Câu hỏi 3

Ta có: \({U_{RL}} \bot {U_{RC}} \to \left| {\tan {\varphi _1}\tan {\varphi _2}} \right| = 1 \to \frac{{{Z_L}}}{R}\frac{{{Z_C}}}{R} = 1 \to {U_L}{U_C} = U_R^2\)

\(\eqalign{
& \left\{ \matrix{
U_R^2 + U_L^2 = {400^2}(1) \hfill \cr
U_R^2 + U_C^2 = {300^2}(2) \hfill \cr} \right. \cr
& (1) + (2) \leftrightarrow 2U_R^2 + U_L^2 + U_C^2 = {400^2} + {300^2} \leftrightarrow 2U_R^2 + {\left( {{U_L} + {U_C}} \right)^2} - 2{U_L}{U_C} = {400^2} + {300^2} \cr
& \to \left( {{U_L} + {U_C}} \right) = 500 \cr
& (1) - (2) \leftrightarrow U_L^2 - U_C^2 = {400^2} - {300^2} \leftrightarrow \left( {{U_L} + {U_C}} \right)\left( {{U_L} - {U_C}} \right) = {400^2} - {300^2} \to \left( {{U_L} - {U_C}} \right) = 140 \cr
& \to \left\{ \matrix{
{U_L} = 320 \hfill \cr
{U_C} = 140 \hfill \cr} \right. \to {U_R} = \sqrt {{{400}^2} - {{320}^2}} = 240V \cr} \)

=> Chọn C