Circuit Analysis and Theorems

Replace everything to the left of a port by one source and one impedance. $V_{th}$ is the open-circuit voltage; $R_{th}$ is what you see with independent sources killed — voltage sources shorted, current sources opened. With no independent source inside, there is nothing to drive the port: $V_{th}=0$ and only $R_{th}$ exists. Dependent sources can make $R_{th}$ come out negative; that is a real answer, not an arithmetic slip.

Power in the load peaks when the load matches the source. For AC the match is the complex conjugate, which cancels the reactance. If the load can only be a resistor, the best it can do is match the magnitude of the source impedance.

With several independent sources, find the response to each alone and add. Only independent sources are switched off; dependent ones stay active throughout. It is the only correct method when sources run at different frequencies: each source gets its own phasor circuit with its own $j\omega$, and the answers are added in the time domain.

$Z$, $Y$, $h$ and $ABCD$ describe the same network in different currencies. Convert with a matrix inverse; pick whichever makes the connection type easy — $Z$ adds in series, $Y$

Used to unlock bridges that are neither series nor parallel. For three equal elements the ratio is simply 3.

Count first and pick the smaller system: $N-1$ node equations for $N$ nodes, or $B-N+1$ mesh equations for $B$