Điện năng được truyền từ trạm phát điện đến nơi tiêu thụ bằng đường dây

* Hướng dẫn giải

Đáp án A

Ta có: $H=\frac{P_{tt}}{P}=\frac{P-\Delta P}{P}\iff \Delta P=P\left(1-H\right)$

$\Rightarrow \frac{\Delta P_2}{\Delta P_1}=\frac{1-H_2}{1-H_1}\stackrel{\Delta P_2=\frac{\Delta P_1}{4}}{\to }\frac{1}{4}=\frac{1-H_2}{1-0,8}\Rightarrow H_2=0,95$

Lại có: $\overrightarrow{U}=\overrightarrow{U_R}+\overrightarrow{U_{tt}}\Rightarrow U^2=U_R^2+U_{tt}^2+2U_{tt}{U}_R\cos \phi$

$\Rightarrow \left\{\begin{array}{l}{U}_1^2=U_{R1}^2+U_{tt1}^2+1,6U_{tt1}{U}_{R1}\left(1\right)\\ U_2^2=U_R^2+U_{tt}^2+1,6U_{tt2}{U}_{R2}\left(2\right)\end{array}\right.$

Mặt khác: $\left\{\begin{array}{l}{h}_1=1-H_1=0,2=\frac{U_{R_1}}{U_{R_1}+0,8U_{tt1}}\Rightarrow U_{tt1}=5U_{R1}\stackrel{\left(1\right)}{\to }{U}_1=\sqrt{34}{U}_{R1}\\ h_2=1-H_2=0,05=\frac{U_{R2}}{U_{R2}+0,8U_{tt2}}\Rightarrow U_{tt2}=23,75U_{R2}\stackrel{\left(2\right)}{\to }{U}_2=\frac{\sqrt{9649}}{4}{U}_{R2}\end{array}\right.$

Suy ra: $\frac{U_2}{U_1}=\frac{\sqrt{9649}}{4\sqrt{34}}\frac{U_{R2}}{U_{R1}}=\frac{\sqrt{9649}}{4\sqrt{34}}\frac{I_{2}R}{I_{1}R}\underset{\frac{I_2}{I_1}=\sqrt{\frac{\Delta P_2}{\Delta P_1}}=\frac{1}{2}}{\overset{\Delta P=I^{2}R}{\to }}n=\frac{U_2}{U_1}=\frac{\sqrt{9649}}{4\sqrt{34}}\frac{1}{2}\approx 2,1$

Chú ý: $\left\{\begin{array}{l}h=\frac{\Delta P}{P}=\frac{\Delta P}{\Delta P+P_{tt}}=\frac{U_{R}I}{U_{R}I+U_{tt}\cos {\phi }_{tt}}\\ h=1-H\end{array}\right.$