From 082276dba034eaab34c1fd0b61917b439ab0a940 Mon Sep 17 00:00:00 2001 From: Matthias Jung Date: Sat, 16 May 2026 13:30:04 +0200 Subject: [PATCH] fix: small typo --- contents/solutions/AD109.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/contents/solutions/AD109.md b/contents/solutions/AD109.md index 94fec5f..edaf92a 100644 --- a/contents/solutions/AD109.md +++ b/contents/solutions/AD109.md @@ -15,7 +15,7 @@ $ R_\mathrm{par} = \frac{\qty{100}{\ohm} \cdot \qty{200}{\ohm}}{\qty{100}{\ohm} Mit dem vorgeschalteten Widerstand ergibt sich: -$ R_\mathrm{cmin} = \qty{200}{\ohm} + \qty{67}{\ohm} = \qty{267}{\ohm} $ +$ R_\mathrm{min} = \qty{200}{\ohm} + \qty{67}{\ohm} = \qty{267}{\ohm} $ Nun betrachten wir den größten Wert. Dazu setzen wir: