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Update contents/solutions/AI610.md

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Matthias Jung
2026-08-02 20:10:07 +02:00
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@@ -16,4 +16,4 @@ $\hat{U} = U \cdot \sqrt{2} \approx \qty{10,1}{\volt}
Die Eingangsspannung wird durch den $2 \cdot \qty{330}{\ohm}$ Spannungsteiler halbiert und dann um $U_F$ verringert:
$U_A = \frac{\hat{U}}{2} - U_F = 5,05 - 0,23 \approx \qty{4,8}{\volt}$
$U_A = \frac{\hat{U}}{2} - U_F = \qty{5,05}{\volt} - \qty{0,23}{\volt} \approx \qty{4,8}{\volt}$