Câu hỏi của Trần Duy Sang

\(R_{tđ}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{20\cdot30}{20+30}=12\Omega\)\(R_1//R_2\Rightarrow U_1=U_2=U_m=36V\)\(I_1=\dfrac{U_1}{R_1}=\dfrac{36}{20}=1,8A\)\(I_2=\dfrac{U_2}{R_2}=\dfrac{36}{30}=1,2A\)Câu trả lời: \(R_{tđ}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{20\cdot30}{20+30}=12\Omega\)\(R_1//R_2\Rightarrow U_1=U_2=U_m=36V\)\(I_1=\dfrac{U_1}{R_1}=\dfrac{36}{20}=1,8A\)\(I_2=\dfrac{U_2}{R_2}=\dfrac{36}{30}=1,2A\)