Câu hỏi 44

Đáp án C

\(\varphi  = {\tan ^{ - 1}}\left( {\frac{{{Z_L} - {Z_C}}}{r}} \right) - {\tan ^{ - 1}}\left( {\frac{{{Z_L} - {Z_C}}}{{R + r}}} \right)\)(1);

theo bài ta có \({\tan ^{ - 1}}\left( {\frac{{{Z_L} - {Z_C}}}{r}} \right) - {\tan ^{ - 1}}\left( {\frac{{{Z_L} - {Z_C}}}{{R + r}}} \right) = {\tan ^{ - 1}}\left( {\frac{{{Z_L} - 2{Z_C}}}{r}} \right) - {\tan ^{ - 1}}\left( {\frac{{{Z_L} - 2{Z_C}}}{{R + r}}} \right)\)

\({\tan ^{ - 1}}\left( {\frac{{7r - {Z_C}}}{r}} \right) - {\tan ^{ - 1}}\left( {\frac{{7r - {Z_C}}}{{4r}}} \right) = {\tan ^{ - 1}}\left( {\frac{{7r - 2{Z_C}}}{r}} \right) - {\tan ^{ - 1}}\left( {\frac{{7r - 2{Z_C}}}{{4r}}} \right)\)=>Z_C=3r; thay vào (1) ta tìm được \(\varphi  = 0,54rad\)