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electrical_engineering_and_electronics_1:block20 [2026/01/20 15:34] – mexleadminelectrical_engineering_and_electronics_1:block20 [2026/01/20 15:39] (aktuell) – [20.3 Exercises] mexleadmin
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 #@HiddenBegin_HTML~4511R,Result~@# #@HiddenBegin_HTML~4511R,Result~@#
 \begin{align*}  \begin{align*} 
-L_1 &= 3.0 ~\rm mH+L_1 &= 0.75 ~\rm mH
 \end{align*} \end{align*}
 #@HiddenEnd_HTML~4511R,Result~@# #@HiddenEnd_HTML~4511R,Result~@#
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 #@HiddenBegin_HTML~4512R,Result~@# #@HiddenBegin_HTML~4512R,Result~@#
 \begin{align*}  \begin{align*} 
-L_1 &= 12 ~\rm mH+L_1 &= 4 ~\rm mH
 \end{align*} \end{align*}
 #@HiddenEnd_HTML~4512R,Result~@# #@HiddenEnd_HTML~4512R,Result~@#
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 #@HiddenBegin_HTML~4513R,Result~@# #@HiddenBegin_HTML~4513R,Result~@#
 \begin{align*}  \begin{align*} 
-L_1 &= 6.0 ~\rm mH+L_1 &= 1.5 ~\rm mH
 \end{align*} \end{align*}
 #@HiddenEnd_HTML~4513R,Result~@# #@HiddenEnd_HTML~4513R,Result~@#
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 #@HiddenBegin_HTML~4514R,Result~@# #@HiddenBegin_HTML~4514R,Result~@#
 \begin{align*}  \begin{align*} 
-L_4 &= 3.0 ~\rm H+L_4 &= 0.75 ~\rm H
 \end{align*} \end{align*}
 #@HiddenEnd_HTML~4514R,Result~@# #@HiddenEnd_HTML~4514R,Result~@#
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 <panel type="info" title="Exercise 4.5.2 Self Induction II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 4.5.2 Self Induction II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-A cylindrical air coil (length $l=40 ~\rm cm$, diameter $d=5.0 ~\rm cm$, and a number of windings $N=300$) passes a current of $30 ~\rm A$. The current shall be reduced linearly in $2.0 ~\rm ms$ down to $0.0 ~\rm A$.+A cylindrical air coil (length $l=40 ~\rm cm$, radius $r=5.0 ~\rm cm$, and a number of windings $N=300$) passes a current of $30 ~\rm A$. The current shall be reduced linearly in $2.0 ~\rm ms$ down to $0.0 ~\rm A$.
  
 What is the amount of the induced voltage $u_{\rm ind}$?  What is the amount of the induced voltage $u_{\rm ind}$?