1)
I=\frac{E}{R_{td}+r}=\frac{12}{0,5+R+2,5}=\frac{12}{R+3}
P=I^2.R
=(\frac{12}{R+3})^2.R
=\frac{12^2}{R^2+6R+9}.R
=\frac{144}{R+6+\frac{9}{R}}
Áp dụng bất đẳng thức AM-GM ,có:
R+\frac{9}{R}\geq6
P_{max}<=>(R+6+\frac{9}{R})_{min}<=>R=\frac{9}{R}<=>R=3
Vậy P_{max}=12(W) khi R=3