Question

Answer 1

a)\(P_N=R_N\cdot I^2=R_N\cdot\left(\dfrac{\xi}{R_N+r}\right)^2=\dfrac{R_N\cdot\xi^2}{R_N^2+2R_N\cdot r+r^2}=\dfrac{\xi^2}{R_N+2r+\dfrac{r^2}{R_N}}\)

Để \(P_Nmax\Leftrightarrow\left(R_N+2r+\dfrac{r^2}{R_N}\right)min\).

\(\Leftrightarrow R_N+2r+\dfrac{r^2}{R_N}\ge2\sqrt{R_N\cdot\dfrac{r^2}{R_N}}+2r=4r=8\Omega\)

Khi đó \(R_N=\dfrac{r^2}{R_N}\Leftrightarrow R_N=r=2\Omega\)

Mặt khác: \(R//R_X\Rightarrow R_N=\dfrac{R\cdot R_X}{R+R_X}=\dfrac{3\cdot R_X}{3+R_X}=2\)

\(\Rightarrow R_X=6\Omega\)

b)Nếu \(\xi=80V\) thì:

\(P_{max}=\dfrac{\xi^2}{8}=\dfrac{80^2}{8}=800W\)