Câu hỏi của Quỳnh Phương Nguyễn

a)\(R_1//R_2\Rightarrow R_{12}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{30\cdot40}{30+40}=\dfrac{120}{7}\Omega\)\(I=\dfrac{U}{R}=\dfrac{15}{\dfrac{120}{7}}=0,875A\)b)\(R_3ntR_{12}\Rightarrow I_3=I_{12}=0,875A\)\(U_3=I_3\cdot R_3=0,875\cdot10=8,75\)Câu trả lời: a)\(R_1//R_2\Rightarrow R_{12}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{30\cdot40}{30+40}=\dfrac{120}{7}\Omega\)\(I=\dfrac{U}{R}=\dfrac{15}{\dfrac{120}{7}}=0,875A\)b)\(R_3ntR_{12}\Rightarrow I_3=I_{12}=0,875A\)\(U_3=I_3\cdot R_3=0,875\cdot10=8,75\)