Câu 19.
Cấu tạo mạch: \(R_3nt\left(R_1//R_2\right)\)
\(U_1=R_1\cdot I_1=6\cdot\dfrac{2}{3}=4V\)
\(R_1//R_2\Rightarrow U_2=U_1=4V\)
\(U_{12}=U_1=U_2=4V\Rightarrow U_3=6-U_{12}=2V\)
\(I_3=\dfrac{U_3}{R_3}=\dfrac{2}{4}=0,5A\)
Mà \(I_3=I_{12}=I_1+I_2\Rightarrow I_2=0,5-\dfrac{2}{3}=-\dfrac{1}{6}A\)




