1) R2//R3
\(\Rightarrow R_{23}=\dfrac{R_2.R_3}{R_2+R_3}=\dfrac{12.6}{12+6}=4\left(\Omega\right)\)
\(R_{tđ}=R_1+R_{23}=8+4=12\left(\Omega\right)\)
2) \(I=I_1=I_{23}=\dfrac{U}{R_{tđ}}=\dfrac{24}{12}=2\left(A\right)\)
\(U_{23}=U_2=U_3=I_{23}.R_{23}=2.4=8\left(V\right)\)
\(\left\{{}\begin{matrix}I_2=\dfrac{U_2}{R_2}=\dfrac{8}{12}=\dfrac{2}{3}\left(A\right)\\I_3=\dfrac{U_3}{R_3}=\dfrac{8}{6}=\dfrac{4}{3}\left(A\right)\end{matrix}\right.\)
Điện trở tương đương: \(R=R1+\left(\dfrac{R2.R3}{R2+R3}\right)=8+\left(\dfrac{12.6}{12+6}\right)=12\Omega\)
\(\left\{{}\begin{matrix}I=U:R=24:12=2V\\I=I1=I23=2V\left(R1ntR23\right)\end{matrix}\right.\)
Hiệu điện thế R23: \(U23=R23.I23=\left(\dfrac{12.6}{12+6}\right).2=8V\)
\(U23=U2=U3=8V\)(R2//R3)
\(\left\{{}\begin{matrix}I2=U2:R2=8:12=\dfrac{2}{3}A\\I3=U3:R3=8:6=\dfrac{4}{3}A\end{matrix}\right.\)
