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184_notes:resistivity [2018/06/19 14:41] – [Conductivity] curdemma | 184_notes:resistivity [2018/10/09 13:39] – dmcpadden | ||
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====Resistance==== | ====Resistance==== | ||
- | Before when we talked about resistors, we said that a resistor was a section or part of the circuit where the passage of electrons requires more energy (conventionally, | + | [[184_notes: |
- | {{ 184_notes:resistorshape.jpg?350}} | + | [{{ 184_notes:resistor_shape.png?350|A piece of a resistor with a potential difference of $\Delta$ V from one end to the other, a length L, and a cross-sectional area of A.}}] |
- | == Derivation of $R$ == | + | |
+ | === Derivation of $R$ === | ||
For example, suppose we have a resistor that has a cross sectional area of $A$, a length $L$, and a potential difference of $\Delta V$ from one end to the other. If we //__assume a steady state current__//, | For example, suppose we have a resistor that has a cross sectional area of $A$, a length $L$, and a potential difference of $\Delta V$ from one end to the other. If we //__assume a steady state current__//, | ||
$$\Delta V =- \int_i^f \vec{E} \cdot \vec{dl}$$ | $$\Delta V =- \int_i^f \vec{E} \cdot \vec{dl}$$ | ||
- | {{184_notes: | + | [{{ 184_notes:resistor_efield_dl.png?300|Electric field direction in a resistor (shown by the red arrow) and the dl vector shown by the blue arrow.}}] |
Because $\vec{E}$ would point along the length of the wire, we would want to integrate along the length of the wire, which would mean that $\vec{E}$ and $\vec{dl}$ would be parallel. This simplifies the dot product to just a multiplication, | Because $\vec{E}$ would point along the length of the wire, we would want to integrate along the length of the wire, which would mean that $\vec{E}$ and $\vec{dl}$ would be parallel. This simplifies the dot product to just a multiplication, | ||
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$$R =\frac{L}{\sigma A}$$ | $$R =\frac{L}{\sigma A}$$ | ||
- | == Making sense of $R$ == | + | === Making sense of $R$ === |
Why does the bottom fraction make sense? A longer, thinner wire should be more resistive, so the geometric properties make sense (directly proportionally to $L$ and inversely proportional to $A$). A wire with higher conductivity should be less resistive, which also make sense (inversely proprtional to $\sigma$). | Why does the bottom fraction make sense? A longer, thinner wire should be more resistive, so the geometric properties make sense (directly proportionally to $L$ and inversely proportional to $A$). A wire with higher conductivity should be less resistive, which also make sense (inversely proprtional to $\sigma$). | ||
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^ Micro ^ Macro ^ | ^ Micro ^ Macro ^ | ||
| $v_{avg}=uE$ | | $v_{avg}=uE$ | ||
- | | $i=nAv_{avg}=nAUE$ | | + | | $i=nAv_{avg}=nAuE$ | |
==== Examples ==== | ==== Examples ==== |