A load resistor R${}_L$ is connected to a battery of voltage E with internal resistance Rᵢ through a resistance R${}_S$ as shown in the figure. For fixed values of R${}_L$ and Rᵢ, the value of R${}_S$ (≥ 0) for maximum power transfer to R${}_L$ is

Analysis:

Finding the voltage across load resistance using voltage division rule, we get:

$V_{RL}=\frac{E\times R_L}{R_i+R_S+R_L}$

Power loss in the R${}_L$ is given as:

$P_{RL}=\frac{V_{RL}^2}{R_L}$

$=\frac{{\left(E{R}_L\right)}^2}{{\left(R_i+R_S+R_L\right)}^2\times R_L}$

We conclude that maximum power is transferred to R${}_L$ when RS = 0



Maximum power theorem states that for maximum power to be transferred to the load resistance RL, RL must equal the Thevenin Equivalent resistance, i.e.

RL = Rth

But here, we are asked to find the value of R, and not RL that will result in the maximum power to be transferred to load RL. So we cannot go by the standard procedure of equating RL with the Thevenin equivalent resistance.