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Update contents/solutions/AI610.md

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Co-authored-by: Matthias Jung <jungma@eit.uni-kl.de>
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zeti2342
2026-06-22 07:01:50 +02:00
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co-authored by Matthias Jung
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@@ -12,7 +12,7 @@ $U = \sqrt{P * R} = \sqrt{1 \cdot 51} \approx \qty{7}{\volt}$
Für die Diode zählt nur die obere Halbwelle, also die Spitzenspannung:
$\hat{U} = U * \sqrt{2} \approx \qty{10,1}{\volt}
$\hat{U} = U \cdot \sqrt{2} \approx \qty{10,1}{\volt}
Die Eingangsspannung wird durch den 2*330 Ohm Spannungsteiler halbiert und dann um $U_F$ verringert: