Câu hỏi 15

Ta có: \(R = r = \sqrt {\frac{L}{C}}  \Rightarrow {R^2} = {r^2} = {Z_L}{Z_C}\)

Lại có:

\(\eqalign{
& {U_{MB}} = \sqrt 3 {U_{AM}} \Leftrightarrow {r^2} + Z_L^2 = 3\left( {{R^2} + Z_C^2} \right) \Leftrightarrow Z_L^2 - 3Z_C^2 - 2{R^2} = 0 \cr
& \Leftrightarrow Z_L^2 - 3Z_C^2 - 2{Z_L}{Z_C} = 0 \Leftrightarrow Z_L^2 - 3Z_C^2 - 3{Z_L}{Z_C} + {Z_L}{Z_C} = 0 \Leftrightarrow \left( {Z_L^2 + {Z_L}{Z_C}} \right) - \left( {3Z_C^2 + 3{Z_L}{Z_C}} \right) = 0 \cr
& \Leftrightarrow {Z_L}\left( {{Z_L} + {Z_C}} \right) - 3{Z_C}\left( {{Z_L} + {Z_C}} \right) = 0 \Leftrightarrow \left( {{Z_L} - 3{Z_C}} \right)\left( {{Z_L} + {Z_C}} \right) = 0 \cr
& \Leftrightarrow \left[ \matrix{
{Z_L} = 3{Z_C} \hfill \cr
{Z_L} = - {Z_C}(loai) \hfill \cr} \right. \cr}\)

\( \Rightarrow \left\{ \matrix{
{Z_L} = 3{Z_C} \hfill \cr
{Z_L}{Z_C} = {R^2} \hfill \cr} \right. \Rightarrow \left\{ \matrix{
{Z_C} = {R \over {\sqrt 3 }} \hfill \cr
{Z_L} = \sqrt 3 R \hfill \cr} \right. \Rightarrow \cos \varphi = {{R + r} \over {\sqrt {{{\left( {R + r} \right)}^2} + {{\left( {\sqrt 3 R - {R \over {\sqrt 3 }}} \right)}^2}} }} = 0,866\)