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ee2:task_ludzwiuhjxitz85b_with_calculation [2024/07/03 02:02] – angelegt mexleadminee2:task_ludzwiuhjxitz85b_with_calculation [2024/07/03 10:03] (aktuell) – mexleadmin
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-{{tag>induction flux induced_voltage exam_ee2_SS2021}}{{include_n>1000}}+{{tag>induction flux induced_voltage exam_ee2_SS2021}}{{include_n>1200}}
    
 #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ effect of induction\\ #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ effect of induction\\
 <fs medium>(written test, approx. 5 % of a 120-minute written test, SS2021)</fs> #@TaskText_HTML@# <fs medium>(written test, approx. 5 % of a 120-minute written test, SS2021)</fs> #@TaskText_HTML@#
  
-A simple conductor loop is penetrated by a changing magnetic flux. \\+A single conductor loop is penetrated by a changing magnetic flux. \\
 The following figure shows the variation of the flux $\Phi(t)$ over time.  \\ \\ The following figure shows the variation of the flux $\Phi(t)$ over time.  \\ \\
-Calculate the variation of the induced voltage $U_{\rm ind}(t)$ over time and draw it in a separate diagram. +Calculate the variation of the induced voltage $u_{\rm ind}(t)$ over time and draw it in a separate diagram. 
  
 {{drawio>ee2:LUdzWiUhjxITZ85B_diagram1.svg}} {{drawio>ee2:LUdzWiUhjxITZ85B_diagram1.svg}}
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 Based on Faraday's Law of Induction the induced voltage is given by: Based on Faraday's Law of Induction the induced voltage is given by:
 \begin{align*} \begin{align*}
-U_{\rm ind} = - {{ {\rm d} }\over{ {\rm d}t}} \Phi(t)\\+u_{\rm ind} =& - {{ {\rm d} }\over{ {\rm d}t}} \Psi(t) \bigg\rvert_{n=1}\\ 
 +            =& - {{ {\rm d} }\over{ {\rm d}t}} \Phi(t) \\
 \end{align*} \end{align*}
  
 For a linear function, the derivative can be substituted by Deltas ($\rm d \rightarrow \Delta$): \\ For a linear function, the derivative can be substituted by Deltas ($\rm d \rightarrow \Delta$): \\
 \begin{align*} \begin{align*}
-U_{\rm ind} = - {{ \Delta \Phi(t)}\over{ \Delta t}} = - { { \Phi(t_{\rm n+1} ) - \Phi(t_{\rm n} ) } \over { t_{\rm n+1} - t_{\rm n} } } \\+u_{\rm ind} = - {{ \Delta \Phi(t)}\over{ \Delta t}} = - { { \Phi(t_{\rm n+1} ) - \Phi(t_{\rm n} ) } \over { t_{\rm n+1} - t_{\rm n} } } \\
 \end{align*} \end{align*}
  
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 {{drawio>ee2:LUdzWiUhjxITZ85B_path1.svg}} {{drawio>ee2:LUdzWiUhjxITZ85B_path1.svg}}
  
-  * For the intervals $\rm I$, $\rm III$, and $\rm V$ , the flux $\Phi(t)$ is constant. Therefore, $\Delta \Phi(t)=0$ and $U_{\rm ind}(t)=0{~\rm V}$ \\ {{drawio>ee2:LUdzWiUhjxITZ85B_path2.svg}}+  * For the intervals $\rm I$, $\rm III$, and $\rm V$ , the flux $\Phi(t)$ is constant. Therefore, $\Delta \Phi(t)=0$ and $u_{\rm ind}(t)=0{~\rm V}$ \\ {{drawio>ee2:LUdzWiUhjxITZ85B_path2.svg}}
   * For the interval  $\rm II$:    * For the interval  $\rm II$: 
     * The change in the flux is: $ \Delta \Phi(t) = 1.5 \cdot 10^{-4} {~\rm Vs} - 4.5 \cdot 10^{-4} {~\rm Vs}= - 3.0 \cdot 10^{-4} {~\rm Vs}$     * The change in the flux is: $ \Delta \Phi(t) = 1.5 \cdot 10^{-4} {~\rm Vs} - 4.5 \cdot 10^{-4} {~\rm Vs}= - 3.0 \cdot 10^{-4} {~\rm Vs}$
     * The time span is: $0.2 ~\rm s$     * The time span is: $0.2 ~\rm s$
-    * Conclusively, the induced voltage is: $U_{\rm ind}(t) = + {{3.0 \cdot 10^{-4} {~\rm Vs}} \over {0.2 ~\rm s} } = 1.5 {~\rm mV}$ \\ {{drawio>ee2:LUdzWiUhjxITZ85B_path3.svg}}+    * Conclusively, the induced voltage is: $u_{\rm ind}(t) = + {{3.0 \cdot 10^{-4} {~\rm Vs}} \over {0.2 ~\rm s} } = 1.5 {~\rm mV}$ \\ {{drawio>ee2:LUdzWiUhjxITZ85B_path3.svg}}
   * For the interval  $\rm IV$:    * For the interval  $\rm IV$: 
     * The change in the flux is: $ \Delta \Phi(t) = 0 \cdot 10^{-4} {~\rm Vs} - 1.5 \cdot 10^{-4} {~\rm Vs}= - 1.5 \cdot 10^{-4} {~\rm Vs}$     * The change in the flux is: $ \Delta \Phi(t) = 0 \cdot 10^{-4} {~\rm Vs} - 1.5 \cdot 10^{-4} {~\rm Vs}= - 1.5 \cdot 10^{-4} {~\rm Vs}$
     * The time span is: $0.2 ~\rm s$     * The time span is: $0.2 ~\rm s$
-    * Conclusively, the induced voltage is: $U_{\rm ind}(t) = + {{1.5 \cdot 10^{-4} {~\rm Vs}} \over {0.2 ~\rm s} } = 0.75 {~\rm mV}$ \\ {{drawio>ee2:LUdzWiUhjxITZ85B_path4.svg}}+    * Conclusively, the induced voltage is: $u_{\rm ind}(t) = + {{1.5 \cdot 10^{-4} {~\rm Vs}} \over {0.2 ~\rm s} } = 0.75 {~\rm mV}$ \\ {{drawio>ee2:LUdzWiUhjxITZ85B_path4.svg}}
  
 #@HiddenEnd_HTML~LUdzWiUhjxITZ85B_11,Path ~@# #@HiddenEnd_HTML~LUdzWiUhjxITZ85B_11,Path ~@#