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Câu hỏi: Cho đoạn mạch điện xoay chiều như hình vẽ: Biết UAB = 30V , f không đổi.
Khi C = C1 thì UAM = 42V, UMB = 54V. Khi C = C2 thì UAM = 2UMB. Tính UMB khi đó.
A. ${{U}_{MB}}=25,2\left( \text{V} \right)$.
B. ${{U}_{MB}}=53,24\left( \text{V} \right)$.
C. ${{U}_{MB}}=23,24\left( \text{V} \right)$.
D. ${{U}_{MB}}=26,24\left( \text{V} \right)$.
A. ${{U}_{MB}}=25,2\left( \text{V} \right)$.
B. ${{U}_{MB}}=53,24\left( \text{V} \right)$.
C. ${{U}_{MB}}=23,24\left( \text{V} \right)$.
D. ${{U}_{MB}}=26,24\left( \text{V} \right)$.
Cách 1: Cách đại số.
*Khi C = C1 ta có: $\left\{ \begin{matrix}
{{30}^{2}}=U_{R}^{2}+{{\left( {{U}_{L}}-42 \right)}^{2}} \\
{{54}^{2}}=U_{R}^{2}+U_{L}^{2}\text{ } \\
\end{matrix}\Rightarrow \left\{ \begin{matrix}
{{U}_{L}}=45\text{ } \\
{{U}_{R}}=9\sqrt{11} \\
\end{matrix}\Rightarrow \dfrac{{{Z}_{L}}}{R}=\dfrac{5}{\sqrt{11}} \right. \right..$
Chuẩn hóa ${{Z}_{L}}=5\Rightarrow R=\sqrt{11}$ (1)
Khi C = C2 thì ${{U}_{C2}}=2{{U}_{RL}}\Rightarrow \dfrac{{{Z}_{C2}}}{\sqrt{{{R}^{2}}+Z_{L}^{2}}}=2\xrightarrow{\left( 1 \right)}{{Z}_{C2}}=12.$
$\begin{aligned}
& \dfrac{{{U}_{RL2}}}{U}=\sqrt{\dfrac{{{R}^{2}}+Z_{L}^{2}}{{{R}^{2}}+{{\left( {{Z}_{L}}-{{Z}_{C2}} \right)}^{2}}}}=\dfrac{\sqrt{15}}{5}. \\
& \Rightarrow {{U}_{RL2}}=\dfrac{U\sqrt{15}}{5}=\dfrac{30\sqrt{15}}{5}=23,24\left( \text{V} \right) \\
\end{aligned}$
Cách 2: Cách dùng đường tròn.
*Đặt ${{\varphi }_{1}}=\angle \left( \text{A}{{\text{M}}_{1}}\text{B} \right)=\angle \left( \text{A}{{\text{M}}_{2}}\text{B} \right)$
$\cos {{\varphi }_{1}}=\dfrac{{{42}^{2}}+{{54}^{2}}-{{30}^{2}}}{2.42.54}=\dfrac{{{x}^{2}}+{{\left( 2x \right)}^{2}}-{{30}^{2}}}{2.x2x}.$
Từ đó tinh được: $x=23,24\left( \text{V} \right)$.
*Khi C = C1 ta có: $\left\{ \begin{matrix}
{{30}^{2}}=U_{R}^{2}+{{\left( {{U}_{L}}-42 \right)}^{2}} \\
{{54}^{2}}=U_{R}^{2}+U_{L}^{2}\text{ } \\
\end{matrix}\Rightarrow \left\{ \begin{matrix}
{{U}_{L}}=45\text{ } \\
{{U}_{R}}=9\sqrt{11} \\
\end{matrix}\Rightarrow \dfrac{{{Z}_{L}}}{R}=\dfrac{5}{\sqrt{11}} \right. \right..$
Chuẩn hóa ${{Z}_{L}}=5\Rightarrow R=\sqrt{11}$ (1)
Khi C = C2 thì ${{U}_{C2}}=2{{U}_{RL}}\Rightarrow \dfrac{{{Z}_{C2}}}{\sqrt{{{R}^{2}}+Z_{L}^{2}}}=2\xrightarrow{\left( 1 \right)}{{Z}_{C2}}=12.$
$\begin{aligned}
& \dfrac{{{U}_{RL2}}}{U}=\sqrt{\dfrac{{{R}^{2}}+Z_{L}^{2}}{{{R}^{2}}+{{\left( {{Z}_{L}}-{{Z}_{C2}} \right)}^{2}}}}=\dfrac{\sqrt{15}}{5}. \\
& \Rightarrow {{U}_{RL2}}=\dfrac{U\sqrt{15}}{5}=\dfrac{30\sqrt{15}}{5}=23,24\left( \text{V} \right) \\
\end{aligned}$
Cách 2: Cách dùng đường tròn.
*Đặt ${{\varphi }_{1}}=\angle \left( \text{A}{{\text{M}}_{1}}\text{B} \right)=\angle \left( \text{A}{{\text{M}}_{2}}\text{B} \right)$
$\cos {{\varphi }_{1}}=\dfrac{{{42}^{2}}+{{54}^{2}}-{{30}^{2}}}{2.42.54}=\dfrac{{{x}^{2}}+{{\left( 2x \right)}^{2}}-{{30}^{2}}}{2.x2x}.$
Từ đó tinh được: $x=23,24\left( \text{V} \right)$.
Đáp án C.