Gọi \(R_b=x\left(\Omega\right)\)
\(RntR_0\)\(\Rightarrow I=\dfrac{\xi}{r+R_0}=\dfrac{10}{4+x}\)
Ta có:
\(P_b=\left(\dfrac{12}{4+x}\right)^2\cdot x=\dfrac{144x}{16+8x+x^2}=\dfrac{144}{\dfrac{16}{x}+8+x}\)
\(P_{max}\Leftrightarrow\left(\dfrac{16}{x}+x\right)min\)
Áp dụng bđt Cô-si:
\(\dfrac{16}{x}+x\ge2\cdot\sqrt{\dfrac{16}{x}\cdot x}=2\cdot4=8\)
\(\Rightarrow\left(\dfrac{16}{x}+x\right)min=8\Leftrightarrow\dfrac{16}{x}=x\Leftrightarrow x=4\)
Vậy \(R_b=4\Omega\)