Câu hỏi 23

U_AM + U_MB max

U_MB = U_C1cos(φ₁ – φ₂)

\(\tan \alpha =\frac{{{Z}_{L}}}{R}=\sqrt{3}\Rightarrow \alpha =\frac{\pi }{3}\)  

\(\frac{U}{\sin \frac{\pi }{6}}=\frac{{{U}_{AM}}}{\sin (\frac{\pi }{2}-{{\varphi }_{2}})}=\frac{{{U}_{MB}}}{\sin ({{\varphi }_{2}}+\frac{\pi }{3})}\Rightarrow \frac{U}{\sin \frac{\pi }{6}}=\frac{{{U}_{AM}}+{{U}_{MB}}}{\sin (\frac{\pi }{2}-{{\varphi }_{2}})+\sin ({{\varphi }_{2}}+\frac{\pi }{3})}=\frac{{{U}_{AM}}+{{U}_{MB}}}{2\sin \frac{\pi }{12}.\sin (2{{\varphi }_{2}}-\frac{\pi }{6})}\)

\(\Rightarrow {{U}_{AM}}+{{U}_{MB}}=\frac{2U\sin \frac{5\pi }{12}}{\sin \frac{\pi }{6}}.\sin (2{{\varphi }_{2}}-\frac{\pi }{6})\Rightarrow {{\varphi }_{2}}=\frac{\pi }{3}\)

u\(\bot \) u_RL nên φ₁ = π/6

\(\text{cos}{{\varphi }_{3}}=c\text{os}{{\varphi }_{1}}.c\text{os}{{\varphi }_{2}}=\frac{\sqrt{3}}{4}\Rightarrow {{Z}_{C}}=\sqrt{13}-3\)

Khi đó: \(R\sqrt{3}={{Z}_{L}}\Rightarrow R=\sqrt{3}\)

Tỉ số R/Z_C = 2,86

Chọn C