Lời giải:
Khi
\(\begin{array}{l}
C = {C_1}\left( {{Z_{C1}} = 160\Omega } \right) \to {P_{\max }} \to MCH\\
\to \left| \begin{array}{l}
{Z_L} = {Z_{C1}} = 160\Omega \\
{P_{\max }} = \frac{{{U^2}}}{{R + r}} \to R + r = 240\Omega \left( 1 \right)
\end{array} \right.
\end{array}\)
Khi \(C={{C}_{2}}\left( {{Z}_{C2}}=90\Omega \right)\to {{u}_{RC}}\bot {{u}_{d}}\)
\(\tan {{\varphi }_{1}}.\tan {{\varphi }_{2}}=1\to \frac{{{Z}_{C2}}}{R}.\frac{{{Z}_{L}}}{r}=1\to \frac{90}{R}.\frac{160}{r}=1\left( 2 \right)\)
Từ (1) và (2) cho \(R=r=120\Omega \)
\(\frac{{{U}_{d}}}{U}=\frac{\sqrt{{{r}^{2}}+Z_{L}^{2}}}{\sqrt{{{\left( R+r \right)}^{2}}+{{\left( {{Z}_{L}}-{{Z}_{C2}} \right)}^{2}}}}\to \frac{{{U}_{d}}}{150}=\frac{4}{5}\to {{U}_{d}}=120V.\)
