Lời giải:
Cách 1:
+ Áp dụng công thức: ${\left( {\frac{{{U_2}}}{{{U_1}}}} \right)^2} = \frac{{(1 - {H_1}){H_1}}}{{(1 - {H_2}){H_2}}}.\left[ {\frac{{1 + {{({H_2}.\tan {\phi _{tt}})}^2}}}{{1 + {{({H_1}.\tan {\phi _{tt}})}^2}}}} \right]$ = $\frac{{(1 - 0,8)0,8}}{{(1 - 0,9)0,9}}.\left[ {\frac{{1 + {{(0,9.0,75)}^2}}}{{1 + {{(0,8.0,75)}^2}}}} \right] = \frac{{137}}{{72}}$
=> $\frac{{{U_2}}}{{{U_1}}} \approx 1,3794$. => Chọn B.
Cách 2: ( do Pt không đổi ) $H = \frac{{{P_t}}}{P} = 1 - \frac{{\Delta P}}{P}$
$\frac{{{H_1}}}{{{H_2}}} = \frac{{{P_2}}}{{{P_1}}} = \frac{{{U_2}{I_2}c{\rm{os}}{\varphi _{\rm{2}}}}}{{{U_1}{I_1}c{\rm{os}}{\varphi _{\rm{1}}}}}$; $\frac{{1 - {H_1}}}{{1 - {H_2}}} = \left( {\frac{{RI_1^2}}{{{U_1}{I_1}c{\rm{os}}{\varphi _{\rm{1}}}}}} \right)\left( {\frac{{{U_2}{I_2}c{\rm{os}}{\varphi _{\rm{2}}}}}{{RI_2^2}}} \right) = \frac{{{U_2}{I_1}c{\rm{os}}{\varphi _{\rm{2}}}}}{{{{\rm{U}}_{\rm{1}}}{I_2}c{\rm{os}}{\varphi _{\rm{1}}}}}$
$\frac{{{H_1}}}{{{H_2}}}.\frac{{1 - {H_1}}}{{1 - {H_2}}} = \frac{{U_2^2c{\rm{o}}{{\rm{s}}^{\rm{2}}}{\varphi _2}}}{{U_1^2c{\rm{o}}{{\rm{s}}^{\rm{2}}}{\varphi _1}}} = > {\left( {\frac{{{U_2}}}{{{U_1}}}} \right)^2} = \frac{{{H_1}(1 - {H_1})}}{{{H_2}(1 - {H_2})}}\frac{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}{\varphi _1}}}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}{\varphi _2}}}$;Ta có Ut1=5Ud1 và Ut2=$\frac{{45}}{4}{U_{d2}} = 11,25{U_{d2}}$
$\tan {\varphi _1} = \frac{{{U_{t1}}\sin {\varphi _{t1}}}}{{{U_{t1}}c{\rm{os}}{\varphi _{t1}} + {U_{d1}}}} = \frac{{5.0,6}}{{5.0,8 + 1}} = 0,6 = > c{\rm{os}}{\varphi _{\rm{1}}} = 0,857$
$\tan {\varphi _2} = \frac{{{U_{t2}}\sin {\varphi _{t2}}}}{{{U_{t2}}c{\rm{os}}{\varphi _{t2}} + {U_{d2}}}} = \frac{{11,25.0,6}}{{11,25.0,8 + 1}} = 0,675 = > c{\rm{os}}{\varphi _{\rm{2}}} = 0,82884$ =>$\frac{{{U_2}}}{{{U_1}}} = 1,378367511 \approx 1,38$