Đáp án:
ZL=210
Giải thích các bước giải:
\({Z_C} = 144\Omega ;{R_1} = 121\Omega ;R{}_2 = 36\Omega ;{\alpha _1} + {\alpha _2} = - \frac{\pi }{2}\)
Ta có:
\(\left\{ \begin{array}{l}
\tan {\alpha _1} = \frac{{\left| {{Z_L} - {Z_C}} \right|}}{{{R_1}}}\\
\tan {\alpha _2} = \frac{{\left| {{Z_L} - {Z_C}} \right|}}{{{R_2}}}
\end{array} \right.\)
ta có:
\(\frac{{\tan {\alpha _1}}}{{\tan {\alpha _2}}} = \frac{{{R_2}}}{{{R_1}}} = \frac{{36}}{{121}}\)
\(\frac{{\tan {\alpha _1}}}{{\tan ( - \frac{\pi }{2} - {\alpha _1})}} = \frac{{36}}{{121}} = > {\alpha _1} = {28^0}36\)
thay vào:
\(\tan {28^0}36 = \frac{{\left| {{Z_L} - 144} \right|}}{{121}} < = > \left| {{Z_L} - 144} \right| = 66 = > \left[ \begin{array}{l}
{Z_L} = 210\Omega \\
{Z_L} = 78\Omega
\end{array} \right.\)
