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184_notes:examples:week7_resistance_wire [2017/10/06 18:25] – [Solution] tallpaul | 184_notes:examples:week7_resistance_wire [2018/06/19 14:54] (current) – curdemma | ||
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=====Resistance of a Wire===== | =====Resistance of a Wire===== | ||
- | Suppose you have a wire whose resistance | + | Suppose you have a wire whose resistance |
===Facts=== | ===Facts=== | ||
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===Representations=== | ===Representations=== | ||
- | * We represent the resistance of a simple wire such as this with: R=LσA | + | * We represent the resistance of a simple wire with: R=LσA |
====Solution==== | ====Solution==== | ||
All we need here is our representation for the resistance of the wire. In the first change to the wire, we triple it's length (2 cm→6 cm). Our new resistance then is found by Rnew=LnewσA=3LσA=3R=180 mΩ | All we need here is our representation for the resistance of the wire. In the first change to the wire, we triple it's length (2 cm→6 cm). Our new resistance then is found by Rnew=LnewσA=3LσA=3R=180 mΩ | ||
- | If instead, we made the other change, we would have tripled the cross sectional area (1 mm2→3 mm2). Our new resistance would then be Rnew=LσAnew=Lσ3A=13R=20 mΩ | + | If instead, we made the other change, we would have tripled the cross-sectional area (1 mm2→3 mm2). Our new resistance would then be Rnew=LσAnew=Lσ3A=13R=20 mΩ |
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+ | These answers should make sense physically. The longer the wire, the more material there is for the electrons to push through, so the resistance is higher. The bigger the area, the more electrons are able to flow through the wire in at a given time so the resistance should be smaller. |