From af41e3f052ba4289c90558b09f9a6d45d06bfc46 Mon Sep 17 00:00:00 2001 From: zeti2342 Date: Thu, 2 Jul 2026 06:31:16 +0200 Subject: [PATCH] Update contents/solutions/AI606.md Co-authored-by: Mario Haustein --- contents/solutions/AI606.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/contents/solutions/AI606.md b/contents/solutions/AI606.md index 2bcafb6..d9f8c54 100644 --- a/contents/solutions/AI606.md +++ b/contents/solutions/AI606.md @@ -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: -$\hat{U} = U_{eff} \cdot \sqrt{2}$ eingesetzt in $P = \frac{U^2}{R}$ +$\hat{U} = U_{\mathrm{eff}} \cdot \sqrt{2}$ eingesetzt in $P = \frac{U^2}{R}$ $P = \frac{(\frac{\hat{U}_{Sender}}{\sqrt{2}})^2}{R} = \frac{\hat{U}_{Sender}^2}{2R} = \frac{78,8^2}{100} = \qty{62,0944}{\watt}$