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184_notes:examples:week7_wire_dimensions [2017/10/04 13:38] – [Solution] tallpaul | 184_notes:examples:week7_wire_dimensions [2017/10/06 18:22] – [Solution] tallpaul | ||
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====Solution==== | ====Solution==== | ||
- | Let's start with segment 1. The power dissipated through the segment is just $$P_1=I_1 \Delta V_1=$$ | + | 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 $$E_1 = \frac{\Delta V_1}{L_1} = 0.5 \text{ V/m}$$ |
+ | 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: | ||
+ | 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: $\Delta V_3=E_3 L_3 = 2 \text{ V}$. 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. |