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electrical_engineering_1:simple_circuits [2023/09/17 00:37] – mexleadminelectrical_engineering_1:simple_circuits [2024/10/24 08:13] (aktuell) – mexleadmin
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- The simulatiok====== 2. Simple DC circuits ======+====== 2 Simple DC circuits ======
  
 So far, only simple circuits consisting of a source and a load connected by wires have been considered. \\  So far, only simple circuits consisting of a source and a load connected by wires have been considered. \\ 
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 +\\ \\
 +The current divider rule shows in which way an incoming current on a node will be divided into two outgoing branches.
 +The rule states that the currents $I_1, ... I_n$ on parallel resistors $R_1, ... R_n$ behave just like their conductances $G_1, ... G_n$ through which the current flows. \\
  
-The current divider rule can also be derived from Kirchhoff's current law. \\ +$\large{{I_1}\over{I_{\rm res}} = {{G_1}\over{G_{\rm res}}}$ 
-This states that, for resistors $R_1, ... R_n$ their currents $I_1, ... I_n$ behave just like the conductances $G_1, ... G_n$ through which they flow. \\ +
- +
-$\large{{I_1}\over{I_g}} = {{G_1}\over{G_g}}$ +
  
 $\large{{I_1}\over{I_2}} = {{G_1}\over{G_2}}$ $\large{{I_1}\over{I_2}} = {{G_1}\over{G_2}}$
  
-This can also be derived by Kirchhoff's current law: The voltage drop $U$ on parallel resistors $R_1, ... R_n$ is the same. When $U_1 = U_2 = ... = U$, then also $R_1 \cdot I_1 = R_2 \cdot I_2 = ... = R_{\rm eq} \cdot I_{\rm res}$. \\ +The rule also be derived from Kirchhoff's current law: \\ 
-Therefore, we get with the conductance: ${{I_1} \over {G_1}} = {{I_2} \over {G_2}}= ... = {{I_{\rm eq}} \over {G_{\rm res}}}$+  - The voltage drop $U$ on parallel resistors $R_1, ... R_n$ is the same.  
 +  - When $U_1 = U_2 = ... = U$, then the following equation is also true: $R_1 \cdot I_1 = R_2 \cdot I_2 = ... = R_{\rm eq} \cdot I_{\rm res}$. \\ 
 +  - Therefore, we get with the conductance: ${{I_1} \over {G_1}} = {{I_2} \over {G_2}}= ... = {{I_{\rm eq}} \over {G_{\rm res}}}$
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
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 \begin{align*} \begin{align*}
-R_g = R || R + R || R || R || R + R || R || R || R + R || R = {{1}\over{2}}\cdot R + {{1}\over{4}}\cdot R + {{1}\over{4}}\cdot R + {{1}\over{2}}\cdot R = 1.5\cdot R+R_{\rm eq} = R || R + R || R || R || R + R || R || R || R + R || R = {{1}\over{2}}\cdot R + {{1}\over{4}}\cdot R + {{1}\over{4}}\cdot R + {{1}\over{2}}\cdot R = 1.5\cdot R
   \end{align*}   \end{align*}
  
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-<panel type="info" title="other Exercises"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> 
  
-More German exercises can be found online on the pages of [[https://www.eit.hs-karlsruhe.de/hertz/teil-b-gleichstromtechnik/zusammenschaltung-von-widerstaenden-und-idealen-quellen/uebungsaufgaben-zusammenschaltung-von-widerstaenden/berechnung-von-ersatzwiderstaenden.html|HErTZ]] (selection on the left in the menu). 
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