[latex]Dane :\R_1 = 3 Omega \R_2 = 6 Omega \U = 12 V \ t = 10 s \\Szukane : \E = ? \\Rozwiazanie : \ frac{1}{R_z} =frac{1}{R_1} + frac{1}{R_2} \\frac{1}{R_z} = frac{1}{3} + frac{1}{6} = frac{3}{6} \\R_z = frac{6}{3} = 2 Omega [/latex] [latex]U = frac{E}{q} \E = U * q\\I = frac{q}{t} \q = I*t \ \ \ U = RI \I = frac{U}{R_z} \\\I = frac{12}{2} = 6 A \\q = 10 * 6 = 60 C \\E = 12 * 60 = 720 J [/latex] E = 720 J
Skoro równolegle, to korzystamy ze wzoru: [latex]frac{1}{R}=frac{1}{R_1}+frac{1}{R_2}+...[/latex] Podstawiamy: [latex]frac{1}{R}=frac{1}{3 Omega} + frac{1}{6 Omega} \ frac{1}{R}=frac{2}{6 Omega} + frac{1}{6 Omega} \ frac{1}{R}=frac{3}{6 Omega} = frac{1}{2 Omega} \ R = 2 Omega[/latex] Opór to 2 omy. Teraz przydatny będzie wzór: [latex]E = W = U cdot I cdot t \ I = frac{U}{R} \ W = frac{U^2 cdot t}{R} \ W = frac{12^2 cdot 10}{2} = frac{1440}{2} = 720 J [/latex]
