Cho R,L,C mac noi tiep. Mac vao mach xoay chieu co U dang: u=Uo cos(wt). Khi C=C1 thi P mach max = 400. Khi C=C2 thi he so cong suat = (can 3) / 2. P mach khi do la?
~O) [tex]C = C_{1}[/tex]: [tex]P_{1} = RI^{2} = R \frac{U_{AB}^{2}}{R^{2}+(Z_{L}-Z_{C_{1}})}[/tex]
Ta thấy: [tex]P_{1}_{max}\Leftrightarrow Z_{L}= Z_{C_{1}}[/tex]: cộng hưởng điện
Công suất cực đại: [tex]P_{1}_{max}= \frac{U_{AB}^{2}}{R} = 400W[/tex]
~O) [tex]C = C_{2}[/tex]: [tex]cos\varphi = \frac{R}{\sqrt{\left(R^{2}+(Z_{L}-Z_{C_{2}})^{2} \right)}}= \frac{\sqrt{3}}{2} \Rightarrow (Z_{L}-Z_{C_{2}})^{2} = \frac{R^{2}}{3}[/tex]
Tổng trở:
[tex]Z_{AB}= \sqrt{(R^{2} + (Z_{L}-Z_{C_{2}})^{2}} = \sqrt{R^{2}+ \frac{R^{2}}{3}} = \frac{2}{\sqrt{3}}R[/tex]
Công suất lúc sau:
[tex]P_{2}= U_{AB}.I_{2}.cos\varphi = \frac{U_{AB}^{2}}{Z_{AB}}. \frac{\sqrt{3}}{2} = \frac{U_{AB}^{2}}{\frac{2}{\sqrt{3}}R}. \frac{\sqrt{3}}{2} = \frac{3}{4}.\frac{U_{AB}^{2}}{R} = \frac{3}{4}. 400 = 300W[/tex]
