Question 2.21: Find the load resistance for maximum power transfer from the...

First, we must find the Thévenin equivalent circuit. Zeroing the voltage source, we find that the resistances R_1 and R_2 are in parallel. Thus, the Thévenin resistance is

R_t = \frac{1}{1/R_1+1/R_2}=\frac{1}{1/20+1/5} = 4 Ω

The Thévenin voltage is equal to the open-circuit voltage. Using the voltage-division principle, we find that

V_{t} = v_{oc}=\frac{R_2}{R_1+R_2}(50) =\frac{5}{5+20}(50) = 10 V

Hence, the load resistance that receives maximum power is

R_L = R_t = 4 Ω
and the maximum power is given by Equation 2.78:

P_{Lmax} = \frac{V_t^2}{4R_t} =\frac{10^2}{4\times 4} = 6.25 W