Câu hỏi 22

Ta có: \(20\% P = {U_d}I\)  và \(80\% P = {U_t}I\)

\( \Rightarrow \dfrac{{{U_d}}}{{{U_t}}} = \dfrac{1}{4}\)

Lại có: \(U = {U_d} + {U_t}\)

\( \Rightarrow \left\{ \begin{array}{l}{U_d} = \dfrac{U}{5}\\{U_t} = \dfrac{{4U}}{5}\end{array} \right.\)

+ Lúc sau: \({P_{hp}} = \dfrac{{{P_{h{p_1}}}}}{4} = \dfrac{{20\% P}}{4} = 5\% P = {U_d}'.I'\)

\( \Rightarrow 95\%  = {U_t}'.I'\)

\( \Rightarrow \dfrac{{{U_d}'}}{{{U_t}'}} = \dfrac{5}{{95}} = \dfrac{1}{{19}}\)

Mặt khác, \(95\% P = {U_t}'.\dfrac{I}{2}\)

\(\begin{array}{l} \Rightarrow \dfrac{{{U_t}}}{{{U_t}'}} = \dfrac{8}{{19}}\\ \Rightarrow {U_t}' = \dfrac{{19}}{8}{U_t} = \dfrac{{19}}{8}.\dfrac{{4U}}{5} = \dfrac{{19U}}{{10}}\\ \Rightarrow {U_d}' = \dfrac{U}{{10}}\end{array}\)

Lại có:

\(\begin{array}{l}nU = {U_t}' + {U_d}' = \dfrac{{19U}}{{10}} + \dfrac{U}{{10}}\\ \Rightarrow n = 2\end{array}\)

Chọn B