Maximum Power Transfer

Problem-Solving Strategy for solving the Maximum Power Transfer

STEP 1. Maximum power transfer takes place when the load resistance $R_L = R$.

Note: This post is based on Irwin's Basic Engineering Circuit Analysis 11th ed.

If there are any errors in solution, please let me know in the comments

At first, we can find $V_x$ from $2mA$ source.

$\frac{V_x}{1k}=-2mA$, $V_x=-2V$

Then we can apply KCL on the top of node.

$\frac{V_{oc}+10}{2k}+\frac{V_{oc}+2}{1k}=2mA$

$V_{oc}=-\frac{10}{3}V$

To find $V_x$, we can apply KCL on top of the node.

$\frac{V_{x}+12}{2k}+\frac{-V_{x}}{1k}+\frac{V_{x}}{1k}=0$

$V_x=-12V$

So $I_{sc}=2mA+\frac{-12}{1k}=-10mA$.

$R_{Th}$ is the ratio of $V_{oc}$ to $I_{sc}$.

$R_{Th}=\frac{V_{oc}}{I_{sc}}=\frac{1}{3}k\Omega$

$R_L=R_{Th}=\frac{1}{3}k\Omega$ for maximum power transfer. Thus, the maximum power transferred to the load is

$P_L=\frac{V_L^2}{R_L}=\frac{25}{3}mW$