$R_{tđ}$ = $\frac{R_{2}.R_{3}}{R_{2}+R_{3}}$ + $R_{1}$⇔$R_{tđ}$ = $\frac{30.60}{30+60}$ + 15⇔$R_{tđ}$ = 35(ôm)
$I_{mc}$ =$\frac{U_{mc}}{R_{tđ}}$⇔$I_{mc}$= $\frac{70}{35}$ =2(A)
Ta có: $I_{mc}$=$I_{1}$=2(A)
⇒$U_{1}$ = $I_{1}$.$R_{1}$⇔$U_{1}$=2.15=30(V)
⇒$U_{23}$=$U_{2}$=$U_{3}$=$U_{mc}$-$U_{1}$⇔$U_{23}$=$U_{2}$=$U_{3}$=70-30=40(V)
⇒$I_{2}$=$\frac{U_{2}}{R_{2}}$ =$\frac{40}{30}$ =1,3(A)
⇒$I_{3}$=$I_{mc}$-$I_{2}$=2-1,3=0,7(A)