Câu hỏi của Khieem Duy

\(R_1nt\left(R_2//R_3\right)\)\(R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{12\cdot12}{12+12}=6\Omega\)\(R_{tđ}=R_1+R_{23}=24+6=30\Omega\)\(I_m=\dfrac{U_{AB}}{R}=\dfrac{12}{30}=0,4A\)\(\Rightarrow I_A=I_m=0,4A\)\(U_V=U_3=U_{23}=I_{23}\cdot R_{23}=I_m\cdot R_{23}=0,4\cdot6=2,4V\)Câu trả lời: \(R_1nt\left(R_2//R_3\right)\)\(R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{12\cdot12}{12+12}=6\Omega\)\(R_{tđ}=R_1+R_{23}=24+6=30\Omega\)\(I_m=\dfrac{U_{AB}}{R}=\dfrac{12}{30}=0,4A\)\(\Rightarrow I_A=I_m=0,4A\)\(U_V=U_3=U_{23}=I_{23}\cdot R_{23}=I_m\cdot R_{23}=0,4\cdot6=2,4V\)