Question

Answer 1

Bài 42:

\(MCD:R4nt\left[\left(R1ntR3\right)//R2\right]\)

\(\Rightarrow R=R4+R123=6+\dfrac{\left(10+5\right)\cdot10}{10+5+10}=12\Omega\)

\(\Rightarrow I=I4=I123=\dfrac{U}{R}=\dfrac{12}{12}=1A\)

\(\rightarrow U123=U13=U2=I123\cdot R123=1\cdot\left(12-6\right)=6V\Rightarrow\left[{}\begin{matrix}I13=I1=I3=\dfrac{U13}{R13}=\dfrac{6}{10+5}=0,4A\\I2=\dfrac{U2}{R2}=\dfrac{6}{10}=0,6A\end{matrix}\right.\)

Answer 2

Bài 43:

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

\(\Rightarrow U2=I2\cdot R2=0,2\cdot32=6,4V\)

Ta có: \(U2=U3=U4=U234=6,4V\)

\(\Rightarrow I=I1=I234=\dfrac{U234}{R234}=\dfrac{6,4}{8}=0,8A\)

Ta có: \(R=R1+R234=4+8=12\Omega\)

\(\Rightarrow U=IR=0,8\cdot12=9,6V\)