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electrical_engineering_and_electronics_1:block14 [2025/11/02 21:10] – mexleadminelectrical_engineering_and_electronics_1:block14 [2025/11/02 21:32] (aktuell) – [Conceptual overview] mexleadmin
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 ===== Conceptual overview ===== ===== Conceptual overview =====
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-  - **Analogy:** Replace *displacement flow* in dielectrics ($\vec{D}=\varepsilon\vec{E}$, charge storage) by **flow density** in conductors ($\vec{J}=\sigma\vec{E}$, charge transport). Driving cause is still the electric field $\vec{E}$; the material parameter changes from $\varepsilon$ to $\sigma=\dfrac{1}{\rho}$. +  - **Analogy:** Replace *displacement flow* in dielectrics ($\vec{D}=\varepsilon\vec{E}$, charge storage) by **flow density** in conductors ($\vec{J}=\sigma\vec{E}$, charge transport). \\ Driving cause is still the electric field $\vec{E}$; the material parameter changes from $\varepsilon$ to $\sigma=\dfrac{1}{\rho}$. 
-  - **Global relations:** Voltage is a line integral $U=\int \vec{E}\cdot{\rm d}\vec{s}$; current is a flux integral $I=\iint_A \vec{J}\cdot{\rm d}\vec{A}$. Their ratio defines $G=\dfrac{I}{U}$ and $R=\dfrac{U}{I}$ for a given geometry and material. +  - **Global relations:** Voltage is a line integral $U=\int \vec{E}\cdot{\rm d}\vec{s}$; current is a flux integral $I=\iint_A \vec{J}\cdot{\rm d}\vec{A}$. \\ Their ratio defines $G=\dfrac{I}{U}$ and $R=\dfrac{U}{I}$ for a given geometry and material. 
-  - **Geometry matters:** Uniform fields (parallel plates) give $E=\text{const}$ and simple $G=\dfrac{\sigma A}{l}$. Curved fields (coax) spread with radius → logarithmic dependence.+  - **Geometry matters:** Uniform fields (parallel plates) give $E=\text{const}$ and simple $G=\dfrac{\sigma A}{l}$. \\ Curved fields (coax) spread with radius → logarithmic dependence.
   - **Checks:** Units ($\sigma$ in $\rm S/m$, $G$ in $\rm S$, $R$ in $\Omega$). Limits: \\ $A\!\to\!\infty \Rightarrow R\!\to\!0$ \\ $l\!\to\!0 \Rightarrow R\!\to\!0$ \\ $r_a\!\downarrow r_i \Rightarrow R\!\to\!0$.   - **Checks:** Units ($\sigma$ in $\rm S/m$, $G$ in $\rm S$, $R$ in $\Omega$). Limits: \\ $A\!\to\!\infty \Rightarrow R\!\to\!0$ \\ $l\!\to\!0 \Rightarrow R\!\to\!0$ \\ $r_a\!\downarrow r_i \Rightarrow R\!\to\!0$.
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 ===== Exercises ===== ===== Exercises =====
-==== Worked examples ==== 
  
-...+<panel type="info" title="Task 2.2.1 Simulation"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
 +The simulation program of [[http://www.falstad.com/emstatic-old/|Falstad]] can show equipotential surfaces, electric field strength, and current density in different objects.
 +
 +  - Open the simulation program via the link
 +  - Select: ''Setup: Wire w/ Current'' and ''Show Current (j)''.
 +  - You will now see a finite conductor with charge carriers starting at the top end and arriving at the bottom end.
 +  - We now want to observe what happens when the conductor is tapered.
 +      - To do this, select ''Mouse = Clear Square''. You can now use the left mouse button to remove parts from the conducting material. The aim should be, that in the middle of the conductor, there is only a one-box wide line, on a length of at least 10 boxes. If you want to add conductive material again, this is possible with ''Mouse = Add - Conductor''.
 +      - Consider why more equipotential lines are now accumulating as the conductor is tapered.
 +      - If you additionally draw in the E-field with ''Show E/j'', you will see that it is stronger along the taper. This can be checked with the slider ''Brightness''. Why is this?
 +  - Select ''Setup: Current in 2D 1'', ''Show E/rho/j''. Why doesn't the cavity behave like a Faraday cage here?
 +
 +</WRAP></WRAP></panel>
 +
 +<panel type="info" title="Task 2.2.2 Water Resistor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 +
 +In transformer stations sometimes water resistors are used as {{wp>Liquid rheostat}}. In this resistor, the water works as a (poor) conductor which can handle a high power loss. 
 +
 +The water resistor consists of a water basin. In the given basin two quadratic plates with the edge length of $l = 80 ~{\rm cm}$ are inserted with the distance $d$ between them. 
 +The resistivity of the water is $\rho = 0.25 ~\Omega {\rm m}$. The resistor shall dissipate the energy of $P = 4 ~{\rm kW}$ and shall exhibit a homogeneous current field. 
 +
 +  - Calculate the required distance of the plates to get a current density of $J = 25 ~{\rm mA/cm^2}$
 +  - What are the values of the current $I$ and the voltage $U$ at the resistor, such as the internal electric field strength $E$ in the setup?
 +
 +</WRAP></WRAP></panel>
 ===== Embedded resources ===== ===== Embedded resources =====
 <WRAP column half> <WRAP column half>