khi $C=C_{0}$ thì mạch xảy ra cộng hưởng $Z_{L}=Z_{C}=Z_{C_{0}}$ tức là $Z=R=25\Omega \Rightarrow I=\frac{U}{R}=\frac{120}{25}=4,8\neq 2,4A\Rightarrow $cuộn cảm có r
$Z=R+r=\frac{U}{I}=\frac{120}{2,4}=50\Omega \Rightarrow r=25\Omega $
khi $C=C_{1}=\frac{C_{0}}{2}$
thì $Z_{C_{1}}=2Z_{C_{0}}$
$P_{1}=\frac{P_{0}}{2}\Leftrightarrow I_{1}^{2}(R+r)=\frac{I^{2}(R+r)}{2}\Leftrightarrow I_{1}=\frac{I}{\sqrt{2}}=\frac{6\sqrt{2}}{5}A$
$\Rightarrow Z_{1}=\frac{U}{I_{1}}=\frac{120}{\frac{6\sqrt{2}}{5}}=50\sqrt{2}\Omega $
$\Leftrightarrow (R+r)^{2}+(Z_{L}-Z_{C_{1}})^{2}=5000$
$\Leftrightarrow 50^{2}+Z^{2}_{C_{0}}=5000$
$\Leftrightarrow Z_{C_{0}}=50\Omega \Rightarrow C=\frac{1}{w.Z_{C_{0}}}=\frac{10^{-3}}{5\Pi }$