Lời giải

Mạch: \({R_0}nt\left( {{R_1}//R{}_x} \right)\)

Đặt R_x = x \(\left( \Omega \right)\)

\({R_{td}} = {R_0} + \frac{{{R_1}.x}}{{{R_1} + x}} = 4 + \frac = \frac\)

\( \Rightarrow I = \frac{U}{{{R_{td}}}} = \frac{{\frac}} = \frac\)

\({U_{Rx}} = U{}_{R1} = U{}_{R1x} = \frac.\frac = \frac\)

Mà \(P = \frac{{U_{Rx}^2}}{} = 9W \Leftrightarrow \frac{{\frac{{144{x^2}}}{{{{\left( {x + 3} \right)}^2}}}}}{x} = 9 \Rightarrow 9{x^2} - 90x + 81 = 0\)

\( \Rightarrow \left[ {\begin{array}{*{20}{c}}{x = 1 \Rightarrow {R_x} = 1\left( \Omega \right)}\\{x = 9 \Rightarrow {R_x} = 9\left( \Omega \right)}\end{array}} \right.\)

Hiệu suất của mạch điện

\(H = \frac{{{P_{1x}}}}{P}.100\% = \frac{{{I^2}.{R_{1x}}}}{{{I^2}.{R_{td}}}}.100\% = \frac{{{R_{1x}}}}{{{R_{td}}}}.100\% = \frac{{3{R_x}}}{{4{R_x} + 12}}.100\% \)

+ R_x = 1\(\left( \Omega \right)\) \( \Rightarrow H = 18,75\% \)

+ R_x = 9\(\left( \Omega \right)\) \( \Rightarrow H = 56,25\% \)

b. Từ câu a

\( \Rightarrow {P_} = \frac{{{U^2}}}{} = \frac{{{{\left( {\frac} \right)}^2}}}{x} = \frac{{{x^2} + 6x + 9}} = \frac\)

Để P_Rx max khi \((x + \frac{9}{x} + 6)\,\,\min \)\( \Rightarrow \left( {x + \frac{9}{x}} \right)\,\,\min \)

Áp dụng BĐT Cô – si: \(x + \frac{9}{x} \ge 2\sqrt {x.\frac{9}{x}} = 6\)