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184_notes:examples:week7_wire_dimensions [2017/10/04 13:51] – [Solution] tallpaul | 184_notes:examples:week7_wire_dimensions [2017/10/04 14:11] – [Solution] tallpaul | ||
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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 find voltage pretty easily: $\Delta V_3=E_3 L_3 = 2 \text{ V}$. |