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184_notes:resistivity [2018/10/09 13:36] – dmcpadden | 184_notes:resistivity [2021/02/24 17:45] – bartonmo | ||
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Section 19.2 in Matter and Interactions (4th edition) | Section 19.2 in Matter and Interactions (4th edition) | ||
- | [[184_notes: | + | /*[[184_notes: |
- | [[184_notes: | + | [[184_notes: |
===== Resistors and Conductivity ===== | ===== Resistors and Conductivity ===== | ||
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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 $\sigma$). | ||
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Resistance has units of volts per amp, which is also called an ohm. An ohm is represented by a capital omega ($\Omega$). Sometimes you may see resistance rewritten in terms of **resistivity**($\rho$), | Resistance has units of volts per amp, which is also called an ohm. An ohm is represented by a capital omega ($\Omega$). Sometimes you may see resistance rewritten in terms of **resistivity**($\rho$), | ||
- | === Ohm's Model === | + | ==== Ohm's Model ==== |
Perhaps equally as important, we can now relate the change in electric potential over a resistor to the resistance and current passing through the resistor. This model of resistance works well for low voltage and currents. This model is also often called " | Perhaps equally as important, we can now relate the change in electric potential over a resistor to the resistance and current passing through the resistor. This model of resistance works well for low voltage and currents. This model is also often called " |