184_notes:resistivity

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184_notes:resistivity [2018/10/09 13:36] – dmcpadden184_notes:resistivity [2018/10/09 13:39] – dmcpadden
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 [{{  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.}}] [{{  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__//, then the electric field inside the resistor would be constant in magnitude and direction and would point along the length of the wire. We could use the relationship between electric potential and electric field then to write: 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__//, then the electric field inside the resistor would be constant in magnitude and direction and would point along the length of the wire. We could use the relationship between electric potential and electric field then to write:
 $$\Delta V =- \int_i^f \vec{E} \cdot \vec{dl}$$ $$\Delta V =- \int_i^f \vec{E} \cdot \vec{dl}$$
  
-[{{  184_notes:resistorefielddl.jpg?300|Electric field direction in a resistor (shown by the red arrow) and the dl vector shown by the blue arrow.}}]+[{{  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, leaving: 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, leaving:
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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 $\sigma$).
  • 184_notes/resistivity.txt
  • Last modified: 2021/02/27 04:07
  • by bartonmo