Решение:
\[\frac{1}{R_{\text{общ}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{10} + \frac{1}{40} = \frac{4 + 1}{40} = \frac{5}{40} = \frac{1}{8}\]
\[R_{\text{общ}} = 8 \,\text{Ом}\]
\[I_1 = \frac{U}{R_1} = \frac{80 \,\text{В}}{10 \,\text{Ом}} = 8 \,\text{А}\]
\[I_2 = \frac{U}{R_2} = \frac{80 \,\text{В}}{40 \,\text{Ом}} = 2 \,\text{А}\]
\[I_{\text{общ}} = I_1 + I_2 = 8 \,\text{А} + 2 \,\text{А} = 10 \,\text{А}\]
Ответ: Б) 8 Ом, В) 8 А и 2 А, общая 10 А.