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184_notes:resistivity [2018/10/09 13:36] – dmcpadden | 184_notes:resistivity [2018/10/09 13:39] – dmcpadden | ||
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[{{ 184_notes: | [{{ 184_notes: | ||
- | == 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:resistorefielddl.jpg? | + | [{{ 184_notes:resistor_efield_dl.png? |
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 $\sigma$). |