Câu hỏi 15

Ta có: \(R = 50\Omega ;{Z_L} = 100\pi .\frac{1}{{2\pi }} = 50\Omega ;{U_X} = 120V;U = 120V\)

Ta có giản đồ vecto

Ta có: \({U_{RL}} = 2.OI = 2.120.\sin 15 = 62V \Rightarrow I = \frac{{{U_{RL}}}}{{{Z_{RL}}}} = \frac{{62}}{{50\sqrt 2 }} = 0,878A\)

\(\begin{array}{l}\beta  = \alpha  + \frac{\pi }{4};\cos \beta  = \frac{{OI}}{{{U_{AB}}}} = \frac{{31}}{{120}} \Rightarrow \beta  = {75^0} \Rightarrow \alpha  = {30^0}\\{\varphi _{{\rm{uX}}}} = \frac{\pi }{6} + \alpha  = {60^0} \Rightarrow {P_X} = {U_X}I.\cos {\varphi _{{\rm{uX}}}} = 120.0,878.\cos 60 \approx 53W\end{array}\)