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Update contents/solutions/AI606.md

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Co-authored-by: Mario Haustein <[email protected]>
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zeti2342
2026-07-02 06:31:16 +02:00
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co-authored by Mario Haustein
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@@ -14,7 +14,7 @@ $\hat{U}_{\mathrm{Sender}} = \frac{50}{5} \cdot \hat{U} = \qty{78,8}{\volt}$
Aus den Formeln für Leistung und Effektivspannung in den Hilfsmitteln lässt sich dann die Antwort errechnen: Aus den Formeln für Leistung und Effektivspannung in den Hilfsmitteln lässt sich dann die Antwort errechnen:
$\hat{U} = U_{eff} \cdot \sqrt{2}$ eingesetzt in $P = \frac{U^2}{R}$ $\hat{U} = U_{\mathrm{eff}} \cdot \sqrt{2}$ eingesetzt in $P = \frac{U^2}{R}$
$P = \frac{(\frac{\hat{U}_{Sender}}{\sqrt{2}})^2}{R} = \frac{\hat{U}_{Sender}^2}{2R} = \frac{78,8^2}{100} = \qty{62,0944}{\watt}$ $P = \frac{(\frac{\hat{U}_{Sender}}{\sqrt{2}})^2}{R} = \frac{\hat{U}_{Sender}^2}{2R} = \frac{78,8^2}{100} = \qty{62,0944}{\watt}$