Câu hỏi 3

Ta có: 

\(L = C{R^2} = C{r^2} \Rightarrow \left\{ \matrix{
R = r \hfill \cr
{L \over C} = {R^2} \Rightarrow {Z_L}.{Z_C} = {R^2} \hfill \cr} \right.\)

Có:

\(\eqalign{
& {U_{RC}} = \sqrt 3 {Z_d} \Leftrightarrow {Z_{RC}} = \sqrt 3 {Z_d} \Leftrightarrow {R^2} + Z_C^2 = 3\left( {{r^2} + Z_L^2} \right) \Leftrightarrow Z_C^2 = 2{R^2} + 3Z_L^2 \cr
& \Leftrightarrow Z_L^2 + 2.{Z_L}{Z_C} + Z_C^2 = 2Z_C^2 - 2Z_L^2 \Rightarrow {Z_L} = 3{Z_C} \cr} \)

Thay Z_L = 3Z_C vào biểu thức  Z_L.Z_C = R² ta được:

\({Z_C} = {R \over {\sqrt 3 }} \Rightarrow {Z_L} = \sqrt 3 R \Rightarrow Z = \sqrt {{{\left( {2R} \right)}^2} + {{\left( {\sqrt 3  - {1 \over {\sqrt 3 }}} \right)}^2}.{R^2}}  = {{4R} \over {\sqrt 3 }} \Rightarrow c{\rm{os}}\varphi  = {{R + r} \over Z} = 0,866\) 

Chọn D