Question

РЕШИТЕ ПО ФИЗИКЕ БАЛОВ.​ только с решением

Answer 1

Объяснение:

$R = \dfrac{R_1R_2}{R_1+R_2} + R_3 + \dfrac{R_4R_5}{R_4+R_5} = \dfrac{2*3}{2+3} + 0.8 + \dfrac{4*4}{4+4} = 4~Om$

$I_{\sum} = \dfrac{U}{R} = \dfrac{4}{4} = 1A\\I_{12} = I = 1A\\U_1 = U_2 = U_{12} = I_{12}*\dfrac{R_1R_2}{R_1+R_2} = 1*1.2 = 1.2~V\\I_1 = \dfrac{U_1}{R_1} = \dfrac{1.2}{2} = 0.6~A\\I_2 = \dfrac{U_2}{R_2} = \dfrac{1.2}{3} = 0.4~A\\I_3 = I = 1~A\\U_3 = I_3*R_3 = 1*0.8 = 0.8~V\\I_{45} = I = 1A\\U_4 = U_5 = U_{45} = I_{45}*\dfrac{R_4R_5}{R_4+R_5} = 1*2 = 2~V\\I_4 = \dfrac{U_4}{R_4} = \dfrac{2}{4} = 0.5~A\\I_5 = \dfrac{U_5}{R_5} = \dfrac{2}{4} = 0.5~A\\$