Câu hỏi 17

ta có

\(\begin{array}{l}
\frac{{{u^2}}}{{U_0^2}} + \frac{{{i^2}}}{{I_0^2}} = 1\\
= > \frac{{{{40}^2}}}{{U_0^2}} + \frac{{{1^2}}}{{I_0^2}} = \frac{{{{50}^2}}}{{U_0^2}} + \frac{{0,{6^2}}}{{I_0^2}}\\
\Leftrightarrow \frac{{{{50}^2} - {{30}^2}}}{{U_0^2}} = \frac{{{1^2} - 0,{6^2}}}{{I_0^2}} \Leftrightarrow \frac{{{{30}^2}}}{{U_0^2}} = \frac{{0,{8^2}}}{{I_0^2}}\\
= > \frac{{U_0^2}}{{I_0^2}} = {\left( {{Z_C}} \right)^2} = \frac{{{{30}^2}}}{{0,{8^2}}} = > {Z_C} = 37,5\Omega
\end{array}\)