Đáp án:
\(u = {U_0}.co{\rm{s(}}\omega {\rm{t - }}\frac{\pi }{{12}})\)
Giải thích các bước giải:R,C:
\({i_1} = {I_0}.co{\rm{s(}}\omega {\rm{t + }}\frac{\pi }{6})\)
R,L,C:
\({i_2} = {I_0}.co{\rm{s(}}\omega {\rm{t - }}\frac{\pi }{3})\)
Ta có:
\(\left\{ \begin{array}{l}
{Z_1} = {Z_2}\\
\overrightarrow {{I_1}} \bot \overrightarrow {{I_2}}
\end{array} \right. < = > \left\{ \begin{array}{l}
\sqrt {{R^2} + Z_C^2} = \sqrt {{R^2} + {{({Z_L} - {Z_C})}^2}} \\
\frac{{{Z_C}}}{R}.\frac{{({Z_L} - {Z_C})}}{R} = 1
\end{array} \right. < = > \left\{ \begin{array}{l}
{Z_L} = {Z_C}\\
R = 2{{\rm{Z}}_C}
\end{array} \right.\)
\(\tan ({\alpha _u} - {\alpha _{i1}}) = \frac{{ - {Z_C}}}{R} = - 1 = > {\alpha _u} = {\alpha _{i1}} - \frac{\pi }{4} = - \frac{\pi }{{12}}\)
BIỂU THỨC:
\(u = {U_0}.co{\rm{s(}}\omega {\rm{t - }}\frac{\pi }{{12}})\)