Câu 4:

1.

Khi khóa K mở, mạch có: $(R_2\ nt\ R_3)\ //\ R_1$

$R_{tđ}=\dfrac{U}{I}=\dfrac{24}{3}=8\ (\Omega)$

$\dfrac{1}{R_{tđ}}=\dfrac{1}{R_1}+\dfrac{1}{R_{23}}$ 
$\Rightarrow\dfrac{1}{8}=\dfrac{1}{12}+\dfrac{1}{R_{23}}$ 
$\Rightarrow R_{23}=24\ (\Omega)$
$R_{23}=R_2+R_3$
$\Rightarrow 24=9+R_3$
$\Rightarrow R_3=15\ (\Omega)$

2.

Khi khóa K đóng, mạch có: $[(R_4\ //\ R_3)\ nt\ R_2]\ //\ R_1$

$R_{34}=\dfrac{R_3R_4}{R_3+R_4}=\dfrac{6.6}{6+6}=3\ (\Omega)$

$R_{234}=R_{34}+R_2=3+9=12\ (\Omega)$

$R_{tđ}=\dfrac{R_{234}R_1}{R_{234}+R_1}=\dfrac{12.12}{12+12}=6\ (\Omega)$

$I_A=I=\dfrac{U}{R_{tđ}}=\dfrac{24}{6}=4\ (A)$

$I_1=\dfrac{U_1}{R_1}=\dfrac{U}{R_1}=\dfrac{24}{12}=2\ (A)$

$I_{234}=I-I_1=4-2=2\ (A)=I_2=I_{34}$

$U_{34}=I_{34}.R_{34}=2.3=6\ (V)=U_3=U_4$

$I_3=\dfrac{U_3}{R_3}=\dfrac{6}{6}=1\ (A)$

$I_4=\dfrac{U_4}{R_4}=\dfrac{6}{6}=1\ (A)$

3.

Khi khóa K đóng, mạch có: $[(R_4\ //\ R_3)\ nt\ R_2]\ //\ R_1$

Đặt $R_3=x$, khi đó:

$R_{34}=\dfrac{R_3R_4}{R_3+R_4}=\dfrac{6x}{6+x}$
$R_{234}=R_{34}+R_2=\dfrac{6x}{6+x}+9=\dfrac{54+15x}{6+x}$
$R_{tđ}=\dfrac{R_{234}R_1}{R_{234}+R_1}=\dfrac{\dfrac{54+15x}{6+x}.12}{\dfrac{54+15x}{6+x}+12}=\dfrac{72+20x}{14+3x}$

$I=\dfrac{U}{R_{tđ}}=\dfrac{24}{\dfrac{72+20x}{14+3x}}=\dfrac{6(14+3x)}{18+5x}$

$I_1=\dfrac{U}{R_1}=\dfrac{24}{12}=2\ (A)$

$I_{234}=I-I_1=\dfrac{6(14+3x)}{18+5x}-2=\dfrac{8(6+x)}{18+5x}=I_{34}$

$U_{34}=I_{34}.R_{34}=\dfrac{8(6+x)}{18+5x}.\dfrac{6x}{6+x}=\dfrac{48x}{18+5x}=U_3$

$I_3=\dfrac{U_3}{R_3}=\dfrac{\dfrac{48x}{18+5x}}{x}=\dfrac{48}{5x+18}$

`\mathcal{P}_3=I_3^2.R_3=(\frac{48}{5x+18})^2.x=48^2/{(5\sqrtx+18/\sqrtx)^2}`

Để `mathcal{P}_3` lớn nhất thì `(5\sqrtx+18/\sqrtx)^2` nhỏ nhất

Áp dụng bất đẳng thức Cô - si:

`(5\sqrtx+18/\sqrtx)^2\ge 4.(\sqrt{5\sqrt{x}.{18}/\sqrtx})^2=360`

Dấu "=" xảy ra khi và chỉ khi:

`5\sqrtx=18/\sqrtx =>x=3,6\ (\Omega)`

Vậy `\mathcal{P}_{3\ max}=48^2/360=6,4\ (W)` khi `R_3=3,6\ (\Omega)`