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Câu hỏi: Đặt điện áp xoay chiều $u=U\sqrt{2}\cos \omega t\left( V \right)$ ( $U$ không đổi, còn $\omega $ thay đổi được) vào mạch nối tiếp RLC biết $C{{R}^{2}}<2L$. Điều chỉnh giá trị $\omega $ để ${{U}_{C\max }}$ khi đó ${{U}_{C\max }}=90 V$ và ${{U}_{RL}}=30\sqrt{5}V$. Giá trị của $U$ là:
A. $60 V.$
B. $80 V.$
C. $60\sqrt{2} V.$
D. $24\sqrt{10} V.$
A. $60 V.$
B. $80 V.$
C. $60\sqrt{2} V.$
D. $24\sqrt{10} V.$
+ Ta có:
${{U}_{C}}={{U}_{C\max }}$ khi $w=\dfrac{1}{L}\sqrt{\dfrac{L}{C}-\dfrac{{{R}^{2}}}{2}}$ (1) và ${{U}_{C\max }}=\dfrac{2UL}{R\sqrt{4LC-{{R}^{2}}{{C}^{2}}}}$ (*)
Khi đó: ${{Z}_{L}}=wL=\sqrt{\dfrac{L}{C}-\dfrac{{{R}^{2}}}{2}};$ ${{Z}_{C}}=\dfrac{1}{\omega C}=\dfrac{L}{C}\dfrac{1}{\sqrt{\dfrac{L}{C}-\dfrac{{{R}^{2}}}{2}}}$
Ta lại có:
${{U}_{RL}}=\dfrac{U\sqrt{{{R}^{2}}+Z_{L}^{2}}}{\sqrt{{{R}^{2}}+{{\left( {{Z}_{L}}-{{Z}_{C}} \right)}^{2}}}}$ và ${{U}_{C\max }}=\dfrac{U{{Z}_{C}}}{\sqrt{{{R}^{2}}+{{\left( {{Z}_{L}}-{{Z}_{C}} \right)}^{2}}}}$
$\Rightarrow \dfrac{{{U}_{RL}}}{{{U}_{C\max }}}=\dfrac{\sqrt{{{R}^{2}}+Z_{L}^{2}}}{{{Z}_{C}}}=\dfrac{\sqrt{5}}{3}$
$\Rightarrow 9\left( {{R}^{2}}+Z_{L}^{2} \right)=5Z_{C}^{2}\Rightarrow 9\left( {{R}^{2}}+\dfrac{L}{C}-\dfrac{{{R}^{2}}}{2} \right)=5\dfrac{{{L}^{2}}}{{{C}^{2}}\left( \dfrac{L}{C}-\dfrac{{{R}^{2}}}{2} \right)}=5$
$\Rightarrow 9\left( \dfrac{{{R}^{2}}}{2}+\dfrac{L}{C} \right)=5\dfrac{{{L}^{2}}}{{{C}^{2}}\left( \dfrac{L}{C}-\dfrac{{{R}^{2}}}{2} \right)}$
$\Rightarrow 9{{C}^{2}}\left( \dfrac{{{L}^{2}}}{{{C}^{2}}}-\dfrac{{{R}^{4}}}{4} \right)=5{{L}^{2}}$
$\Rightarrow 4{{L}^{2}}=\dfrac{9{{R}^{4}}{{C}^{2}}}{4}\Rightarrow 4L=3{{R}^{2}}C$ (**)
Thay vào ${{U}_{C\max }}:$
${{U}_{C\max }}=\dfrac{2UL}{R\sqrt{4LC-{{R}^{2}}{{C}^{2}}}}=\dfrac{2UL}{R\sqrt{C\left( 4L-{{R}^{2}}C \right)}}=\dfrac{2UL}{R\sqrt{C.2{{R}^{2}}C)}}$
$=\dfrac{2U}{\sqrt{2}}\dfrac{L}{{{R}^{2}}C}=\dfrac{2U}{\sqrt{2}}\dfrac{3}{4}=\dfrac{3U}{2\sqrt{2}}=90 V$
$\Rightarrow U=60\sqrt{2} V.$
${{U}_{C}}={{U}_{C\max }}$ khi $w=\dfrac{1}{L}\sqrt{\dfrac{L}{C}-\dfrac{{{R}^{2}}}{2}}$ (1) và ${{U}_{C\max }}=\dfrac{2UL}{R\sqrt{4LC-{{R}^{2}}{{C}^{2}}}}$ (*)
Khi đó: ${{Z}_{L}}=wL=\sqrt{\dfrac{L}{C}-\dfrac{{{R}^{2}}}{2}};$ ${{Z}_{C}}=\dfrac{1}{\omega C}=\dfrac{L}{C}\dfrac{1}{\sqrt{\dfrac{L}{C}-\dfrac{{{R}^{2}}}{2}}}$
Ta lại có:
${{U}_{RL}}=\dfrac{U\sqrt{{{R}^{2}}+Z_{L}^{2}}}{\sqrt{{{R}^{2}}+{{\left( {{Z}_{L}}-{{Z}_{C}} \right)}^{2}}}}$ và ${{U}_{C\max }}=\dfrac{U{{Z}_{C}}}{\sqrt{{{R}^{2}}+{{\left( {{Z}_{L}}-{{Z}_{C}} \right)}^{2}}}}$
$\Rightarrow \dfrac{{{U}_{RL}}}{{{U}_{C\max }}}=\dfrac{\sqrt{{{R}^{2}}+Z_{L}^{2}}}{{{Z}_{C}}}=\dfrac{\sqrt{5}}{3}$
$\Rightarrow 9\left( {{R}^{2}}+Z_{L}^{2} \right)=5Z_{C}^{2}\Rightarrow 9\left( {{R}^{2}}+\dfrac{L}{C}-\dfrac{{{R}^{2}}}{2} \right)=5\dfrac{{{L}^{2}}}{{{C}^{2}}\left( \dfrac{L}{C}-\dfrac{{{R}^{2}}}{2} \right)}=5$
$\Rightarrow 9\left( \dfrac{{{R}^{2}}}{2}+\dfrac{L}{C} \right)=5\dfrac{{{L}^{2}}}{{{C}^{2}}\left( \dfrac{L}{C}-\dfrac{{{R}^{2}}}{2} \right)}$
$\Rightarrow 9{{C}^{2}}\left( \dfrac{{{L}^{2}}}{{{C}^{2}}}-\dfrac{{{R}^{4}}}{4} \right)=5{{L}^{2}}$
$\Rightarrow 4{{L}^{2}}=\dfrac{9{{R}^{4}}{{C}^{2}}}{4}\Rightarrow 4L=3{{R}^{2}}C$ (**)
Thay vào ${{U}_{C\max }}:$
${{U}_{C\max }}=\dfrac{2UL}{R\sqrt{4LC-{{R}^{2}}{{C}^{2}}}}=\dfrac{2UL}{R\sqrt{C\left( 4L-{{R}^{2}}C \right)}}=\dfrac{2UL}{R\sqrt{C.2{{R}^{2}}C)}}$
$=\dfrac{2U}{\sqrt{2}}\dfrac{L}{{{R}^{2}}C}=\dfrac{2U}{\sqrt{2}}\dfrac{3}{4}=\dfrac{3U}{2\sqrt{2}}=90 V$
$\Rightarrow U=60\sqrt{2} V.$
Đáp án C.