Phương pháp giải bài tập mạch xoay chiều RLC - có R thay đổi

1. Mạch RLC có cuộn dây thuần cảm (r=0)

\(P = UIc{\rm{os}}\varphi {\rm{ = }}{{\rm{I}}^2}R = \frac{{{U^2}}}{{{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}}}R = \frac{{{U^2}}}{{R + \frac{{{{\left( {{Z_L} - {Z_C}} \right)}^2}}}{R}}}\)

Để \({P_{max}} \to {\left( {R + \frac{{{{\left( {{Z_L} - {Z_C}} \right)}^2}}}{R}} \right)_{\min }}\)

Ta có: \(R + \frac{{{{\left( {{Z_L} - {Z_C}} \right)}^2}}}{R} \ge 2\sqrt {R\frac{{{{\left( {{Z_L} - {Z_C}} \right)}^2}}}{R}}  = 2\left| {{Z_L} - {Z_C}} \right|\)

Dấu “=” xảy ra \( \leftrightarrow {R^2} = {({Z_L} - {Z_C})^2} \to R = \left| {{Z_L} - {Z_C}} \right|\)

2. Mạch RLC có cuộn dây không thuần cảm (r≠0)

- Công suất trên toàn mạch:

\(P{\rm{ = }}{{\rm{I}}^2}(R + r) = \frac{{{U^2}}}{{{{(R + r)}^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}}}(R + r) = \frac{{{U^2}}}{{R + r + \frac{{{{\left( {{Z_L} - {Z_C}} \right)}^2}}}{{R + r}}}}\)

Để \({P_{max}} \to {\left( {R + r + \frac{{{{\left( {{Z_L} - {Z_C}} \right)}^2}}}{{R + r}}} \right)_{\min }}\)

Ta có: \((R + r) + \frac{{{{\left( {{Z_L} - {Z_C}} \right)}^2}}}{{R + r}} \ge 2\sqrt {(R + r)\frac{{{{\left( {{Z_L} - {Z_C}} \right)}^2}}}{{R + r}}}  = 2\left| {{Z_L} - {Z_C}} \right|\)

Dấu “=” xảy ra \( \leftrightarrow {(R + r)^2} = {({Z_L} - {Z_C})^2} \to R + r = \left| {{Z_L} - {Z_C}} \right| \to R = \left| {{Z_L} - {Z_C}} \right| - r\)

Chú ý: Nếu \(r > {Z_L} - {Z_C} \to {P_{{\rm{max}}}} \leftrightarrow R = 0,{P_{{\rm{max}}}} = \frac{{{U^2}}}{{{r^2} + {{({Z_L} - {Z_C})}^2}}}r\)

- Công suất trên R: \(P{\rm{ = }}\frac{{{U^2}}}{{{{(R + r)}^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}}}R = \frac{{{U^2}}}{{R + 2r + \frac{{{r^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}}}{R}}}\)

\(\begin{array}{l}A = R + 2r + \frac{{{r^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}}}{R}\\{A_{\min }} \leftrightarrow {\left( {R + \frac{{{r^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}}}{R}} \right)_{\min }}\\R + \frac{{{r^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}}}{R} \ge 2\sqrt {R.\frac{{{r^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}}}{R}}  = 2\sqrt {{r^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} \end{array}\)

Dấu “=” xảy ra: \( \leftrightarrow {R^2} = {r^2} + {\left( {{Z_L} - {Z_C}} \right)^2},{P_{{\rm{max}}}} = \frac{{{U^2}}}{{2{\rm{r}} + 2\sqrt {{r^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} }}\)

- Công suất trên r: \({P_r}{\rm{ = }}\frac{{{U^2}r}}{{{{(R + r)}^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}}}\)

\({P_{r{\rm{ }}max}} = \frac{{{U^2}r}}{{{r^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}}}\) xảy ra khi R=0

II- KHI R=R₁ HOẶC R=R₂ THÌ P CÓ CÙNG 1 GIÁ TRỊ (P<P_MAX) (CUỘN DÂY THUẦN CẢM)

\(\begin{array}{l}P = \frac{{{U^2}}}{{{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}}}R\\ \to P({R^2} + {\left( {{Z_L} - {Z_C}} \right)^2}) = {U^2}R\\ \leftrightarrow P{R^2} - {U^2}R + {\left( {{Z_L} - {Z_C}} \right)^2}P = 0\\ \leftrightarrow {R^2} - \frac{{{U^2}R}}{P} + {\left( {{Z_L} - {Z_C}} \right)^2} = 0{\rm{        }}(1)\end{array}\)

PT (1) có 2 nghiệm: R₁, R₂ : \(\left\{ \begin{array}{l}{R_1} + {R_2} = \frac{{{U^2}}}{P}\\{R_1}{R_2} = {\left( {{Z_L} - {Z_C}} \right)^2} \end{array} \right.\)