ece140: simplify thevenin

docs/1b/ece140.md

@@ -173,17 +173,28 @@ The arrow of the current source must point in the positive direction of the volt
 
 Any part of a circuit including an independent source can be replaced with exactly one voltage source and a resistor in series. Two circuits are **Thevenin equivalent** if their $\lambda$ are equal in $V=\lambda I$.
 
-1. Cut off the load.
-2. Group the rest of the circuit together, removing all independent sources (short / open).
-3. The Thevenin resistance of the new resistor is the same as the load $R_{Th}=R_L$.
+If there are no dependent sources, all independent sources should be removed to determine the resistance across points $AB$:
 
-If dependent sources exist, the load should be replaced with an independent source of arbitrary value (e.g., 1 V) and the other variable determined to find $R_{Th}=V_{Th}/I$, where $V_{Th}$ is the voltage drop across the load.
+$$R_{Th}=R_{AB}$$
+
+Otherwise, $V_{AB}$ and $I_{AB}$ should be found by repeating these steps:
+
+1. Cut off the load (open if finding voltage, short if finding current)
+  - If dependent sources depend on elements inside the load branch, zero them
+2. Use analysis to determine the desired quantity
 
 Across the load:
 
 $$
 I_L=\frac{V_{Th}}{R_{Th}+R_L} \\
 V_L=R_LI_L = \frac{R_L}{R_{Th}+R_L}V_{Th}
+$$
 
 !!! warning
     A negative resistance $R_{L}$ indicates that the load supplies power.
+
+### Maximum power transfer
+
+To maximise the power transferred from the circuit to the load, $R_L$ should be equal to $R_{Th}$.
+
+$$P_L=v_Li_L$$