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184_notes:examples:week7_wire_dimensions [2017/10/04 13:11] – [Example: Changing the Dimensions of a Wire] tallpaul | 184_notes:examples:week7_wire_dimensions [2017/10/04 14:11] – [Solution] tallpaul | ||
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* Segment diameters: $d_1=3 \text{ mm}$, $d_2=1 \text{ mm}$, and $d_3=3 \text{ mm}$. | * Segment diameters: $d_1=3 \text{ mm}$, $d_2=1 \text{ mm}$, and $d_3=3 \text{ mm}$. | ||
* Current: $I_1 = 5 \text{ A}$. | * Current: $I_1 = 5 \text{ A}$. | ||
- | * Voltage: $\Delta V_1 = 1 \text{ V}$. | + | * Voltage: $\Delta V_1 = 1 \text{ V}$, $\Delta V_{battery} = 12 \text{ V}$. |
===Lacking=== | ===Lacking=== | ||
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===Representations=== | ===Representations=== | ||
- | * We represent the situation with diagram below. We number the segments for simplicity of representing the quantities we are interested in. | + | * We represent the situation with diagram below. We number the segments for simplicity of representing the quantities we are interested in (see above in " |
{{ 184_notes: | {{ 184_notes: | ||
====Solution==== | ====Solution==== | ||
- | Let's start with node $A$. Incoming current | + | 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 $$E_1 = \frac{\Delta V_1}{L_1} = 0.5 \text{ |
- | $$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 | + | The power dissipated through |
- | We do a similar analysis | + | Now, for segment 2. We can use [[184_notes: |
- | $$I_{B\rightarrow D} = I_{out}-I_3 | + | A simple application of the [[184_notes: |
- | For node $C$, incoming current is $I_2$ and $I_3$. There is no outgoing current defined yet! $I_{C\rightarrow D}$ must be outgoing to balance. To satisfy the Node Rule, we set | + | For segment 3, we can reason based on the thicknesses of the segments that $E_3=E_1$. This yields |
- | $$I_{C\rightarrow D} = I_{out} = I_{in} = I_2+I_3 = 7 \text{ | + | We can use the same reasoning as before to say that $I_3=I_2=I_1$. We can also find voltage pretty easily: |
- | + | ||
- | 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} | + | |
- | + | ||
- | 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, | + | |
- | + | ||
- | {{ 184_notes: | + |