Câu hỏi 49

Suất điện động xoay chiều:  

\(\left\{ \begin{array}{l}
{e_1} = {E_0}.\cos \left( {\omega t} \right)\\
{e_2} = {E_0}.\cos \left( {\omega t + \frac{{2\pi }}{3}} \right)\\
{e_3} = {E_0}.\cos \left( {\omega t - \frac{{2\pi }}{3}} \right)
\end{array} \right.\)

Tích:

\(\begin{array}{l}
{e_2}.{e_3} = E_0^2.\frac{1}{2}.\left[ {\cos \left( {2\omega t} \right) + \cos \left( {\frac{{4\pi }}{3}} \right)} \right] = E_0^2.\frac{1}{2}.\left[ {2{{\cos }^2}\omega t - 1 - \frac{1}{2}} \right]\\
\Leftrightarrow {e_2}.{e_3} = E_0^2.{\cos ^2}\omega t - \frac{3}{4}E_0^2 = {\left( {{E_0}.\cos \omega t} \right)^2} - \frac{3}{4}E_0^2 = - 1275\\
\Leftrightarrow {\left( {35} \right)^2} - \frac{3}{4}E_0^2 = - 1275 \Rightarrow {E_0} = \frac{{100}}{{\sqrt 3 }}V \approx 57,{7_{}}V
\end{array}\)

Chọn A.