Cho mạch điện (Hình 2): các điện trở R1 = R2 = R3

* Hướng dẫn giải

1. Ta có:$\left\{\begin{array}{l}{I}_A=\frac{U}{R_1+R_D}=\frac{U}{4R}=1A\\ P=\frac{3U^2}{4R}=36W\end{array}\right.\Rightarrow \left\{\begin{array}{l}R=3\Omega \\ U=12V\end{array}\right.$

$P_2=8W\Rightarrow \left\{\begin{array}{l}{I}_2=\sqrt{\frac{P_2}{R_2}}=\sqrt{\frac{8}{3}}A\\ U_{CB}=\sqrt{P_2{R}_2}=2\sqrt{6}V\end{array}\right.\Rightarrow I_1=\frac{U_{AC}}{R_1}=\frac{U-U_{CB}}{R_1}=4-\sqrt{\frac{8}{3}}V$

Ta có : $\left\{\begin{array}{l}{I}_X=I_1-I_2=I_3-I_D\\ U=R_D{I}_D+R_3{I}_3\end{array}\right.\iff \left\{\begin{array}{l}{I}_3-I_D=4-\sqrt{\frac{32}{3}}\\ 4=3I_D+I_3\end{array}\right.$

$I_D=\sqrt{\frac{2}{3}}A\Rightarrow P_D=R_D.I_D^2=6W$

2b.

Ta có: $\left\{\begin{array}{l}{I}_2=I_1-I_X\\ U_{AB}=U=R_1{I}_1+R_2{I}_2\end{array}\right.\Rightarrow U=R(2I_1-I_X)\left(1\right)$

Và: $\left\{\begin{array}{l}{I}_3=I_X+I_D\\ U=R_D{I}_D+R_3{I}_3\end{array}\right.\Rightarrow U=R(4I_D+I_X)\left(2\right)$

$\stackrel{\left(1\right),\left(2\right)}{\to }{I}_1=2I_D+I_X\left(3\right)$

 Lại có:

$U_{AD}=R_D{I}_D=R_1{I}_1+R_X{I}_X\left(4\right)$

$$\begin{gathered} \stackrel{\left(3\right);\left(4\right)}{\to }R{I}_D=(R+R_X)I_X\left(5\right) \\ \stackrel{\left(2\right);\left(5\right)}{\to }{I}_X=\frac{\text{U}}{5R+4R_X}\Rightarrow P_X=R_X{I}_X^2=\frac{{\text{U}}^2{R}_X}{{\left(5R+4R_X\right)}^2}\Rightarrow P_{\max }=\frac{U^2}{80R}=0,6W\iff R_X=\frac{5}{4}R=3,75\Omega \end{gathered}$$

 2c. $I_D=\frac{U}{R\left(4+\frac{R}{R+R_X}\right)}$ => R_X tăng => I_D tăng hay độ sáng tăng.