From 87982bd9cbbc56b4f797e53e9e441b392ea52217 Mon Sep 17 00:00:00 2001 From: zeti2342 Date: Mon, 22 Jun 2026 07:00:09 +0200 Subject: [PATCH] Update contents/solutions/AI606.md Co-authored-by: Matthias Jung --- 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 08f0d02..c74a151 100644 --- a/contents/solutions/AI606.md +++ b/contents/solutions/AI606.md @@ -14,7 +14,7 @@ $\hat{U}_{Sender} = \frac{50}{5} \cdot \hat{U} = \qty{78,8}{\volt}$ Aus den Formeln für Leistung und Effektivspannung in den Hilfsmitteln lässt sich dann die Antwort errechnen: -$\hat{U} = U_{eff} * \sqrt{2}$ eingesetzt in $P = \frac{U^2}{R}$ +$\hat{U} = U_{eff} \cdot \sqrt{2}$ eingesetzt in $P = \frac{U^2}{R}$ $P = \frac{(\frac{\hat{U}_{Sender}}{\sqrt{2}})^2}{R} = \frac{\hat{U}_{Sender}^2}{2R} = \frac{78,8^2}{100} = \qty{62,0944}{\watt}$