Dane: d=0.5mm=0,0005m ro=4,8*10^-7 P=800W U=220V Oblicz l=? Rozwiązanie [latex]r=frac{1}{2}d[/latex] [latex]r=frac{1}{2}0,0005[/latex] [latex]r=0,00025[/latex] [latex]S=pi r^{2}[/latex] [latex]S=3,14*0,00025^{2}[/latex] [latex]S=3,14*0,0000000625[/latex] [latex]S=0,00000019625m^{2}[/latex] [latex]Rw= hofrac{l}{S}[/latex] [latex]Rw=4,8*10^{-7}frac{1}{0,00000019625}[/latex] [latex]Rw=4,8*10^{-7}*5095541,40[/latex] [latex]Rw=24458598,72*10^{-7}Omegaapprox 2,44Omega [/latex] <--- Opór naszzego drutu dla jednego metra [latex]P=frac{U^{2}}{R}[/latex] [latex]R=frac{U^{2}}{P}[/latex] [latex]R=frac{220^{2}}{800}[/latex] [latex]R=frac{48400}{800}[/latex] [latex]R=60,5Omega[/latex]<--Opór naszej grzałki [latex]Rw*l=R[/latex] [latex]l= frac{R}{Rw}[/latex] [latex]l= frac{60,5}{2,44}[/latex] [latex]l=24,79m[/latex]
