$R_{2}=12$ ôm
$R_{0}=2$ ôm
$R_{b}=r$
$R_{1}$
$U=24V$
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Xét tại lúc công suất trên biến trở lớn nhất
$R_{tđ}=\dfrac{R_{2}.r}{R_{2}+r}+R_{1}+R_{0}$
$=\dfrac{12r}{12+r}+R_{1}+2$
$=\dfrac{(R_{1}+14)r+24+12R_{1}}{12+r}$
$⇒I=\dfrac{24(12+r)}{(R_{1}+14)r+24+12R_{1}}$
Mà $I=\dfrac{12,6}{R_{1}}$
$⇒\dfrac{24(12+r)}{(R_{1}+14)r+24+12R_{1}}=\dfrac{12,6}{R_{1}}$
$⇔24(12+r).R_{1}=12,6[(R_{1}+14)r+24+12R_{1}]$
$⇔288R_{1}+24r.R_{1}=12,6.R_{1}.r+176,4r+302,4+151,2R_{1}$
$⇔136,8R_{1}+11,4r.R_{1}-176,4r+302,4=0$
$⇒R_{1}=\dfrac{176,4.r-302,4}{136,8+11,4r}$
Ta có: $\dfrac{I_{2}}{r}=\dfrac{I_{b}}{R_{2}}$
$⇒I_{b}=\dfrac{R_{2}.I}{R_{2}+r}=\dfrac{12.I}{12+r}=\dfrac{12.24}{(R_{1}+14)r+24+12R_{1}}$
$=\dfrac{288}{(R_{1}+14)r+24+12R_{1}}$
$⇒P_{b}=I_{b}².r=\dfrac{82944}{[(R_{1}+14)r+12(2+R_{1})]²}.r$
$≥\dfrac{82944}{4.(R_{1}+14).r.12(2+R_{1})}.r$
$=\dfrac{1728}{(R_{1}+14)(2+R_{1})}$
Dấu $=$ xảy ra khi $(R_{1}+14)r=12(2+R_{1})$
$⇔R_{1}.r+14r=24+12R_{1}$
$⇔R_{1}=\dfrac{14r-24}{12-r}$
$⇒\dfrac{176,4.r-302,4}{136,8+11,4r}=\dfrac{14r-24}{12-r}$
$⇔(176,4.r-302,4)(12-r)=(14r-24)(136,8+11,4r)$
$⇔2116,8r-176,4r²-3628,8+302,4r=1915,2r+159,6r²-3283,2-273,6r$
$⇔336r²-777,6r+345,6=0$
$⇔r=\dfrac{12}{7}$ ôm
hoặc $r=0,6$ ôm
Bạn tìm $R_{1}$ rồi tìm lại $P_{max}$ nha
