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Để UAM & UNB lệch nhau một góc $\frac{\pi}{2}$
$\Rightarrow \tan \varphi_{AM}.\tan \varphi_{NB}=1\\\Leftrightarrow\frac{Z_{L}}{R}.\frac{Z_{C}}{R}=1\\\Leftrightarrow Z_{C}=\frac{R^{2}}{Z_{L}}=\frac{100}{\sqrt{3}}(\Omega)$
$\Rightarrow C=\frac{1}{\omega.Z_{C}}=\frac{\sqrt{3}}{10000\pi}(C)\approx 5,51.10^{-5}(C)$
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