* Đáp án
* Hướng dẫn giải
Ta có ${P_{\max }} = \frac{{{U^2}}}{R} \to U = \sqrt {{P_{\max }}.R} = 100\sqrt 3 V$.
${P_{th}} = \frac{{{U^2}.R}}{{Z_{th}^2}} = \frac{{{U^2}.R}}{{{{\left( {\sqrt {{R^2} + Z_C^2} } \right)}^2}}} \to 100 = \frac{{{{\left( {100\sqrt 3 } \right)}^2}.100}}{{{{\left( {\sqrt {{{100}^2} + Z_C^2} } \right)}^2}}} \to {Z_C} = 100\sqrt 2 \Omega .$