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Update contents/solutions/AK106.md

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Co-authored-by: Christoph Fischer <dc6gf@darc.de>
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Ralf Mahler
2026-06-16 20:41:55 +02:00
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co-authored by Christoph Fischer
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@@ -26,7 +26,7 @@ $P_\mathrm{EIRP} = \qty{100}{\watt} \cdot 10^{\frac{2{,}15 - 0{,}0}{10}} = \qty{
2. Berechnung des Sicherheitsabstands $d$:
$d = \frac{\sqrt{\qty{30}{\ohm} \cdot \qty{164,1}{\watt}}}{\qty{28}{\volt/\meter}} = \frac{\qty{70,2}{\volt}}{\qty{28}{\volt/\meter}} \approx \qty{2,50}{\meter}$
$d = \frac{\sqrt{\qty{30}{\ohm} \cdot \qty{164,1}{\watt}}}{\qty{28}{\volt\per\meter}} = \frac{\qty{70,2}{\volt}}{\qty{28}{\volt\per\meter}} \approx \qty{2,50}{\meter}$
3. Prüfung der Fernfeldbedingung: