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ee2:task_7el8zlzjglaazxtw_with_calculation [2024/07/05 02:18]
mexleadmin angelegt
ee2:task_7el8zlzjglaazxtw_with_calculation [2024/07/05 02:21] (aktuell)
mexleadmin
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 With the real and imaginary parts, we can derive the magnitude and phase: With the real and imaginary parts, we can derive the magnitude and phase:
 \begin{align*} \begin{align*}
-X_{RLC}  &=\sqrt{R^2 + {X}_{LC}^2}  \\+Z_{RLC}  &=\sqrt{R^2 + {X}_{LC}^2}  \\
          &=\sqrt{(88 ~\rm m\Omega)^2 + (-121.45 ~\rm m\Omega)^2}  \\          &=\sqrt{(88 ~\rm m\Omega)^2 + (-121.45 ~\rm m\Omega)^2}  \\
          &= 150.0... ~\rm m\Omega \\          &= 150.0... ~\rm m\Omega \\
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 #@HiddenBegin_HTML~7El8zLZJglAazXTw_12,Result~@# #@HiddenBegin_HTML~7El8zLZJglAazXTw_12,Result~@#
-  * $X_{RLC} = 150.0~\rm m\Omega $+  * $Z_{RLC} = 150.0~\rm m\Omega $
   * $\varphi = -54.07° $   * $\varphi = -54.07° $
 #@HiddenEnd_HTML~7El8zLZJglAazXTw_12,Result~@# #@HiddenEnd_HTML~7El8zLZJglAazXTw_12,Result~@#
  
 2. The impedance magnitude of task 1. can be interpreted as the impedance magnitude of an effective ideal capacity $C_0$. \\  2. The impedance magnitude of task 1. can be interpreted as the impedance magnitude of an effective ideal capacity $C_0$. \\ 
-In this case, the magnitude of the impedance of $C_0$ would be $X_{C0}=X_{RLC}$. \\+In this case, the magnitude of the impedance of $C_0$ would be $X_{C0}=Z_{RLC}$. \\
 Which value would $C_0$ have for the given $f_0$? Which value would $C_0$ have for the given $f_0$?
  
 #@HiddenBegin_HTML~7El8zLZJglAazXTw_21,Path~@# #@HiddenBegin_HTML~7El8zLZJglAazXTw_21,Path~@#
-The calculated impedance of $X_{RLC}$ has to be set equal to $X_{C0}$+The calculated impedance of $Z_{RLC}$ has to be set equal to $X_{C0}$
 \begin{align*} \begin{align*}
-X_{RLC} &= X_{C0}  \\+Z_{RLC} &= X_{C0}  \\
         &= {{1}\over{2\pi f \cdot C}} \\         &= {{1}\over{2\pi f \cdot C}} \\
-\rightarrow C &= {{1}\over{2\pi f \cdot X_{RLC}}}+\rightarrow C &= {{1}\over{2\pi f \cdot Z_{RLC}}}
 \end{align*} \end{align*}