Vật lý

\(MCD:R1nt\left(R2//R3//R4\right)\)

\(\Rightarrow R=R1+R_{234}=12+\left(\dfrac{1}{24}+\dfrac{1}{24}+\dfrac{1}{24}\right)=12+\dfrac{3}{24}=12+8=20\Omega\)

\(\Rightarrow I=I1=I_{234}=\dfrac{U}{R}=\dfrac{30}{20}=\dfrac{3}{2}A\)

\(\Rightarrow U_{234}=U2=U3=U4=I_{234}\cdot R_{234}=\dfrac{3}{2}\cdot8=12V\)

\(\Rightarrow I2=I3=I4=\dfrac{U2}{R2}=\dfrac{12}{24}=0,5A\)

Sơ đồ mạch điện đâu bạn?

Ta có \(R_{12}//R_3\) 

\(\Leftrightarrow I=I_{12}+I_3\\ \Leftrightarrow1,5=I_{12}+1\\ \Rightarrow I_{12}=0,5A\)

mà R1ntR2 ( theo đề bài ) nên 

\(\Leftrightarrow I_1=I_2=0,5A\)

\(\Rightarrow U_3=I_3R_3=1.12=12V\\ \Leftrightarrow U_{12}=U_3=12V\\ \Rightarrow U_1=I_1.R_1=0,5.6=3V\\ \Leftrightarrow U_1+U_2=U_{12}\\ \Leftrightarrow3+U_2=12\Rightarrow U_2=9V\\ \Rightarrow R_2=\dfrac{U_2}{I_2}=\dfrac{9}{0,5}=18\Omega\)

\(MCD:R1//R2\)

\(\Rightarrow R=\dfrac{R1\cdot R2}{R1+R2}=\dfrac{15\cdot10}{15+10}=6\Omega\)

Ta có: \(U=U1=U2=18V\Rightarrow\left[{}\begin{matrix}I1=\dfrac{U1}{R1}=\dfrac{18}{15}=1,2A\\I2=\dfrac{U2}{R2}=\dfrac{18}{10}=1,8A\end{matrix}\right.\)