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--- electrical_engineering_2:the_stationary_electric_flow [2023/03/17 08:37] – mexleadmin
+++ electrical_engineering_2:the_stationary_electric_flow [2024/03/29 19:57] (aktuell) – mexleadmin
@@ Zeile 1: / Zeile 1: @@
-====== 2. The stationary Current Density and Flux ======
+====== 2 The stationary Current Density and Flux ======
 <callout>
@@ Zeile 35: / Zeile 35: @@
 <WRAP> <imgcaption ImgNr01 | part of a Conductor> </imgcaption> {{drawio>charges_in_conductor.svg}} </WRAP>
-For this purpose, a packet of charges ${\rm d}Q$ is considered, which will pass the area $A$ during the period ${\rm d}t$ (see <imgref imageNo02>). These charges are located in a partial volume element ${\rm d}V$, which is given by the area $A$ to be traversed and a partial section ${\rm d}x$: ${\rm d}V = A \cdot {\rm d}x$. The amount of charges per volume is given by the charge carrier density, specifically for metals by the electron density $n_\rm{e}$. The electron density $n_\rm{e}$ gives the number of free electrons per unit volume a. For copper, for example, this is approximately $n_{\rm e}({\rm Cu})=8.47 \cdot 10^{19} ~\rm{{1}\over{mm^3}}$.
+For this purpose, a packet of charges ${\rm d}Q$ is considered, which will pass the area $A$ during the period ${\rm d}t$ (see <imgref imageNo02>). These charges are located in a partial volume element ${\rm d}V$, which is given by the area $A$ to be traversed and a partial section ${\rm d}x$: ${\rm d}V = A \cdot {\rm d}x$. The amount of charges per volume is given by the charge carrier density, specifically for metals by the electron density $n_{\rm e}$. The electron density $n_{\rm e}$ gives the number of free electrons per unit volume a. For copper, for example, this is approximately $n_{\rm e}({\rm Cu})=8.47 \cdot 10^{19} ~{\rm {1}\over{mm^3}}$.
 ~~PAGEBREAK~~ ~~CLEARFIX~~
@@ Zeile 43: / Zeile 43: @@
 The flowing charges contained in the partial volume element $dV$ are then (with elementary charge $e_0$):
-\begin{align*} dQ = n_e \cdot e_0 \cdot A \cdot dx \end{align*}
+\begin{align*} {\rm d} Q = n_{\rm e} \cdot e_0 \cdot A \cdot  {\rm d}x \end{align*}
 The current is then given by $I={{{\rm d}Q}\over{{\rm d}t}}$:
-\begin{align*} I = n_e \cdot e_0 \cdot A \cdot {{{\rm d}x}\over{{\rm d}t}} = n_e \cdot e_0 \cdot A \cdot v_e \end{align*}
+\begin{align*} I = n_{\rm e} \cdot e_0 \cdot A \cdot {{{\rm d}x}\over{{\rm d}t}} = n_e \cdot e_0 \cdot A \cdot v_{\rm e} \end{align*}
 This leads to an electron velocity $v_e$ of:
-\begin{align*} v_\rm{e} = {{{\rm d}x}\over{{\rm d}t}} = {{I}\over{n_e \cdot e_0 \cdot A }} \end{align*}
+\begin{align*} v_{\rm e} = {{{\rm d}x}\over{{\rm d}t}} = {{I}\over{n_{\rm e} \cdot e_0 \cdot A }} \end{align*}
 In contrast to the considerations in electrostatics, the charges now travel with finite velocities.
-With regard to the electron velocity $v_\rm{e} \sim {{I}\over{A}}$ it is obvious to determine a current density $S$ (related to the area):
+With regard to the electron velocity $v_{\rm e} \sim {{I}\over{A}}$ it is obvious to determine a current density $S$ (related to the area):
 \begin{align*}
@@ Zeile 149: / Zeile 149: @@
 <panel type="info" title="Task 2.1.2 Electron velocity in copper"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-A current of $I = 2\{\rm A}$ flows in a conductor made of copper with cross-sectional area $A$. \\ The electron density in copper is $n_{\rm e}({\rm Cu})=8.47 \cdot 10^{19} ~\rm{{1}\over{mm^3}}$ and the magnitude of the elementary charge $e_0 = 1.602 \cdot 10^{-19} ~\rm{As}$
+A current of $I = 2~{\rm A}$ flows in a conductor made of copper with cross-sectional area $A$. \\ The electron density in copper is $n_{\rm e}({\rm Cu})=8.47 \cdot 10^{19} ~{\rm {1}\over{mm^3}}$ and the magnitude of the elementary charge $e_0 = 1.602 \cdot 10^{-19} ~{\rm As}$
-  - What is the average electron velocity $v_\rm{e,1}$ of electrons when the cross-sectional area of the conductor is $A = 1.5~\rm{mm}^2$?
+  - What is the average electron velocity $v_{\rm e,1}$ of electrons when the cross-sectional area of the conductor is $A = 1.5~{\rm mm}^2$?
-  - What is the average electron velocity $v_\rm{e,1}$ of electrons when the cross-sectional area of the conductor is $A = 1.0~\rm{mm}^2$?
+  - What is the average electron velocity $v_{\rm e,1}$ of electrons when the cross-sectional area of the conductor is $A = 1.0~{\rm mm}^2$?
 </WRAP></WRAP></panel>
@@ Zeile 158: / Zeile 158: @@
 <panel type="info" title="Task 2.1.3 Electron velocity in Semiconductors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-Similar to task 2.1.2 a current of $I = 2~\rm{A}$ shall flow through a cross-sectional area $A = 1.5~\rm{mm}^2$ of semiconductors. \\
+Similar to task 2.1.2 a current of $I = 2~{\rm A}$ shall flow through a cross-sectional area $A = 1.5~{\rm mm}^2$ of semiconductors. \\
-(magnitude of the elementary charge $e_0 = 1.602 \cdot 10^{-19} ~\rm{As}$)
+(magnitude of the elementary charge $e_0 = 1.602 \cdot 10^{-19} ~{\rm As}$)
-  - What is the average electron velocity $v_\rm{e,1}$ of electrons for undoped silicon (electron density $n_{\rm e}({\rm Si})=9.65 \cdot 10^{9} ~\rm{{1}\over{cm^3}}$ at room temperature)?
+  - What is the average electron velocity $v_{\rm e,1}$ of electrons for undoped silicon (electron density $n_{\rm e}({\rm Si})=9.65 \cdot 10^{9} ~{\rm {1}\over{cm^3}}$ at room temperature)?
-  - The electron density of doped silicon $n_\rm{e}(Si\; doped)$ is (mainly) given by the number of dopant atoms per volume. The doping shall provide 1 donator atom per $10^7$ silicon atoms (here: one donator atom shall release one electron). The molar volume of silicon is $12 \cdot 10^{-6}~\rm{{m^3}\over{mol}}$ with $6.022 \cdot 10^{23}$ silicon atoms per $\rm{mol}$. What is the average electron velocity $v_\rm{e,1}$ of electrons for doped silicon?
+  - The electron density of doped silicon $n_{\rm e}(Si\; doped)$ is (mainly) given by the number of dopant atoms per volume. The doping shall provide 1 donator atom per $10^7$ silicon atoms (here: one donator atom shall release one electron). The molar volume of silicon is $12 \cdot 10^{-6}~{\rm {m^3}\over{mol}}$ with $6.022 \cdot 10^{23}$ silicon atoms per ${\rm mol}$. What is the average electron velocity $v_{\rm e,1}$ of electrons for doped silicon?
@@ Zeile 171: / Zeile 171: @@
 <callout>
-==== Targets ====
+==== Goals ====
 After this lesson, you should:
@@ Zeile 240: / Zeile 240: @@
 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 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.
+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 $S = 25 ~\rm{mA/cm^2}$
+  - Calculate the required distance of the plates to get a current density of $S = 25 ~{\rm mA/cm^2}$
   - 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>