 = I_{in}-I_3 = I_{A\rightarrow B}-I_3 = 1 \text{ A}$$ | ||
+ | |||
+ | 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 |
+ | $$I_{D\rightarrow battery} = I_{out} = I_{in} = I_{B\rightarrow D}+I_{B\rightarrow D} = 8 \text{ A}$$ | ||
- | We could write this as a comparison between the left and right side of the cylinder. The $\text{d}\vec{A}$-vectors are mirrored | + | 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 |
- | $$\Phi_{left}=-\Phi_{right}$$ | + | |
- | Putting it together, we tentatively write: | + | {{ 184_notes:6_nodes_with_arrows.png? |
- | $$\Phi_{\text{cylinder}}=\Phi_{left}+\Phi_{right}=0$$ | + | |
- | We gain more confidence when we read the [[184_notes:gauss_ex|next section of notes]], where we define " | + | |
- | $$\Phi_{\text{total}}=\int \vec{E} \cdot \text{d}\vec{A}=\frac{Q_{\text{enclosed}}}{\epsilon_0}$$ | + | |
- | 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. | + |