Question

Answer 1

a. Khoá K mở:

\(MCD:R_1ntR_2\)

\(\Rightarrow R_{td}=R_1+R_3=9+10=19\Omega\)

\(\Rightarrow I=I_1=I_3=\dfrac{U}{R_{td}}=\dfrac{24}{19}A\)

b. Khoá K đóng:

\(MCD:R_1nt\left(R_2//R_3\right)\)

\(\Rightarrow R_{td}=R_1+R_{23}=9+\dfrac{15\cdot10}{15+10}=15\Omega\)

\(\Rightarrow I=I_1=I_{23}=\dfrac{U}{R_{td}}=\dfrac{24}{15}=1,6A\)

Ta có: \(U_{23}=U_2=U_3=I_{23}\cdot R_{23}=1,6\cdot\left(15-9\right)=9,6V\)

\(\Rightarrow\left\{{}\begin{matrix}I_2=\dfrac{U_2}{R_2}=\dfrac{9,6}{15}=0,64A\\I_3=\dfrac{U_3}{R_3}=\dfrac{9,6}{10}=0,96A\end{matrix}\right.\)