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184_notes:examples:week7_wire_dimensions [2017/10/04 13:51] – [Solution] tallpaul | 184_notes:examples:week7_wire_dimensions [2017/10/09 15:11] – dmcpadden | ||
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===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
- | * The circuit is in a steady state. | + | * The circuit is in a steady state - which means the current should be the same in all three sections. |
* Approximating the battery as a mechanical battery. | * Approximating the battery as a mechanical battery. | ||
* The wire has a circular cross-section. | * The wire has a circular cross-section. | ||
Line 27: | Line 27: | ||
The power dissipated through the segment is just $$P_1=I_1 \Delta V_1 = 5 \text{ W}$$ | The power dissipated through the segment is just $$P_1=I_1 \Delta V_1 = 5 \text{ W}$$ | ||
- | Now, for segment 2. We can use [[184_notes: | + | Now, for segment 2. We can use [[184_notes: |
+ | A simple application of the [[184_notes: | ||
+ | |||
+ | For segment 3, we can reason based on the thicknesses of the segments that $E_3=E_1$. This yields $$E_3 = 0.5 \text{ V/m}$$ | ||
+ | We can use the same reasoning as before to say that $I_3=I_2=I_1$. We can also use the same equation to find voltage: $\Delta V_3=E_3 L_3 = 2 \text{ V}$. The power is calculated as before. $$P_3=I_3 \Delta V_3 = 10 \text{ W}$$ | ||
+ | |||
+ | Notice that we could have also used [[184_notes: | ||
+ | $$\Delta V_{battery}+\Delta V_1+\Delta V_2+\Delta V_3 = 12 \text{ V} - 1 \text{ V} - 9 \text{ V} - 2 \text{ V} = 0$$ | ||
+ | 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. |