\(\tan (\varphi -{{\varphi }_{{{R}_{2}}C}})=\frac{\tan \varphi -\tan {{\varphi }_{{{R}_{2}}C}}}{1+\tan \varphi \tan {{\varphi }_{{{R}_{2}}C}}}=\frac{\frac{{{Z}_{C}}}{{{R}_{2}}}-\frac{{{Z}_{C}}}{{{R}_{1}}+{{R}_{2}}}}{1+\frac{Z_{C}^{2}}{{{R}_{2}}({{R}_{1}}+{{R}_{2}})}}=\frac{\frac{1}{{{R}_{2}}}-\frac{1}{{{R}_{1}}+{{R}_{2}}}}{\frac{1}{{{Z}_{C}}}+\frac{{{Z}_{C}}}{{{R}_{2}}({{R}_{1}}+{{R}_{2}})}}\)

Đặt \(y=\frac{1}{{{Z}_{C}}}+\frac{{{Z}_{C}}}{{{R}_{2}}({{R}_{1}}+{{R}_{2}})}\)

Để \(\tan (\varphi -{{\varphi }_{{{R}_{2}}C}})\)lớn nhất thì y phải nhỏ nhất, theo bất đẳng thức Cosi y nhỏ nhất khi

\({{Z}_{C}}=\sqrt{{{R}_{2}}({{R}_{1}}+{{R}_{2}})}=\sqrt{25\sqrt{3}(50\sqrt{3}+25\sqrt{3})}=75\Omega \)

*Chứng minh: \(y=\frac{1}{{{Z}_{C}}}+\frac{{{Z}_{C}}}{{{R}_{2}}({{R}_{1}}+{{R}_{2}})}\ge 2\sqrt{\frac{1}{{{R}_{2}}({{R}_{1}}+{{R}_{2}})}}\)

y nhỏ nhất khi dấu “=” xảy ra nghĩa là:

\(\frac{1}{{{Z}_{C}}}+\frac{{{Z}_{C}}}{{{R}_{2}}({{R}_{1}}+{{R}_{2}})}=2\sqrt{\frac{1}{{{R}_{2}}({{R}_{1}}+{{R}_{2}})}}\)

Bình phương 2 vế ta được:

\(\begin{align} & \frac{1}{Z_{C}^{2}}+\frac{2}{{{R}_{2}}({{R}_{1}}+{{R}_{2}})}+\frac{Z_{C}^{2}}{R_{2}^{2}{{({{R}_{1}}+{{R}_{2}})}^{2}}}=\frac{4}{{{R}_{2}}({{R}_{1}}+{{R}_{2}})} \\ & \Leftrightarrow \frac{1}{Z_{C}^{2}}-\frac{2}{{{R}_{2}}({{R}_{1}}+{{R}_{2}})}+\frac{Z_{C}^{2}}{R_{2}^{2}{{({{R}_{1}}+{{R}_{2}})}^{2}}}=0 \\ & \Leftrightarrow {{(\frac{1}{{{Z}_{C}}}-\frac{{{Z}_{C}}}{{{R}_{2}}({{R}_{1}}+{{R}_{2}})})}^{2}}=0\Leftrightarrow \frac{1}{{{Z}_{C}}}=\frac{{{Z}_{C}}}{{{R}_{2}}({{R}_{1}}+{{R}_{2}})} \\ & \Leftrightarrow Z_{C}^{2}={{R}_{2}}({{R}_{1}}+{{R}_{2}})\Leftrightarrow {{Z}_{C}}=\sqrt{{{R}_{2}}({{R}_{1}}+{{R}_{2}})} \\ \end{align}\)

→ Chọn C