Biểu thức hiệu điện thế giữa hai đầu đoạn mạch MB là:

A.  

uMB=60.cos(100πt5π12)Vu_{M B} = 60 . cos \left( 100 \pi t - \dfrac{5 \pi}{12} \right) V

B.  

uMB=60.cos(100πt+7π12)Vu_{M B} = 60 . cos \left( 100 \pi t + \dfrac{7 \pi}{12} \right) V

C.  

uMB=602.cos(100πt5π12)Vu_{M B} = 60 \sqrt{2} . cos \left( 100 \pi t - \dfrac{5 \pi}{12} \right) V

D.  

uMB=602.cos(100πt+7π12)Vu_{M B} = 60 \sqrt{2} . cos \left( 100 \pi t + \dfrac{7 \pi}{12} \right) V

Đáp án đúng là: B

Sử dụng máy tính Casio:  (u)MB=i.(Z)MBˉ=(I)0(φ)i.((Z)L(Z)C)i\left(\text{u}\right)_{\text{MB}} = \text{i} . \bar{\left(\text{Z}\right)_{\text{MB}}} = \left(\text{I}\right)_{0} \angle \left(\varphi\right)_{i} . \left( \left(\text{Z}\right)_{\text{L}} - \left(\text{Z}\right)_{\text{C}} \right) \text{i}
Giải chi tiết:
(u)MB=i.(Z)MBˉ=(I)0(φ)i.((Z)L(Z)C)i\left(\text{u}\right)_{\text{MB}} = \text{i} . \bar{\left(\text{Z}\right)_{\text{MB}}} = \left(\text{I}\right)_{0} \angle \left(\varphi\right)_{i} . \left( \left(\text{Z}\right)_{\text{L}} - \left(\text{Z}\right)_{\text{C}} \right) \text{i}
uMB=6π12.(2010)i=607π12\Rightarrow u_{M B} = 6 \angle \dfrac{\pi}{12} . \left( 20 - 10 \right) i = 60 \angle \dfrac{7 \pi}{12}
(u)MB=60.cos(100πt+7π12)V\Rightarrow \left(\text{u}\right)_{\text{MB}} = 60 . cos \left( 100 \pi\text{t} + \dfrac{7 \pi}{12} \right) \text{V}


 

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