Question

Answer 1

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

Điện trở tương đương mạch AB:

\(R_{AB}=R1+R23=6+\dfrac{4\cdot12}{4+12}=9\Omega\)

Ta có: \(I=I1=I23=\dfrac{U}{R}=\dfrac{12}{9}=\dfrac{4}{3}A\)

\(\rightarrow U23=U2=U3=I23\cdot R23=\dfrac{4}{3}\cdot\left(9-6\right)=4V\)\(\Rightarrow\left[{}\begin{matrix}I2=\dfrac{U2}{R2}=\dfrac{4}{4}=1A\\I3=\dfrac{U3}{R3}=\dfrac{4}{12}=\dfrac{1}{3}A\end{matrix}\right.\)