Sđmđ: $R_{1}//R_{5}//[(R_{3}//R_{2})ntR_{4}]$
a, $R_{23}=\dfrac{R_{2}.R_{3}}{R_{2}+R_{3}}=\dfrac{18.9}{18+9}=6$ ôm
$R_{234}=R_{23}+R_{4}=6+6=12$ ôm
$⇒R=\dfrac{1}{\dfrac{1}{R_{1}}+\dfrac{1}{R_{5}}+\dfrac{1}{R_{234}}}=\dfrac{1}{\dfrac{1}{24}+\dfrac{1}{12}+\dfrac{1}{12}}=4,8$ ôm
b, $I=\dfrac{U}{R}=\dfrac{12}{4,8}=2,5A$
$I_{1}=\dfrac{U}{R_{1}}=\dfrac{12}{24}=0,5A$
$⇒I_{A}=I-I_{1}=2,5-0,5=2A$
c, $R_{234}=\dfrac{r.9}{r+9}+6=\dfrac{15r+54}{r+9}$
$⇒I_{2}+I_{3}=I_{4}=\dfrac{U}{R_{234}}=\dfrac{24(r+9)}{15r+54}=\dfrac{8r+72}{5r+18}$
$⇒U_{4}=I_{4}.R_{4}=\dfrac{8r+72}{5r+18}.6=\dfrac{48r+432}{5r+18}V$
$⇒U_{2}=24-U_{4}=24-\dfrac{48r+432}{5r+18}=\dfrac{72r}{5r+18}V$
$⇒P_{2}=\dfrac{U_{2}}{r²}=\dfrac{72}{r(5r+18)}$
$P_{2}$ lớn nhất $⇒r(5r+18)$ nhỏ nhất
$⇒r(5r+18)=1$
$⇒r=\dfrac{-9+\sqrt{86}}{5}$ ôm
Khi đó $P_{2_{max}}=72W$
