Question

Answer 1

R1 nt r2 nt R0  (gọi R2 là x(ôm))

\(=>Pab=P12=I12^2R12=\left(\dfrac{U}{Rtd}\right)^2\left(R1+x\right)\)

\(=>Pab=\left(\dfrac{24}{R1+R2+Ro}\right)^2\left(4+x\right)\)

\(=>Pab=\dfrac{24^2\left(4+x\right)}{\left(10+x\right)^2}=\dfrac{24}{\dfrac{\left(10+x\right)^2}{24\left(4+x\right)}}\)

xét \(\dfrac{\left(10+x\right)^2}{24\left(4+x\right)}=\dfrac{x^2+20x+100}{24x+96}=1+\dfrac{x^2-4x+4}{24x+96}=1+\dfrac{\left(x-2\right)^2}{24x+96}\ge1\)

\(=>Pab\le\dfrac{24}{1}=24\) dấu"=" xảy ra<=>x=2=R2=>R2=2(om)

Vậy với R2=2(om) thì công suất Pab cực đại =24W