Theorems formulas

$I_0$ = DC component; $I_k$ = rms value of the component at frequency ${\omega }_k$, with all the ${\omega }_k$ non-zero and distinct. Here average powers do add, because the cross terms average to zero over time. Two components at the same frequency do not: combine them as phasors first. Example: $i=3+4\sqrt{2}\sin \omega t$ A gives $I_{\text{rms}}=\sqrt{9+16}=5$ A.

$V_{oc}=V_a-V_b$ with a–b open; $I_{sc}$ flows from a to b through a short across a–b. The ratio cannot be used when both are zero. That is always the case for a network with no independent source, and it can also happen when the sources inside produce nothing at a–b (a balanced bridge, for example). Then use the test-source method.

The arrow of $I_s$ points towards the terminal on the $+$ side of $V_s$. The two forms are equivalent only as seen from the terminals: the current and power inside $Z$ differ. A dependent source can be transformed the same way, provided its controlling variable is not inside the $Z$ being moved. An ideal voltage source ($Z=0$) or an ideal current source ($Z\to \infty$) cannot be transformed.

Short the independent voltage sources, open the independent current sources, and leave every dependent source active. Apply $V_t$ across a–b ($+$ at a) and find the current $I_t$ entering terminal a from the test source (or inject $I_t$ into a and find $V_t$). This works whenever the equivalent exists, and it is the method to use when the network has no independent source.

$R_{th}>0$ is fixed and $R_L$ varies. If the network has dependent sources, use the $R_{th}$ found with them active.

$P_L$ rises for $R_L<R_{th}$ and falls for $R_L>R_{th}$. So if $R_L$ is restricted to a range that excludes $R_{th}$, the best choice is the allowed value nearest $R_{th}$.

This is the efficiency of the Thevenin equivalent circuit only. The power drawn from the sources inside the original network is different, so the real efficiency at matching is generally not 50%.

Use this when both $R_L$ and $X_L$ are free (and $R_{th}>0$). The reactances cancel, so the loop is resonant. $|V_{th}|$ is an rms value; if the phasor is a peak value $V_m$, then $P_{max}=|V_m{|}^2/(8R_{th})$.

Use this when the load must be a pure resistor ($X_L=0$). Choosing $R_L=R_{th}$ here is a common error.

$X_L$ is fixed and $R_L$ varies; $P_{max}$ is evaluated at that $R_L$. The pure-resistor case is $X_L=0$, which gives back $R_L=|Z_{th}|$.

Use this for a load $Z_L=|Z_L|\angle {\theta }_L$ whose angle ${\theta }_L$ is fixed (a fixed power factor) and whose magnitude alone varies; ${\theta }_{th}$ is the angle of $Z_{th}$. Setting ${\theta }_L=0$ gives back the pure-resistor result, and ${\theta }_L=-{\theta }_{th}$ gives back the conjugate match.

Convention: $I_1$ and $I_2$ both flow INTO their ports, and ABCD is defined by $V_1=A{V}_2-B{I}_2$, $I_1=C{V}_2-D{I}_2$. Where the parameter sets exist, any one condition implies the others. Symmetry ($z_{11}=z_{22}$, $A=D$) is a separate, stronger property.

The sums run over all $b$ branches. The first form is conservation of power: the power delivered by the sources equals the power absorbed by everything else. The second form pairs voltages from one network with currents from another that has the same graph and the same reference directions. It also holds for one network at two different instants.

$I_j$ = ideal current sources between the two nodes, directed into a; they add nothing to the denominator. A passive branch enters with $V_k=0$. An ideal voltage source with $Z_k=0$ cannot be included, because it fixes $V_{ab}$ directly.

$Z$ is removed and replaced by $Z_1$ from node 1 to ground and $Z_2$ from node 2 to ground. $K$ must be the gain with $Z$ connected. For $K>1$, $Z_1$ is negative; for $0<K<1$ (a follower), $Z_2$ is negative.

This is the capacitor version (impedance $1/(j\omega C)$). For an inverting stage, $K=-A$ with $A>0$. Then $C_1=C(1+A)$ is large (the Miller effect), and $C_2\approx C$ when $A\gg 1$. A bridging resistor $R$ becomes $R/(1-K)$ at the input.

$I$ = original branch current. $Z_{th}$ = impedance seen by the branch looking into the rest of the network, with independent sources off. The minus sign says that increasing the impedance reduces the branch current. A typical use is the error caused by inserting an ammeter of resistance $\Delta Z$.