$R_{MN}=16\Omega$

$R_{MC}=x$

$R_{NC}=16-x$

$R_{1}=2\Omega$

$U_{AB}=12V$

a, $R_{MCNC}=\dfrac{R_{MC}.R_{NC}}{R_{MN}}=\dfrac{x(16-x)}{16}$

$R_{tđ}=R_{1}+R_{MCNC}=2+\dfrac{x(16-x)}{16}=\dfrac{x(16-x)+32}{16}$

$⇒I_{1}=I=\dfrac{U}{R_{tđ}}=\dfrac{12}{\dfrac{x(16-x)+32}{16}}=\dfrac{192}{x(16-x)+32}$

b, $P_{MN}=I^{2}.R_{MCNC}=\dfrac{192^{2}}{[x(16-x)+32]^{2}}.\dfrac{x(16-x)}{16}$

$=\dfrac{2304}{[x(16-x)+32]^{2}}.x(16-x)$

Đặt $x(16-x)=y$

$⇒P_{MN}=\dfrac{2304}{(y+32)^{2}}.y$

$=\dfrac{2304}{(\sqrt{y}+\dfrac{32}{\sqrt{y}})^{2}}$

$⇒P_{MN}≥\dfrac{2304}{4.1.32}=18W$

Dấu $=$ xảy ra khi $\sqrt{y}=\dfrac{32}{\sqrt{y}}$

$⇔y=32$

$⇒x(16-x)=32$

$⇔x^{2}-16x+32=0$

$⇔x≈13,66\Omega$

hoặc $x≈2,34\Omega$