Mỗi ngày một đề thi ĐH môn Lí

[king_wang.bbang]

Câu 35:
$\begin{array}{l}
{P_1} = 3{P_2} \to {I_1}\cos {\varphi _1} = 3\sqrt 3 {I_2}\cos {\varphi _2}\\
\to \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{3\cos {\varphi _2}}}{{\cos {\varphi _1}}} = \sqrt 3
\end{array}$
$\begin{array}{l}
\left| {{\varphi _1}} \right| + \left| {{\varphi _2}} \right| = \dfrac{\pi }{2} \to \dfrac{{3\sin {\varphi _1}}}{{\cos {\varphi _1}}} = 3\tan {\varphi _1} = \sqrt 3 \\
\to {\varphi _1} = \dfrac{\pi }{6} \to {\varphi _2} = \dfrac{\pi }{3}\\
\to B
\end{array}$

 

[king_wang.bbang]

Câu 38:
Khi ${U_{C\max }}$ thì $\overrightarrow {{u_{RL}}} \bot \overrightarrow u $
Suy ra:
${\left( {\dfrac{u}{{{U_0}}}} \right)^2} + {\left( {\dfrac{{{u_{RL}}}}{{{U_{0RL}}}}} \right)^2} = 1$
Đồng thời áp dụng:
$\dfrac{1}{{U_{0R}^2}} = \dfrac{1}{{U_0^2}} + \dfrac{1}{{U_{0LR}^2}}$
Đáp án C

 

[king_wang.bbang]

Câu 45:
$\begin{array}{l}
{Z_{C1}} = 160\Omega \\
{Z_{C2}} = 90\Omega
\end{array}$
$\begin{array}{l}
{I_1} = \dfrac{{{P_{\max }}}}{U} = 0,625A\\
\to R + r = 240\Omega \\
{Z_L} = {Z_{C1}}
\end{array}$
Ta có:
$\begin{array}{l}
\overrightarrow {{U_{RC2}}} \bot \overrightarrow {{U_{rL}}} \to Rr = {Z_L}{Z_{C2}} \to R = r = 120\Omega \\
\to {U_{Lr}} = {I_2}{Z_{rL}} = \dfrac{{150}}{{\sqrt {{{240}^2} + {{\left( {90 - 160} \right)}^2}} }}.200 = 120V\\
\to B
\end{array}$

 

[king_wang.bbang]

Câu 46:
$\begin{array}{l}
{u_{{M_1}}} = 8\cos \dfrac{{\Delta {d_1}}}{\lambda }\cos \left( {\omega t - \dfrac{{\pi \left( {{d_1} + {d_2}} \right)}}{\lambda }} \right)\\
{u_{{M_2}}} = 8\cos \dfrac{{\Delta {d_2}}}{\lambda }\cos \left( {\omega t - \dfrac{{\pi \left( {{d_1}' + {d_2}'} \right)}}{\lambda }} \right)
\end{array}$
Ta có: ${d_1} + {d_2} = {d_1}' + {d_2}'$
$\begin{array}{l}
\to \dfrac{{{u_{{M_1}}}}}{{{u_{{M_2}}}}} = - \sqrt 3 \to {u_{{M_2}}} = - 3\sqrt 3 mm\\
\to D
\end{array}$