maximum power transfer........

It is textbook's stuff that the maximum power transferred occurred when the load resistance $R_L$ is equal to that of the resistance of the series resistor $R_S$ (sometimes the internal resistance of the voltage source), $100\Omega$ . When $R_L = 100\Omega$ , then the current

$I=\cfrac{10V}{100\Omega+100\Omega}=0.05A$ .

Therefore the maximum power transferred is:

$P_{max}=I^2R_L=0.05^2\times 100= \boxed{2.5W}$ .

Proof:

$P_{R_L}=I^2R_L =\left(\cfrac{V}{R_L+R_S}\right)^2 R_L$

$\cfrac{dP_{R_L}}{dR_L}=\cfrac{d}{dR_L} \left(\cfrac{V^2R_L}{(R_L+R_S)^2}\right)$

$\quad \quad\quad= V^2\left(\cfrac{1}{(R_L+R_S)^2}-\cfrac{2R_L}{(R_L+R_S)^3}\right)$

$\quad \quad\quad= V^2\left(\cfrac{R_L+R_S-2R_L}{(R_L+R_S)^3}\right)$

$\quad \quad\quad = V^2\left(\cfrac{R_S-R_L}{(R_L+R_S)^3}\right)$

$P_{R_L}=P_{max}$ when $\cfrac{dP_{R_L}}{dR_L}=0$ or when $\boxed{R_L=R_S}$ .