các bạn giúp mình với

   $\textit{Tóm tắt:}$

$R_{x_1}=1~\Omega$

$R_{x_2}=4~\Omega$

$\mathscr{P}_x=25W$

$\underline{\mathscr{P_1}=225W\hspace{40pt}}$

$U=~?~V$

$R=R_1=R_2=~?~\Omega$

$r=R_{Đ_1}=R_{Đ_2}=~?~\Omega$

                        $\textit{Giải:}$

Đặt $R_1=R_2=R$

Gọi $r~(\Omega)$ là điện trở của đèn $Đ_1$ và $Đ_2$.

Mđtđ: $Đ_1$ // $(R_1$ nt $(Đ_2$ // $R_x)$ nt $R_2)$

$*$ Tại $R_x=1~\Omega$:

$U_x=U_{Đ_2}=\sqrt{\mathscr{P}_xR_{x_1}}=\sqrt{25\cdot1}=5~(V)$

$I_{2x}=\dfrac{U_x}{R_{2x}}=\dfrac{5}{\dfrac{rR_{x_1}}{r+R_{x_1}}}=\dfrac{5r+5}{r}~(A)$

$U=I_{2x}R_{12x2}=\dfrac{5r+5}{r}\cdot\left(2R+\dfrac{r}{r+1}\right)=\dfrac{5r+5}{r}\cdot\dfrac{2Rr+2R+r}{r+1}=\dfrac{10Rr+10R+5r}{r}~(V)~~^{1)}$

Tại $R_x=4~\Omega$:

$U_x'=U_{Đ_2}'=\sqrt{\mathscr{P}_xR_{x_2}}=\sqrt{25\cdot4}=10~(V)$

$I_{2x}'=\dfrac{U_x'}{R_{2x}'}=\dfrac{10}{\dfrac{rR_{x_2}}{r+R_{x_2}}}=\dfrac{10r+4}{4r}~(A)$

$U=I_{2x}'R_{12x2}'=\dfrac{10r+4}{4r}\cdot\left(2R+\dfrac{4r}{r+4}\right)=\dfrac{10r+4}{4r}\cdot\dfrac{2Rr+8R+4r}{r+4}=\dfrac{5Rr+20R+10r}{r}~(V)~~^{2)}$

Từ $^{1)}$ và $^{2)}$, suy ra:

      $\dfrac{10Rr+10R+5r}{r}=\dfrac{5Rr+20R+10r}{r}$

$\Leftrightarrow10Rr+10R+5r=5Rr+20R+10r$

$\Leftrightarrow Rr-2R=r$

$\Leftrightarrow R=\dfrac{r}{r-2}~\left(R>0\Rightarrow r>2\right)$

Lại có: $U=\sqrt{\mathscr{P_1}r}=15\sqrt r$

$\Rightarrow 15\sqrt{r}=\dfrac{10Rr+10R+5r}{r}$

$\Leftrightarrow 15\sqrt{r}=\dfrac{10\cdot\dfrac{r}{r-2}\cdot r+10\cdot\dfrac{r}{r-2}+5r}{r}$

$\Leftrightarrow 15\sqrt{r}=\dfrac{10r}{r-2}+\dfrac{10}{r-2}+5$

$\Leftrightarrow 15\sqrt{r}=\dfrac{15r}{r-2}$

$\Leftrightarrow\dfrac{\sqrt r}{r-2}=1$

$\Rightarrow r=(r-2)^2$

$\Leftrightarrow r=r^2-4r+4$

$\Leftrightarrow(r-1)(r-4)=0$

$\Leftrightarrow\left[\begin{array}{l}r=4~(TM)\\r=1~\rm(Loại)\end{array}\\\right.$

$\Rightarrow R=\dfrac{r}{r-2}=\dfrac{4}{4-2}=2~(\Omega)$

$U=15\sqrt{r}=15\sqrt{4}=30~(V)$

Vậy $U=30V$.

        $R_1=R_2=2~\Omega$.

        điện trở của đèn là $4~\Omega$.