Merge pull request #179 from hamarituc/transformer
fix transformer slides
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@@ -20,11 +20,11 @@ $ü = \frac{N_P}{N_S} = \frac{U_P}{U_S} = \frac{I_S}{I_P}$
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<fragment>
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Das Verhältnis der Primär- zur Sekundärimpedanz ist wie die obigen Verhältnisse zum Quadrat:
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$ü = \frac{Z_P}{Z_S} = \left(\frac{N_P}{N_S}\right)^2 = \left(\frac{U_P}{U_S}\right)^2 = \left(\frac{I_S}{I_P}\right)^2$
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$ü^2 = \frac{Z_P}{Z_S} = \left(\frac{N_P}{N_S}\right)^2 = \left(\frac{U_P}{U_S}\right)^2 = \left(\frac{I_S}{I_P}\right)^2$
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</fragment>
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<fragment>
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Oder nach Ziehung der Wurzel:
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Oder nach ziehen der Wurzel:
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$ü = \frac{N_P}{N_S} = \frac{U_P}{U_S} = \frac{I_S}{I_P} = \sqrt{\frac{Z_P}{Z_S}}$
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</fragment>
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@@ -53,7 +53,7 @@ $\begin{split} \frac{U_P}{U_S} &= \frac{I_S}{I_P} \\ \Rightarrow I_P &= \frac{I_
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* gesucht: $Z_P$
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<fragment>
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$\begin{split} ü &= \sqrt{\frac{Z_P}{Z_S}} \\ \Rightarrow Z_P &= ü^2 \cdot Z_S = \frac{1^2}{4^2} \cdot 16k\Omega \\ &= \frac{\qty{16}{\kilo\ohm}}{16} = \qty{1}{\kilo\ohm} \end{split}$
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$\begin{split} ü &= \sqrt{\frac{Z_P}{Z_S}} \\ \Rightarrow Z_P &= ü^2 \cdot Z_S = \frac{1^2}{4^2} \cdot \qty{16}{\kilo\ohm} \\ &= \frac{\qty{16}{\kilo\ohm}}{16} = \qty{1}{\kilo\ohm} \end{split}$
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</fragment>
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---
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