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Update contents/solutions/AI606.md

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Co-authored-by: Matthias Jung <[email protected]>
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zeti2342
2026-06-22 07:00:19 +02:00
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co-authored by Matthias Jung
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@@ -18,4 +18,4 @@ $\hat{U} = U_{eff} \cdot \sqrt{2}$ eingesetzt in $P = \frac{U^2}{R}$
$P = \frac{(\frac{\hat{U}_{Sender}}{\sqrt{2}})^2}{R} = \frac{\hat{U}_{Sender}^2}{2R} = \frac{78,8^2}{100} = \qty{62,0944}{\watt}$
Die richtige Antwort ist daher "Zirka 60 W".
Die richtige Antwort ist daher "Zirka $\qty{60}{\watt}$".