Đặt điện áp \(u={{U}_{0}}\cos 100\pi t(V)\) vào hai đầu đoạn mạch AB

Ta có: \(\varphi ={{\varphi }_{d}}-{{\varphi }_{AB}}\)

\(\tan \varphi =\tan \left( {{\varphi }_{d}}-{{\varphi }_{AB}} \right)=\frac{\tan {{\varphi }_{d}}-\tan {{\varphi }_{AB}}}{1-\tan {{\varphi }_{d}}.\tan {{\varphi }_{AB}}}\)

Lại có: 

\(\begin{array}{l}
\left\{ {\begin{array}{*{20}{l}}
{\tan {\varphi _d} = \frac{{{Z_L}}}{r}}\\
{\tan {\varphi _{AB}} = \frac{{{Z_L}}}{{R + r}}}
\end{array}} \right.\\
 \Rightarrow \tan \varphi  = \frac{{\frac{{{Z_L}}}{r} - \frac{{{Z_L}}}{{R + r}}}}{{1 + \frac{{Z_L^2}}{{r(R + r)}}}} = \frac{{{Z_L}R}}{{r(R + r) + Z_L^2}}\\
 \Rightarrow \tan \varphi  = \frac{R}{{\frac{{r(R + r)}}{{{Z_L}}} + {Z_L}}}
\end{array}\)

Ta có: \(\frac{r(R+r)}{{{Z}_{L}}}+{{Z}_{L}}\ge 2\sqrt{r(R+r)}\Rightarrow \tan \varphi \le \frac{R}{2\sqrt{r(R+r)}}\)

\(\Rightarrow \tan {{\varphi }_{\max }}\) khi \(\frac{r(R+r)}{{{Z}_{L}}}={{Z}_{L}}\) (*) và \(\tan {{\varphi }_{\max }}=0,65\Rightarrow \frac{R}{2\sqrt{50(50+R)}}=0,65\Rightarrow R=119,77\Omega \)

Thay vào (*) ta suy ra: \({{Z}_{L}}=92,13\Omega =\omega {{L}_{0}}\Rightarrow {{L}_{0}}=0,29H\)