Câu hỏi 4

Theo các dữ kiện bài cho ta có:

 \(\left\{ \begin{array}{l}\dfrac{{{U_1}}}{{200}} = \dfrac{{{N_1}}}{{{N_2}}}\,\,\,\left( 1 \right)\\\dfrac{{{U_1}}}{U} = \dfrac{{{N_1} - n}}{{{N_2}}}\,\,\,\left( 2 \right)\\\dfrac{{{U_1}}}{{0,5U}} = \dfrac{{{N_1} + n}}{{{N_2}}}\,\,\,\left( 3 \right)\end{array} \right.\)

Lấy \(\left\{ \begin{array}{l}\dfrac{{\left( 1 \right)}}{{\left( 2 \right)}} \Leftrightarrow \dfrac{U}{{200}} = \dfrac{{{N_1}}}{{{N_1} - n}}\,\,\,\\\dfrac{{\left( 2 \right)}}{{\left( 3 \right)}} \Leftrightarrow 0,5 = \dfrac{{{N_1} - n}}{{{N_1} + n}} \Rightarrow {N_1} = 3n\,\,\,\end{array} \right.\)

\( \Rightarrow \dfrac{U}{{200}} = \dfrac{{3n}}{{3n - n}} \Rightarrow U = 300V\)

Chọn D.