The whole Networks (Electric Circuits) formula sheet, to print

166 formulas in 7 topics, each with when to use it. Each is explained in the topic notes.

Circuit laws, sources, and node and mesh analysis

RESISTOR (OHM'S LAW)

v = R i, p = v i = i^{2} R = \frac{v ^{2}}{R} \geq 0

Linear resistor, passive sign convention. Conductance $G = 1/ R$ (siemens), so $i = G v$ .

INDUCTOR

v i (t) w_{L} = L \frac{d i}{d t} = i (t_{0}) + \frac{1}{L} \int_{t_{0}}^{t} v d τ = \frac{1}{2} L i^{2}

Passive sign convention. The stored energy depends only on the present current. In DC steady state $d i / d t = 0$ , so an inductor behaves as a short circuit.

CAPACITOR

i v (t) q = C \frac{d v}{d t} = v (t_{0}) + \frac{1}{C} \int_{t_{0}}^{t} i d τ = C v, w_{C} = \frac{1}{2} C v^{2} = \frac{q ^{2}}{2 C}

Passive sign convention. In DC steady state $d v / d t = 0$ , so a capacitor behaves as an open circuit.

CONTINUITY OF CAPACITOR VOLTAGE AND INDUCTOR CURRENT

v_{C} (0^{+}) = v_{C} (0^{-}), i_{L} (0^{+}) = i_{L} (0^{-})

$0^{-}$ is the instant just before the switching and $0^{+}$ the instant just after. The rule holds whenever the capacitor current and the inductor voltage stay finite (no impulses). That is guaranteed when the sources are finite and the switching creates no loop made only of capacitors and ideal voltage sources and no cut-set made only of inductors and ideal current sources. If such a loop is formed, the capacitor voltages are forced to jump (solve with charge conservation); if such a cut-set is formed, the inductor currents are forced to jump (solve with flux-linkage conservation). Capacitor current, inductor voltage and resistor quantities CAN jump at $t = 0$ .

SERIES AND PARALLEL COMBINATIONS

Series: R = R_{1} + R_{2}, L = L_{1} + L_{2}, \frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}} Parallel: \frac{1}{R} = \frac{1}{R _{1}} + \frac{1}{R _{2}}, \frac{1}{L} = \frac{1}{L _{1}} + \frac{1}{L _{2}}, C = C_{1} + C_{2}

Inductors without mutual coupling. Two resistors in parallel give $R_{1} R_{2} / (R_{1} + R_{2})$ , and two capacitors in series give $C_{1} C_{2} / (C_{1} + C_{2})$ . Series capacitors that start uncharged carry equal charge.

COUPLED INDUCTORS (DOT CONVENTION)

v_{1} v_{2} M = L_{1} \frac{d i _{1}}{d t} \pm M \frac{d i _{2}}{d t} = L_{2} \frac{d i _{2}}{d t} \pm M \frac{d i _{1}}{d t} = k L_{1} L_{2}, 0 \leq k \leq 1

Each coil uses the passive sign convention for its own $v$ and $i$ . Take +M when both currents enter (or both leave) the dotted terminals, and −M when one enters a dot and the other leaves one. $k = 1$ means perfect coupling.

COUPLED INDUCTORS IN SERIES AND PARALLEL; STORED ENERGY

Series: L_{e q} = L_{1} + L_{2} \pm 2 M Parallel: L_{e q} = \frac{L _{1} L _{2} - M ^{2}}{L _{1} + L _{2} \mp 2 M} Energy: w = \frac{1}{2} L_{1} i_{1}^{2} + \frac{1}{2} L_{2} i_{2}^{2} \pm M i_{1} i_{2}

For $L_{e q}$ : upper sign = aiding (in series, the current enters both dotted terminals in turn; in parallel, the dotted terminals are joined together), lower sign = opposing. For the energy: $+ M i_{1} i_{2}$ when $i_{1}$ and $i_{2}$ both enter (or both leave) the dotted terminals, $- M i_{1} i_{2}$ otherwise. $w \geq 0$ for all currents, which is why $M \leq L_{1} L_{2}$ .

COUPLED COILS WITH A COMMON TERMINAL (T-EQUIVALENT)

Like terminals joined: L_{A} = L_{1} - M, L_{B} = L_{2} - M, L_{C} = M Unlike terminals joined: L_{A} = L_{1} + M, L_{B} = L_{2} + M, L_{C} = - M

Two coupled coils that share one terminal can be replaced by an uncoupled T network: series arm $L_{A}$ on coil 1's side, series arm $L_{B}$ on coil 2's side, and shunt arm $L_{C}$ to the common terminal. "Like terminals joined" means both dotted ends (or both undotted ends) meet at the common terminal. A negative inductance in the model is only a mathematical element; the behaviour at the three terminals is exact.

CHARGE SHARING BETWEEN TWO CAPACITORS

V_{f} W_{lost} = \frac{C _{1} V _{1} + C _{2} V _{2}}{C _{1} + C _{2}} = \frac{C _{1} C _{2} ( V _{1} - V _{2} ) ^{2}}{2 ( C _{1} + C _{2} )}

Capacitors charged to $V_{1}$ and $V_{2}$ are connected in parallel, + plate to + plate. If they are joined with opposite polarities, replace $V_{2}$ by $- V_{2}$ . The total charge is conserved; the energy $W_{lost}$ is dissipated in the connecting resistance, whatever its value.

SOURCE TRANSFORMATION

I_{s} = \frac{V _{s}}{R}, V_{s} = I_{s} R (same R)

$V_{s}$ in series with $R$ is equivalent at its two terminals to $I_{s}$ in parallel with the same $R$ . The current-source arrow points toward the terminal that was + on the voltage source. The two are equivalent only at the terminals: the power dissipated inside differs. An ideal source (no $R$ ) cannot be transformed. A dependent source can be transformed the same way, provided its controlling variable is not the current or voltage of the resistor $R$ being transformed, because that quantity changes in the transformation.

KIRCHHOFF'S CURRENT LAW

k \sum i_{k} = 0 (currents leaving a node or any closed surface)

Equivalently, total current entering = total current leaving. It also holds for a closed surface enclosing several nodes, which is the basis of the supernode.

KIRCHHOFF'S VOLTAGE LAW

k \sum v_{k} = 0 (voltage drops around any closed loop)

Count a drop as positive when the loop crosses an element from + to −. It holds for any closed path, including one through a current source or across an open gap (the voltage there is then an unknown).

VOLTAGE DIVISION

V_{k} = V \frac{R _{k}}{R _{1} + R _{2} + \dots + R _{N}}

Resistors in series carrying the same current, with nothing else drawing current from the tap. For impedances, replace $R$ by $Z$ .

CURRENT DIVISION

I_{1} = I \frac{R _{2}}{R _{1} + R _{2}}, I_{k} = I \frac{G _{k}}{G _{1} + G _{2} + \dots + G _{N}}

Branches in parallel across the same voltage. With two branches, the current in one is proportional to the OTHER branch's resistance.

BALANCED WHEATSTONE BRIDGE

\frac{R _{A C}}{R _{C B}} = \frac{R _{A D}}{R _{D B}} \Rightarrow V_{C} = V_{D}, I_{C D} = 0

A bridge with arms A–C–B and A–D–B and an element between C and D. When the bridge is balanced and the C–D element contains no source, that element carries no current and can be removed or replaced by a short without changing anything else.

CUBE OF TWELVE EQUAL RESISTORS

R_{body diagonal} = \frac{5}{6} R, R_{face diagonal} = \frac{3}{4} R, R_{edge} = \frac{7}{12} R

Resistance between two corners of a cube whose twelve edges are each $R$ : opposite corners of the cube, opposite corners of one face, and the two ends of one edge. Each follows from symmetry: join the corners that are at equal potential, then reduce.

INFINITE LADDER NETWORK

R_{in} = \frac{R _{1} + R _{1}^{2} + 4 R _{1} R _{2}}{2} (R_{1} = R_{2} = R : R_{in} = \frac{1 + 5}{2} R \approx 1.618 R)

A ladder that starts at the input with a series resistor $R_{1}$ , then a shunt resistor $R_{2}$ , repeated without end. Adding one more section does not change the input resistance, so $R_{in} = R_{1} + (R_{2} ∥ R_{in})$ , and the formula is the positive root. If the ladder starts with the shunt arm instead, its input resistance is $R_{2} ∥ R_{in}$ with $R_{in}$ from the formula.

DELTA TO STAR

R_{a} R_{b} R_{c} = \frac{R _{ab} R _{c a}}{R _{ab} + R _{b c} + R _{c a}} = \frac{R _{ab} R _{b c}}{R _{ab} + R _{b c} + R _{c a}} = \frac{R _{b c} R _{c a}}{R _{ab} + R _{b c} + R _{c a}}

Star arm at a node = product of the two delta resistors touching that node ÷ sum of all three delta resistors.

STAR TO DELTA

R_{ab} S = \frac{S}{R _{c}}, R_{b c} = \frac{S}{R _{a}}, R_{c a} = \frac{S}{R _{b}} = R_{a} R_{b} + R_{b} R_{c} + R_{c} R_{a}

Delta resistor between two nodes = sum of the pairwise products of the star arms ÷ the star arm at the OPPOSITE (third) node. Equivalently, $R_{ab} = R_{a} + R_{b} + R_{a} R_{b} / R_{c}$ .

BALANCED (EQUAL-ARM) CASE

R_{Y} = \frac{R _{Δ}}{3}, R_{Δ} = 3 R_{Y}

All three arms equal. The same relation holds for equal impedances at a single frequency, for example a balanced three-phase load.

NODAL EQUATIONS BY INSPECTION

[G] [V] = [I] G_{k k} = sum of conductances connected to node k G_{k j} = - (sum of conductances between nodes k and j)

Valid for circuits with only resistors and independent current sources. $I_{k}$ = net source current injected INTO node $k$ . The matrix is symmetric; a dependent source breaks the symmetry and has to be added by hand.

MESH EQUATIONS BY INSPECTION

[R] [I] = [V] R_{k k} = sum of resistances in mesh k R_{k j} = - (sum of resistances shared by meshes k and j)

Valid when all mesh currents are clockwise and the circuit has only resistors and independent voltage sources. $V_{k}$ = algebraic sum of the source voltages driving current in mesh $k$ 's direction (rises positive). The matrix is symmetric unless dependent sources are present.

NUMBER OF KCL OR KVL EQUATIONS TO WRITE

N_{nodal} N_{mesh} = (n - 1) - n_{V} = (b - n + 1) - n_{I}

$n$ = nodes and $b$ = branches of a connected circuit, counting every element as a separate branch. $n_{V}$ = voltage sources (independent or dependent); each fixes a node voltage or supplies a supernode constraint, so one KCL equation fewer is needed. $n_{I}$ = current sources (independent or dependent); each fixes a mesh current or supplies a supermesh constraint. A dependent source's constraint is still expressed in the unknowns, so it has to be substituted, not just read off. Assumes no loop made only of voltage sources and no cut-set made only of current sources. Choose the method that needs fewer equations.

POWER, ENERGY AND POWER BALANCE

p = v i, W = \int_{t_{1}}^{t_{2}} p d t, all elements \sum p_{absorbed} = 0

Passive sign convention. $1 kWh = 3.6 \times 1 0^{6}$ J.

SAFE CURRENT AND VOLTAGE OF A RATED RESISTOR

I_{m a x} = \frac{P _{rated}}{R}, V_{m a x} = P_{rated} R

The power rating is the largest power a component can dissipate continuously without overheating. In series, the smallest $I_{m a x}$ limits the whole string; in parallel, the smallest $V_{m a x}$ limits.

EFFICIENCY

η = \frac{P _{out}}{P _{in}} = \frac{P _{out}}{P _{out} + P _{loss}}

Losses are the power dissipated in resistances, for example $I^{2} R$ in conductors and in a source's internal resistance.

PRACTICAL SOURCE FEEDING A RESISTIVE LOAD

I η = \frac{V _{s}}{R _{s} + R _{L}}, P_{L} = \frac{V _{s}^{2} R _{L}}{( R _{s} + R _{L} ) ^{2}} = \frac{R _{L}}{R _{s} + R _{L}}

$η$ is the fraction of the power generated by the ideal part, $V_{s} I$ , that reaches $R_{L}$ . It approaches 100% as $R_{L} ≫ R_{s}$ .

MAXIMUM POWER FROM A PRACTICAL SOURCE

R_{L} = R_{s} \Rightarrow P_{L, m a x} = \frac{V _{s}^{2}}{4 R _{s}}, η = 50%

Fixed $V_{s}$ and $R_{s}$ with a variable resistive load. Maximum power transfer and maximum efficiency happen at different operating points. If instead $R_{L}$ is fixed and $R_{s}$ can be chosen, $P_{L} = V_{s}^{2} R_{L} / (R_{s} + R_{L})^{2}$ is largest at $R_{s} = 0$ , not at $R_{s} = R_{L}$ : the matching condition applies only when the load is the variable.

BATTERY WITH INTERNAL RESISTANCE

V_{t} V_{t} = E - I r (discharging) = E + I r (charging)

$E$ = EMF (the open-circuit voltage, which depends on the state of charge), $r$ = internal resistance, $I$ = magnitude of the current. Short-circuit current is $E / r$ . Stored energy ≈ nominal voltage × capacity in Ah, which gives Wh.

SOLAR CELL (IDEAL SINGLE-DIODE MODEL)

I I_{sc} V_{oc} = I_{p h} - I_{0} (e^{V / (n V_{T})} - 1) = I_{p h} = n V_{T} ln (1 + \frac{I _{p h}}{I _{0}})

Generator convention: $I$ is the current delivered out of the + terminal. $I_{p h}$ is the photocurrent (proportional to irradiance), $I_{0}$ the diode saturation current, $n$ the ideality factor, and $V_{T} = k T / q \approx 25.9$ mV at 300 K. Series and shunt resistances are neglected.

TWIGS AND LINKS OF A CONNECTED GRAPH

twigs = n - 1, links = b - n + 1

For a graph with $p$ separate parts: twigs $= n - p$ and links $= b - n + p$ .

INDEPENDENT EQUATIONS

N_{KCL} = n - 1, N_{KVL} = b - n + 1

Connected graph. For a planar graph, $b - n + 1$ is also the number of meshes (windows).

Network theorems

Read the notes

POWER AND RMS VALUE WITH COMPONENTS AT DIFFERENT FREQUENCIES

P = I_{0}^{2} R + k \sum I_{k}^{2} R, I_{rms} = I_{0}^{2} + k \sum I_{k}^{2}

$I_{0}$ = DC component; $I_{k}$ = rms value of the component at frequency $ω_{k}$ , with all the $ω_{k}$ 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: $i = 3 + 42 sin ω t$ A gives $I_{rms} = 9 + 16 = 5$ A.

THEVENIN–NORTON RELATIONS

V_{t h} = V_{oc}, I_{N} = I_{sc} Z_{t h} = Z_{N} = \frac{V _{oc}}{I _{sc}}, V_{t h} = I_{N} Z_{t h}

$V_{oc} = V_{a} - V_{b}$ 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.

SOURCE TRANSFORMATION

V_{s} in series with Z \leftrightarrow I_{s} = \frac{V _{s}}{Z} in parallel with Z

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.

TEST-SOURCE METHOD

Z_{t h} = \frac{V _{t}}{I _{t}}_{independent sources off}

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.

MAXIMUM POWER TRANSFER, RESISTIVE LOAD

R_{L} = R_{t h}, P_{ma x} = \frac{V _{t h}^{2}}{4 R _{t h}} = \frac{I _{N}^{2} R _{t h}}{4}

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

LOAD POWER FOR ANY LOAD RESISTANCE

P_{L} = \frac{V _{t h}^{2} R _{L}}{( R _{t h} + R _{L} ) ^{2}}

$P_{L}$ rises for $R_{L} < R_{t h}$ and falls for $R_{L} > R_{t h}$ . So if $R_{L}$ is restricted to a range that excludes $R_{t h}$ , the best choice is the allowed value nearest $R_{t h}$ .

EFFICIENCY OF THE THEVENIN EQUIVALENT

η = \frac{R _{L}}{R _{L} + R _{t h}}, η = 50% at R_{L} = R_{t h}

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%.

AC MAXIMUM POWER TRANSFER, CONJUGATE MATCH

Z_{L} = Z_{t h}^{*} = R_{t h} - j X_{t h}, P_{ma x} = \frac{∣ V _{t h} ∣ ^{2}}{4 R _{t h}}

Use this when both $R_{L}$ and $X_{L}$ are free (and $R_{t h} > 0$ ). The reactances cancel, so the loop is resonant. $∣ V_{t h} ∣$ is an rms value; if the phasor is a peak value $V_{m}$ , then $P_{ma x} = ∣ V_{m} ∣^{2} / (8 R_{t h})$ .

AC SOURCE, PURELY RESISTIVE LOAD

R_{L} = ∣ Z_{t h} ∣ = R_{t h}^{2} + X_{t h}^{2} P_{ma x} = \frac{∣ V _{t h} ∣ ^{2}}{2 ( R _{t h} + ∣ Z _{t h} ∣ )}

Use this when the load must be a pure resistor ( $X_{L} = 0$ ). Choosing $R_{L} = R_{t h}$ here is a common error.

FIXED LOAD REACTANCE, VARIABLE LOAD RESISTANCE

R_{L} = R_{t h}^{2} + (X_{t h} + X_{L})^{2} P_{ma x} = \frac{∣ V _{t h} ∣ ^{2}}{2 ( R _{t h} + R _{L} )}

$X_{L}$ is fixed and $R_{L}$ varies; $P_{ma x}$ is evaluated at that $R_{L}$ . The pure-resistor case is $X_{L} = 0$ , which gives back $R_{L} = ∣ Z_{t h} ∣$ .

FIXED LOAD ANGLE, VARIABLE MAGNITUDE

∣ Z_{L} ∣ = ∣ Z_{t h} ∣ P_{ma x} = \frac{∣ V _{t h} ∣ ^{2} cos θ _{L}}{2∣ Z _{t h} ∣ [ 1 + cos ( θ _{t h} - θ _{L} ) ]}

Use this for a load $Z_{L} = ∣ Z_{L} ∣∠ θ_{L}$ whose angle $θ_{L}$ is fixed (a fixed power factor) and whose magnitude alone varies; $θ_{t h}$ is the angle of $Z_{t h}$ . Setting $θ_{L} = 0$ gives back the pure-resistor result, and $θ_{L} = - θ_{t h}$ gives back the conjugate match.

RECIPROCITY CONDITIONS FOR A TWO-PORT

z_{12} = z_{21}, y_{12} = y_{21}, A D - B C = 1 h_{12} = - h_{21}, g_{12} = - g_{21}

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.

TELLEGEN'S THEOREM

k = 1 \sum b v_{k} i_{k} = 0, k = 1 \sum b v_{k}^{'} i_{k}^{''} = 0

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.

MILLMAN'S THEOREM

V_{ab} = \frac{\sum _{k} V _{k} Y _{k} + \sum _{j} I _{j}}{\sum _{k} Y _{k}}, Z_{e q} = \frac{1}{\sum _{k} Y _{k}}

$I_{j}$ = 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.

MILLER'S THEOREM

Z_{1} = \frac{Z}{1 - K}, Z_{2} = \frac{Z}{1 - 1/ K} = \frac{Z K}{K - 1} K = \frac{V _{2}}{V _{1}}

$Z$ 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.

MILLER CAPACITANCE

C_{1} = C (1 - K), C_{2} = C (1 - \frac{1}{K})

This is the capacitor version (impedance $1/ (j ω C)$ ). 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 $A ≫ 1$ . A bridging resistor $R$ becomes $R / (1 - K)$ at the input.

COMPENSATION THEOREM

V_{c} = I Δ Z, Δ I_{branch} = - \frac{I Δ Z}{Z _{t h} + Z + Δ Z}

$I$ = original branch current. $Z_{t h}$ = 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 $Δ Z$ .

Transients in RL, RC and RLC circuits

Read the notes

CONTINUITY OF THE STATE VARIABLES

v_{C} (0^{+}) = v_{C} (0^{-}), i_{L} (0^{+}) = i_{L} (0^{-})

Holds whenever $i_{C}$ and $v_{L}$ contain no impulse. It can fail in two ways: a source that itself contains an impulse (such as $δ (t)$ ), or a switching that creates a loop of capacitors and ideal voltage sources whose voltages do not already satisfy KVL, or a cut-set of inductors and ideal current sources whose currents do not already satisfy KCL (see the tip below). Put in terms of charge and flux linkage: $q_{C} = C v_{C}$ and $λ_{L} = L i_{L}$ are continuous.

FIRST DERIVATIVES AT 0+

\frac{d v _{C}}{d t}_{0^{+}} = \frac{i _{C} ( 0 ^{+} )}{C}, \frac{d i _{L}}{d t}_{0^{+}} = \frac{v _{L} ( 0 ^{+} )}{L}

Find $i_{C} (0^{+})$ and $v_{L} (0^{+})$ from the $t = 0^{+}$ circuit, using KCL and KVL with capacitors replaced by voltage sources and inductors by current sources. A second-order circuit needs this as its second initial condition. For the derivative of a resistor quantity, differentiate its KVL or KCL equation and substitute these values.

TIME CONSTANT

τ = R_{t h} C (RC), τ = \frac{L}{R _{t h}} (RL)

$R_{t h}$ is the resistance seen from the capacitor or inductor terminals with independent voltage sources shorted and independent current sources opened. Keep dependent sources and find $R_{t h}$ with a test source, $R_{t h} = v_{t es t} / i_{t es t}$ . Use the circuit that exists after the switching.

COMPLETE RESPONSE OF A FIRST-ORDER CIRCUIT WITH DC SOURCES

x (t) = x (\infty) + [x (t_{0}^{+}) - x (\infty)] e^{- (t - t_{0}) / τ}, t \geq t_{0}

Valid for any voltage or current in a first-order circuit with constant sources. $t_{0}$ is the switching instant (usually $0$ ). Use the $0^{+}$ value, not the $0^{-}$ value, for anything other than $v_{C}$ and $i_{L}$ .

SERIES RC ON A DC SOURCE

v_{C} (t) = V_{s} + (V_{0} - V_{s}) e^{- t / R C}, i (t) = \frac{V _{s} - V _{0}}{R} e^{- t / R C}

DC source $V_{s}$ switched on at $t = 0$ with $v_{C} (0^{-}) = V_{0}$ . $i$ flows from the source into the + terminal of the capacitor. $V_{0} = 0$ gives charging, $V_{s} (1 - e^{- t / R C})$ ; $V_{s} = 0$ gives discharge, $V_{0} e^{- t / R C}$ .

SERIES RL ON A DC SOURCE

i_{L} (t) = \frac{V _{s}}{R} + (I_{0} - \frac{V _{s}}{R}) e^{- R t / L}, v_{L} (t) = (V_{s} - R I_{0}) e^{- R t / L}

DC source $V_{s}$ switched on at $t = 0$ with $i_{L} (0^{-}) = I_{0}$ . $I_{0} = 0$ gives build-up, $\frac{V _{s}}{R} (1 - e^{- R t / L})$ ; $V_{s} = 0$ gives decay, $I_{0} e^{- R t / L}$ . $v_{L}$ jumps to $V_{s} - R I_{0}$ at $t = 0^{+}$ ; if the inductor was in DC steady state before switching, $v_{L} (0^{-}) = 0$ .

INITIAL SLOPE AND TIME TO REACH A VALUE

\frac{d x}{d t}_{t_{0}^{+}} = \frac{x ( \infty ) - x ( t _{0}^{+} )}{τ}, t - t_{0} = τ ln \frac{x ( t _{0}^{+} ) - x ( \infty )}{x ( t ) - x ( \infty )}

The tangent at the start reaches the final value after exactly one time constant. The second form gives the instant at which $x$ reaches a stated value between $x (t_{0}^{+})$ and $x (\infty)$ .

RISE TIME AND HALF-CHANGE TIME

t_{10 - 90} = τ ln 9 \approx 2.2 τ, t_{50} = τ ln 2 \approx 0.693 τ

For a first-order step response. The rise time is the time to go from 10 to 90 percent of the change; $t_{50}$ is the time for half the change.

SERIES RLC: CHARACTERISTIC EQUATION

s^{2} + \frac{R}{L} s + \frac{1}{L C} = 0, α = \frac{R}{2 L}, ω_{0} = \frac{1}{L C}

For the source-free series loop (a voltage source is replaced by a short). $α$ is the damping coefficient (neper frequency, in s $^{- 1}$ ) and $ω_{0}$ is the undamped natural frequency (rad/s).

PARALLEL RLC: CHARACTERISTIC EQUATION

s^{2} + \frac{1}{R C} s + \frac{1}{L C} = 0, α = \frac{1}{2 R C}, ω_{0} = \frac{1}{L C}

For the source-free parallel circuit (a current source is replaced by an open). It has the same $ω_{0}$ as the series case but a different $α$ : here a larger $R$ means less damping.

ROOTS, DAMPING RATIO AND DAMPED FREQUENCY

s_{1, 2} = - α \pm α^{2} - ω_{0}^{2}, ζ = \frac{α}{ω _{0}}, ω_{d} = ω_{0}^{2} - α^{2} = ω_{0} 1 - ζ^{2}

Both cases have the form $s^{2} + 2 α s + ω_{0}^{2} = 0$ . $ω_{d}$ is real only when $α < ω_{0}$ . The roots multiply to $s_{1} s_{2} = ω_{0}^{2}$ and add to $- 2 α$ .

CRITICAL RESISTANCE

R_{c, series} = 2 \frac{L}{C}, R_{c, parallel} = \frac{1}{2} \frac{L}{C}

A series RLC is overdamped for $R > 2 L / C$ and underdamped for $R < 2 L / C$ . A parallel RLC is the other way round: overdamped for $R < \frac{1}{2} L / C$ , underdamped for $R > \frac{1}{2} L / C$ .

SERIES RLC SWITCHED ONTO A DC SOURCE: THE CURRENT

overdamped: critically damped: underdamped: i (t) = \frac{V _{s} - V _{0}}{L ( s _{1} - s _{2} )} (e^{s_{1} t} - e^{s_{2} t}) i (t) = \frac{V _{s} - V _{0}}{L} t e^{- α t} i (t) = \frac{V _{s} - V _{0}}{ω _{d} L} e^{- α t} sin ω_{d} t

DC source $V_{s}$ applied at $t = 0$ , with $i (0^{-}) = 0$ and $v_{C} (0^{-}) = V_{0}$ ; $i$ flows from the source into the + terminal of $C$ . All three cases follow from $i (0^{+}) = 0$ and $\frac{d i}{d t} (0^{+}) = \frac{V _{s} - V _{0}}{L}$ . With $R = 0$ : $i (t) = (V_{s} - V_{0}) C / L sin ω_{0} t$ .

SERIES RLC SWITCHED ONTO A DC SOURCE: THE CAPACITOR VOLTAGE

overdamped: critically damped: underdamped: v_{C} (t) = V_{s} + (V_{s} - V_{0}) \frac{s _{2} e ^{s_{1} t} - s _{1} e ^{s_{2} t}}{s _{1} - s _{2}} v_{C} (t) = V_{s} - (V_{s} - V_{0}) (1 + α t) e^{- α t} v_{C} (t) = V_{s} - (V_{s} - V_{0}) e^{- α t} (cos ω_{d} t + \frac{α}{ω _{d}} sin ω_{d} t)

Same conditions as the current formulas above. Each form satisfies $v_{C} (0^{+}) = V_{0}$ and $\frac{d v _{C}}{d t} (0^{+}) = i (0^{+}) / C = 0$ , and $i = C \frac{d v _{C}}{d t}$ gives back the matching current. With $R = 0$ : $v_{C} (t) = V_{s} - (V_{s} - V_{0}) cos ω_{0} t$ .

UNDERDAMPED CAPACITOR VOLTAGE: FIRST PEAK AND OVERSHOOT

t_{p} = \frac{π}{ω _{d}}, v_{C} (t_{p}) = V_{s} + (V_{s} - V_{0}) e^{- α π / ω_{d}}, \frac{v _{C} ( t _{p} ) - V _{s}}{V _{s} - V _{0}} = e^{- α π / ω_{d}} = e^{- ζ π / 1 - ζ^{2}}

Same conditions, with $V_{s} > V_{0}$ , so the first extremum is the maximum (for $V_{s} < V_{0}$ the same instant gives the lowest point, below $V_{s}$ ). The extrema of $v_{C}$ fall at $t = k π / ω_{d}$ ( $k = 1, 2, \dots$ ), the instants where $i = 0$ . Successive maxima are one damped period $2 π / ω_{d}$ apart, and the overshoot above $V_{s}$ shrinks by the factor $e^{- 2 π α / ω_{d}}$ from one maximum to the next. The fractional overshoot is the peak-overshoot formula of the standard second-order system.

LC CIRCUIT (UNDAMPED OSCILLATION)

v_{C} (t) = V_{0} cos ω_{0} t, i (t) = V_{0} \frac{C}{L} sin ω_{0} t, I_{m a x} = V_{0} \frac{C}{L}

A capacitor charged to $V_{0}$ is connected across an uncharged inductor at $t = 0$ ; $i$ is the current leaving the + terminal of $C$ and entering $L$ . The energy swings between $\frac{1}{2} C V_{0}^{2}$ and $\frac{1}{2} L I_{m a x}^{2}$ every quarter period. An uncharged LC switched onto a DC source $V_{s}$ gives $v_{C} = V_{s} (1 - cos ω_{0} t)$ , which peaks at $2 V_{s}$ .

INDUCTOR WITH INITIAL CURRENT

V_{L} (s) = s L I_{L} (s) - L i_{L} (0^{-}) ⟺ I_{L} (s) = \frac{V _{L} ( s )}{s L} + \frac{i _{L} ( 0 ^{-} )}{s}

Series model: impedance $s L$ in series with a voltage source $L i_{L} (0^{-})$ that drives current in the direction of $i_{L}$ (its + terminal is on the side where $i_{L}$ leaves). Parallel model: impedance $s L$ in parallel with a current source $i_{L} (0^{-}) / s$ pointing in the direction of $i_{L}$ .

CAPACITOR WITH INITIAL VOLTAGE

V_{C} (s) = \frac{I _{C} ( s )}{s C} + \frac{v _{C} ( 0 ^{-} )}{s} ⟺ I_{C} (s) = s C V_{C} (s) - C v_{C} (0^{-})

Series model: impedance $\frac{1}{s C}$ in series with a voltage source $v_{C} (0^{-}) / s$ of the same polarity as $v_{C}$ . Parallel model: impedance $\frac{1}{s C}$ in parallel with a current source $C v_{C} (0^{-})$ whose arrow points toward the + terminal of $v_{C}$ (the charged capacitor pushes current out of its + terminal).

INITIAL- AND FINAL-VALUE THEOREMS

x (0^{+}) = s \to \infty lim s X (s), x (\infty) = s \to 0 lim s X (s)

For a rational $X (s)$ , the initial-value theorem needs $X (s)$ to be strictly proper; otherwise $x (t)$ has an impulse at $t = 0$ . The final-value theorem is valid only when every pole of $s X (s)$ lies in the open left half-plane. It gives a wrong answer for a sinusoid or a growing response.

NETWORK FUNCTION, IMPULSE RESPONSE AND STEP RESPONSE

H (s) = \frac{Y ( s )}{X ( s )}_{zero initial state}, h (t) = L^{- 1} {H (s)}, y_{step} (t) = L^{- 1} {\frac{H ( s )}{s}}, h (t) = \frac{d y _{step} ( t )}{d t}

Impulse and step responses are defined with every initial condition zero. The zero-state response to any input is $L^{- 1} {H (s) X (s)}$ , the convolution of $h (t)$ with the input. The complete response adds the zero-input response produced by the initial-condition sources. Every pole of $H (s)$ is a natural frequency of the circuit; a natural frequency can be missing from a particular $H (s)$ when it cancels against a zero, but it still appears in the zero-input response.

DELAYED INPUTS AND RECTANGULAR PULSES

f (t - T) u (t - T) \leftrightarrow e^{- s T} F (s), x (t) = V [u (t) - u (t - T)] \Rightarrow y (t) = V [y_{step} (t) - y_{step} (t - T)]

Time-shift property, with $T > 0$ . A pulse of height $V$ and width $T$ applied at zero initial state gives, by linearity and time invariance, the step response minus a copy delayed by $T$ (each copy is zero before its own step). For the series RC with output $v_{C}$ : $v_{C} = V (1 - e^{- t / R C})$ for $0 \leq t \leq T$ , and $v_{C} = V (1 - e^{- T / R C}) e^{- (t - T) / R C}$ for $t \geq T$ .

IMPULSES CHANGE THE STATE INSTANTLY

v_{C} (0^{+}) = v_{C} (0^{-}) + \frac{1}{C} \int_{0^{-}}^{0^{+}} i_{C} d t, i_{L} (0^{+}) = i_{L} (0^{-}) + \frac{1}{L} \int_{0^{-}}^{0^{+}} v_{L} d t

A current impulse $Q δ (t)$ into a capacitor raises its voltage by $Q / C$ . A voltage impulse $Λ δ (t)$ across an inductor raises its current by $Λ/ L$ . When the impulse is the only source, the circuit is source-free after $0^{+}$ , so the impulse response is the zero-input response from these new initial values.

FIRST-ORDER CIRCUIT WITH ANY SOURCE

x (t) = x_{ss} (t) + [x (0^{+}) - x_{ss} (0^{+})] e^{- t / τ}

$x_{ss} (t)$ is the steady-state response (from phasors, for a sinusoid). This generalises the DC formula, where $x_{ss} = x (\infty)$ . The transient term vanishes when $x_{ss} (0^{+}) = x (0^{+})$ .

SERIES RL SWITCHED ONTO A SINUSOID

i (t) = \frac{V _{m}}{∣ Z ∣} [sin (ω t + θ - ϕ) - sin (θ - ϕ) e^{- R t / L}], ∣ Z ∣ = R^{2} + ω^{2} L^{2}, ϕ = tan^{- 1} \frac{ω L}{R}

With $i (0^{-}) = 0$ . The steady-state current lags the voltage by $ϕ$ . The exponential (DC offset) term starts at $- \frac{V _{m}}{∣ Z ∣} sin (θ - ϕ)$ .

SERIES RC SWITCHED ONTO A SINUSOID

v_{C} (t) = \frac{V _{m}}{ω C ∣ Z ∣} [- cos (ω t + θ + ϕ) + cos (θ + ϕ) e^{- t / R C}], ∣ Z ∣ = R^{2} + \frac{1}{ω ^{2} C ^{2}}, ϕ = tan^{- 1} \frac{1}{ω R C}

With $v_{C} (0^{-}) = 0$ . The steady-state current $\frac{V _{m}}{∣ Z ∣} sin (ω t + θ + ϕ)$ leads the voltage by $ϕ$ . The current's own transient term is $- \frac{V _{m}}{∣ Z ∣} \frac{c o s ( θ + ϕ )}{ω R C} e^{- t / R C}$ , which gives $i (0^{+}) = \frac{V _{m} s i n θ}{R}$ .

SWITCHING ANGLE FOR NO TRANSIENT

RL: θ = ϕ = tan^{- 1} \frac{ω L}{R}, RC: θ = 9 0^{\circ} - ϕ = tan^{- 1} (ω R C) (or 18 0^{\circ} more)

Both come from one rule: there is no transient when the steady-state value of the state variable ( $i_{L}$ or $v_{C}$ ) at the switching instant equals its actual initial value, which is zero here. For RL, switch when the steady-state current passes through zero; for RC, switch when the steady-state capacitor voltage does.

WORST-CASE SWITCHING FOR RL

θ - ϕ = \pm 9 0^{\circ} \Rightarrow offset = \mp \frac{V _{m}}{∣ Z ∣} e^{- R t / L}, i_{peak} \to \frac{2 V _{m}}{∣ Z ∣} when ω L ≫ R

The transient is largest when the switch closes at a peak of the steady-state current. About half a cycle later, the total current's first peak approaches twice the steady-state amplitude, provided the time constant is long compared with the period.

STORED ENERGY

w_{C} = \frac{1}{2} C v_{C}^{2}, w_{L} = \frac{1}{2} L i_{L}^{2}

These depend only on the present $v_{C}$ or $i_{L}$ , which is why those two quantities are the continuous ones.

CHARGING A CAPACITOR FROM A DC SOURCE THROUGH A RESISTOR

W_{source} = C V_{s} (V_{s} - V_{0}), Δ W_{C} = \frac{1}{2} C (V_{s}^{2} - V_{0}^{2}), W_{R} = \frac{1}{2} C (V_{s} - V_{0})^{2}

From $v_{C} = V_{0}$ to the final value $V_{s}$ , through a resistor of any value. Starting uncharged, the source delivers $C V_{s}^{2}$ and exactly half of it is dissipated. If $V_{0} > V_{s}$ the capacitor discharges toward $V_{s}$ , and the same $W_{R}$ holds.

ENERGY DISSIPATED IN AN EXPONENTIAL DECAY

i (t) = I_{0} e^{- t / τ} \Rightarrow W_{R} (0 \to t) = \frac{1}{2} I_{0}^{2} R τ (1 - e^{- 2 t / τ})

The power $i^{2} R$ decays with time constant $τ /2$ . For an RL decay ( $τ = L / R$ ) the total is $\frac{1}{2} L I_{0}^{2}$ , and for an RC discharge from $V_{0}$ it is $\frac{1}{2} C V_{0}^{2}$ . After one time constant, $1 - e^{- 2} = 0.865$ of the stored energy has been dissipated.

CHARGE SHARING BETWEEN TWO CAPACITORS

V_{f} = \frac{C _{1} V _{1} + C _{2} V _{2}}{C _{1} + C _{2}}, W_{lost} = \frac{1}{2} \frac{C _{1} C _{2}}{C _{1} + C _{2}} (V_{1} - V_{2})^{2}

For $C_{1}$ at $V_{1}$ and $C_{2}$ at $V_{2}$ connected in parallel, through a switch or a resistor. $V_{1}$ and $V_{2}$ are measured across the same pair of joined terminals, so a capacitor connected with reversed polarity enters with a negative voltage. The loss does not depend on the connecting resistance $R$ , which only sets the time constant $R \frac{C _{1} C _{2}}{C _{1} + C _{2}}$ .

Sinusoidal steady state, AC power and resonance

Read the notes

AVERAGE VALUE

V_{avg} = \frac{1}{T} \int_{t_{0}}^{t_{0} + T} v (t) d t

Any periodic waveform; integrate over any one full period.

RMS VALUE

V_{rms} = \frac{1}{T} \int_{t_{0}}^{t_{0} + T} v^{2} (t) d t

Any periodic waveform. Square first, then average, then take the root.

SINUSOID

V_{rms} = \frac{V _{m}}{2} \approx 0.707 V_{m}, V_{avg, half cycle} = \frac{2 V _{m}}{π} \approx 0.637 V_{m}

$v = V_{m} sin (ω t + θ)$ . The full-cycle average is zero; the rms value does not depend on the phase $θ$ .

FORM FACTOR AND CREST (PEAK) FACTOR

k_{f} = \frac{V _{rms}}{V _{avg}}, k_{c} = \frac{V _{peak}}{V _{rms}}

$V_{avg}$ is the half-cycle average for a waveform whose full-cycle average is zero; $V_{peak}$ is the largest absolute value. For a sine, $k_{f} = π / (22) \approx 1.11$ and $k_{c} = 2 \approx 1.414$ .

RMS OF A SUM OF DIFFERENT FREQUENCIES

V_{rms} = V_{0}^{2} + n \sum V_{n, rms}^{2} = V_{0}^{2} + n \sum \frac{V _{mn}^{2}}{2}

Only when the components have different frequencies (dc counts as frequency zero): products of different frequencies average to zero over the common period. Add same-frequency terms as phasors first.

ELEMENT IMPEDANCES

Z_{R} = R, Z_{L} = j ω L, Z_{C} = \frac{1}{j ω C} = - \frac{j}{ω C}

Inductor: voltage leads current by $9 0^{\circ}$ . Capacitor: current leads voltage by $9 0^{\circ}$ . Resistor: in phase.

IMPEDANCE

Z = R + j X = ∣ Z ∣∠ ϕ, ∣ Z ∣ = R^{2} + X^{2}, ϕ = tan^{- 1} \frac{X}{R}

$X > 0$ : inductive, current lags. $X < 0$ : capacitive, current leads. $ϕ$ is the angle by which voltage leads current. The arctangent form assumes $R > 0$ (a passive impedance); for $R = 0$ , $ϕ = + 9 0^{\circ}$ or $- 9 0^{\circ}$ with the sign of $X$ .

ADMITTANCE

Y = \frac{1}{Z} = G + j B, G = \frac{R}{R ^{2} + X ^{2}}, B = \frac{- X}{R ^{2} + X ^{2}}

$G = 1/ R$ only when $X = 0$ . Susceptances: capacitor $B_{C} = ω C > 0$ , inductor $B_{L} = - 1/ (ω L) < 0$ .

COMBINATIONS AND DIVIDERS

Z_{series} = k \sum Z_{k}, Y_{parallel} = k \sum Y_{k}, V_{k} = V \frac{Z _{k}}{\sum Z}, I_{1} = I \frac{Z _{2}}{Z _{1} + Z _{2}}

Same as dc, with complex numbers: the voltage divider is for elements in series, the current divider for two parallel branches. Magnitudes add only when the phasors are in phase.

ACTIVE, REACTIVE AND APPARENT POWER

P = ∣ V ∣∣ I ∣ cos ϕ (W), Q = ∣ V ∣∣ I ∣ sin ϕ (var), ∣ S ∣ = ∣ V ∣∣ I ∣ = P^{2} + Q^{2} (VA)

rms values. $P$ is the average of $p (t)$ . $Q$ is the amplitude of the part of $p (t)$ that flows back and forth between the source and the fields of L and C, with zero average.

COMPLEX POWER

S = V I^{*} = P + j Q = ∣ I ∣^{2} Z = \frac{∣ V ∣ ^{2}}{Z ^{*}}

rms phasors, power absorbed by the load. With peak phasors, $S = \frac{1}{2} V_{m} I_{m}^{*}$ . The conjugate goes on the current; writing $V^{*} I$ instead flips the sign of $Q$ .

POWER IN EACH ELEMENT

P_{R} = ∣ I ∣^{2} R = \frac{∣ V _{R} ∣ ^{2}}{R}, Q_{L} = ∣ I ∣^{2} X_{L} = \frac{∣ V _{L} ∣ ^{2}}{X _{L}} > 0, Q_{C} = - ∣ I ∣^{2} X_{C} = - \frac{∣ V _{C} ∣ ^{2}}{X _{C}} < 0

$X_{L} = ω L$ and $X_{C} = 1/ (ω C)$ are positive magnitudes. An inductor absorbs reactive power, a capacitor delivers it, and neither absorbs average power.

POWER FACTOR

pf = cos ϕ = \frac{P}{∣ S ∣} = \frac{R}{∣ Z ∣}

Lagging when current lags voltage ( $ϕ > 0$ , $Q > 0$ , inductive); leading when current leads ( $ϕ < 0$ , $Q < 0$ , capacitive); unity at $ϕ = 0$ . Always state lagging or leading with the number.

CONSERVATION OF COMPLEX POWER

all elements \sum S_{k} = 0 \Rightarrow \sum P_{delivered} = \sum P_{absorbed}, \sum Q_{delivered} = \sum Q_{absorbed}

Every element taken with the passive sign convention. $P$ and $Q$ of loads add whether they are in series or parallel; apparent powers $∣ S ∣$ do NOT add in general (only when every load has the same power-factor angle).

SHUNT CAPACITOR FOR PF CORRECTION

Q_{C} = P (tan ϕ_{1} - tan ϕ_{2}), C = \frac{P ( tan ϕ _{1} - tan ϕ _{2} )}{ω ∣ V ∣ ^{2}}

$ϕ_{1}$ , $ϕ_{2}$ = pf angles before and after (lagging taken positive), $∣ V ∣$ = rms voltage across the capacitor, $P$ in W. For a final leading pf take $ϕ_{2}$ negative.

AVERAGE POWER WITH HARMONICS

P = V_{0} I_{0} + n = 1 \sum \infty V_{n} I_{n} cos ϕ_{n}

$V_{n}$ , $I_{n}$ = rms values of the $n$ -th harmonic; $ϕ_{n}$ = angle by which $V_{n}$ leads $I_{n}$ . A harmonic present in only one of $v$ , $i$ contributes nothing. Power factor $= P / (V_{rms} I_{rms})$ using the full rms values. If $v$ is a pure sinusoid and only $i$ is distorted, this becomes $pf = (I_{1} / I_{rms}) cos ϕ_{1}$ : distortion factor times displacement factor.

LOAD POWER

P_{L} = \frac{∣ V _{T h} ∣ ^{2} R _{L}}{( R _{T h} + R _{L} ) ^{2} + ( X _{T h} + X _{L} ) ^{2}}

$V_{T h}$ rms. Start every restricted-load problem from this expression.

CONJUGATE MATCH

Z_{L} = Z_{T h}^{*} \Rightarrow P_{L, m a x} = \frac{∣ V _{T h} ∣ ^{2}}{4 R _{T h}}

rms $V_{T h}$ (with a peak value $V_{T h, m}$ it is $V_{T h, m}^{2} / (8 R_{T h})$ ). The Thevenin source then delivers twice this power, so the efficiency measured against the Thevenin equivalent is 50%.

PURELY RESISTIVE LOAD

R_{L} = ∣ Z_{T h} ∣ = R_{T h}^{2} + X_{T h}^{2}

When the load cannot have reactance. Gives less power than the conjugate match unless $X_{T h} = 0$ .

RESONANT FREQUENCY

ω_{0} = \frac{1}{L C}, f_{0} = \frac{1}{2 π L C}

Ideal series RLC and ideal parallel RLC (frequency-independent R). The value of R does not shift it.

QUALITY FACTOR, SERIES RLC

Q = \frac{ω _{0} L}{R} = \frac{1}{ω _{0} C R} = \frac{1}{R} \frac{L}{C}

$R$ is the total series resistance, including the coil's own resistance and the source's internal resistance.

VOLTAGE MAGNIFICATION (SERIES)

∣ V_{L} ∣ = ∣ V_{C} ∣ = Q ∣ V ∣ at ω_{0}

The L and C voltages are $18 0^{\circ}$ apart and cancel, but each is $Q$ times the supply voltage.

HALF-POWER FREQUENCIES (SERIES)

ω_{1, 2} = \mp \frac{R}{2 L} + (\frac{R}{2 L})^{2} + \frac{1}{L C}, ω_{0} = ω_{1} ω_{2}

Constant-amplitude voltage source. At $ω_{1}$ and $ω_{2}$ : $∣ I ∣ = I_{m a x} / 2$ , $P = P_{m a x} /2$ , $∣ Z ∣ = 2 R$ , impedance angle $- 4 5^{\circ}$ at $ω_{1}$ (capacitive) and $+ 4 5^{\circ}$ at $ω_{2}$ (inductive). $ω_{0}$ is their exact geometric mean.

BANDWIDTH (SERIES)

BW = ω_{2} - ω_{1} = \frac{R}{L} rad/s = \frac{ω _{0}}{Q}, in Hz: \frac{R}{2 π L} = \frac{f _{0}}{Q}

Exact for any $Q$ . For high $Q$ (roughly $Q \geq 10$ ), $ω_{1, 2} \approx ω_{0} \mp BW /2$ ; the arithmetic symmetry about $ω_{0}$ is only approximate, the geometric one is exact.

QUALITY FACTOR AND BANDWIDTH, PARALLEL RLC

Q = \frac{R}{ω _{0} L} = ω_{0} C R = R \frac{C}{L}, BW = \frac{1}{R C} rad/s = \frac{ω _{0}}{Q}

Ideal parallel RLC. Current magnification: at $ω_{0}$ , $∣ I_{L} ∣ = ∣ I_{C} ∣ = Q ∣ I_{source} ∣$ . A current source with internal resistance $R_{s}$ puts $R_{s}$ in parallel with $R$ : use $R ∥ R_{s}$ in $Q$ and BW.

SERIES-PARALLEL EQUIVALENT AT ONE FREQUENCY

R_{p} = \frac{R _{s}^{2} + X _{s}^{2}}{R _{s}} = R_{s} (1 + Q^{2}), X_{p} = \frac{R _{s}^{2} + X _{s}^{2}}{X _{s}} = X_{s} (1 + \frac{1}{Q ^{2}}), Q = \frac{∣ X _{s} ∣}{R _{s}} = \frac{R _{p}}{∣ X _{p} ∣}

A series $R_{s} + j X_{s}$ and a parallel combination of $R_{p}$ and $j X_{p}$ have the same impedance at the frequency where the values are computed, and only there. For $Q ≫ 1$ , $R_{p} \approx Q^{2} R_{s}$ and $X_{p} \approx X_{s}$ , which is where the tank result $Z_{D} \approx Q^{2} R$ below comes from.

PRACTICAL TANK (COIL R, L IN PARALLEL WITH C)

ω_{0} = \frac{1}{L C} - \frac{R ^{2}}{L ^{2}}, Z_{D} = \frac{L}{C R}

$ω_{0}$ = frequency of unity pf; $Z_{D}$ = dynamic impedance (purely resistive) at that frequency. A real $ω_{0}$ exists only if $R < L / C$ . Exactly, $Z_{D} = R (1 + Q^{2})$ with the coil $Q = ω_{0} L / R$ taken at this $ω_{0}$ ; for $Q ≫ 1$ , $ω_{0} \approx 1/ L C$ and $Z_{D} \approx Q^{2} R$ .

FIRST-ORDER LOW-PASS

H (j ω) = \frac{1}{1 + j ω / ω _{c}}, ω_{c} = \frac{1}{R C} (series R, output across C) or \frac{R}{L} (series L, output across R)

At $ω_{c}$ : $∣ H ∣ = 1/ 2$ ( $- 3$ dB), phase $- 4 5^{\circ}$ . Asymptotic roll-off $- 20$ dB/decade above $ω_{c}$ .

FIRST-ORDER HIGH-PASS

H (j ω) = \frac{j ω / ω _{c}}{1 + j ω / ω _{c}}, ω_{c} = \frac{1}{R C} (series C, output across R) or \frac{R}{L} (series R, output across L)

At $ω_{c}$ : $∣ H ∣ = 1/ 2$ , phase $+ 4 5^{\circ}$ . Asymptotic slope $+ 20$ dB/decade below $ω_{c}$ .

STANDARD SECOND-ORDER FORMS

H_{L P} (s) H_{H P} (s) H_{B P} (s) H_{B S} (s) = \frac{ω _{0}^{2}}{s ^{2} + ( ω _{0} / Q ) s + ω _{0}^{2}} = \frac{s ^{2}}{s ^{2} + ( ω _{0} / Q ) s + ω _{0}^{2}} = \frac{( ω _{0} / Q ) s}{s ^{2} + ( ω _{0} / Q ) s + ω _{0}^{2}} = \frac{s ^{2} + ω _{0}^{2}}{s ^{2} + ( ω _{0} / Q ) s + ω _{0}^{2}}

Match the denominator to read $ω_{0}$ and $Q$ directly: series RLC has $ω_{0} / Q = R / L$ , parallel RLC has $ω_{0} / Q = 1/ (R C)$ . Damping ratio $ζ = 1/ (2 Q)$ . The band-pass and band-stop bandwidth is $ω_{0} / Q$ . Asymptotic slopes: low-pass $- 40$ dB/decade above $ω_{0}$ , high-pass $+ 40$ dB/decade below it, band-pass $\pm 20$ dB/decade on its two skirts.

Magnetically coupled circuits

Read the notes

TERMINAL EQUATIONS OF TWO COUPLED COILS (TIME DOMAIN)

v_{1} = L_{1} \frac{d i _{1}}{d t} \pm M \frac{d i _{2}}{d t}, v_{2} = L_{2} \frac{d i _{2}}{d t} \pm M \frac{d i _{1}}{d t}

Each coil's voltage follows the passive sign convention for that coil (+ at the terminal where its own current enters). Use $+ M$ when both currents enter (or both leave) their dotted terminals and $- M$ when one enters and the other leaves.

PHASOR (SINUSOIDAL STEADY-STATE) FORM

V_{1} = j ω L_{1} I_{1} \pm j ω M I_{2}, V_{2} = j ω L_{2} I_{2} \pm j ω M I_{1}

Same sign rule as the time-domain equations. $ω M$ is the mutual reactance in ohms.

COUPLING COEFFICIENT

k = \frac{M}{L _{1} L _{2}}, 0 \leq k \leq 1 \Rightarrow M \leq L_{1} L_{2}

$k = 0$ : no coupling; $k = 1$ : perfect coupling (all the flux of each coil links the other). Coils wound on a common iron core are typically tightly coupled ( $k$ close to 1); air-cored coils are usually more loosely coupled.

SERIES CONNECTION

L_{eq} = L_{1} + L_{2} \pm 2 M

$+ 2 M$ for series-aiding (the junction joins an undotted terminal to a dotted one); $- 2 M$ for series-opposing (dotted to dotted, or undotted to undotted). Both values are non-negative because $k \leq 1$ .

THREE COUPLED COILS IN SERIES

L_{eq} = L_{1} + L_{2} + L_{3} \pm 2 M_{12} \pm 2 M_{23} \pm 2 M_{13}

The same current flows through all three coils. Decide each pair separately with that pair's own marks: $+ 2 M_{ij}$ if the current enters both of the pair's marked terminals (or leaves both), $- 2 M_{ij}$ otherwise.

MUTUAL INDUCTANCE FROM TWO SERIES MEASUREMENTS

M = \frac{L _{aid} - L _{opp}}{4}, L_{1} + L_{2} = \frac{L _{aid} + L _{opp}}{2}

Measure the series pair, reverse one coil and measure again. The larger reading is always the aiding one.

PARALLEL CONNECTION

L_{eq} = \frac{L _{1} L _{2} - M ^{2}}{L _{1} + L _{2} \mp 2 M}

$- 2 M$ in the denominator when like terminals are joined (both dots on the same node: parallel-aiding); $+ 2 M$ when unlike terminals are joined (parallel-opposing). Coil resistances neglected. With $k = 1$ the numerator is zero, so $L_{eq} = 0$ , except for two equal coils ( $L_{1} = L_{2}$ ) in parallel-aiding, where the formula reads $0/0$ and $L_{eq} = L_{1}$ .

ENERGY STORED IN TWO COUPLED COILS

w = \frac{1}{2} L_{1} i_{1}^{2} + \frac{1}{2} L_{2} i_{2}^{2} \pm M i_{1} i_{2}

Instantaneous currents. $+ M$ when both currents enter (or both leave) their dotted terminals, $- M$ otherwise: the same rule as for the voltages.

T-EQUIVALENT OF COUPLED COILS

dots on the same side: dots on opposite sides: L_{a} = L_{1} - M, L_{b} = L_{2} - M, L_{c} = + M L_{a} = L_{1} + M, L_{b} = L_{2} + M, L_{c} = - M

$L_{a}$ and $L_{b}$ are the series arms on the coil-1 and coil-2 sides, $L_{c}$ the shunt arm to the common terminal. Valid only when the two coils share a common terminal (or their lower terminals are joined in the circuit), because the T adds that connection. "Same side" means both dots away from the common terminal or both on it. A negative inductance in the model is acceptable for analysis, although it is not a physical element.

INPUT IMPEDANCE OF A LINEAR TRANSFORMER (REFLECTED IMPEDANCE)

Z_{in} = Z_{11} + Z_{r}, Z_{r} = \frac{ω ^{2} M ^{2}}{Z _{22}}

$Z_{22}$ is the whole secondary loop impedance, load included. $Z_{r}$ does not depend on the dot positions ( $M$ enters squared), and its reactance has the opposite sign to that of $Z_{22}$ : an inductive secondary reflects as a capacitive reactance.

INPUT INDUCTANCE WITH THE SECONDARY SHORT-CIRCUITED

L_{sc} = L_{1} - \frac{M ^{2}}{L _{2}} = L_{1} (1 - k^{2})

Lossless coils, from $Z_{r}$ with $Z_{22} = j ω L_{2}$ . With the secondary open, the input is simply $j ω L_{1}$ ; comparing the two measurements gives $k$ .

IDEAL TRANSFORMER: VOLTAGE AND CURRENT RATIOS

\frac{V _{1}}{V _{2}} = \frac{N _{1}}{N _{2}} = n, \frac{I _{1}}{I _{2}} = \frac{N _{2}}{N _{1}} = \frac{1}{n}

With the convention above. If one dot is moved to the other end of its winding while the voltage and current references stay on the same physical terminals, both ratios change sign: $V_{1} / V_{2} = - n$ and $I_{1} / I_{2} = - 1/ n$ . In the two-port convention ( $I_{2}$ entering the secondary dot) the current relation reads $I_{1} = - I_{2} / n$ .

IMPEDANCE REFLECTION THROUGH AN IDEAL TRANSFORMER

Z_{in} = \frac{V _{1}}{I _{1}} = n^{2} Z_{L}

$Z_{L}$ connected across the secondary; independent of the dot positions. In the other direction, an impedance on the primary side appears on the secondary side as $Z / n^{2}$ .

POWER CONSERVATION IN AN IDEAL TRANSFORMER

V_{1} I_{1}^{*} = V_{2} I_{2}^{*}

It neither stores nor dissipates energy: complex power in equals complex power out (convention above).

MAXIMUM POWER TRANSFER THROUGH AN IDEAL TRANSFORMER

n^{2} R_{L} = R_{s} \Rightarrow n = \frac{R _{s}}{R _{L}}

Resistive source $R_{s}$ on the primary, resistive load $R_{L}$ on the secondary, $n = N_{1} / N_{2}$ . The load then receives $∣ V_{s} ∣^{2} / (4 R_{s})$ , with $V_{s}$ the rms source voltage.

Two-port networks

Read the notes

Z (OPEN-CIRCUIT IMPEDANCE) PARAMETERS

[V_{1} V_{2}] = [z_{11} z_{21} z_{12} z_{22}] [I_{1} I_{2}], z_{11} z_{21} = \frac{V _{1}}{I _{1}}_{I_{2} = 0}, = \frac{V _{2}}{I _{1}}_{I_{2} = 0}, z_{12} z_{22} = \frac{V _{1}}{I _{2}}_{I_{1} = 0} = \frac{V _{2}}{I _{2}}_{I_{1} = 0}

All four are in ohms, and each is measured with the other port open. $z_{11}$ and $z_{22}$ are driving-point (input and output) impedances. $z_{12}$ and $z_{21}$ are transfer impedances.

Y (SHORT-CIRCUIT ADMITTANCE) PARAMETERS

[I_{1} I_{2}] = [y_{11} y_{21} y_{12} y_{22}] [V_{1} V_{2}], y_{11} y_{21} = \frac{I _{1}}{V _{1}}_{V_{2} = 0}, = \frac{I _{2}}{V _{1}}_{V_{2} = 0}, y_{12} y_{22} = \frac{I _{1}}{V _{2}}_{V_{1} = 0} = \frac{I _{2}}{V _{2}}_{V_{1} = 0}

All four are in siemens, and each is measured with the other port shorted. $[y] = [z]^{- 1}$ whenever both exist. In general $y_{11} \neq = 1/ z_{11}$ , because the two are measured under different terminations.

H (HYBRID) PARAMETERS

[V_{1} I_{2}] = [h_{11} h_{21} h_{12} h_{22}] [I_{1} V_{2}], h_{11} h_{21} = \frac{V _{1}}{I _{1}}_{V_{2} = 0}, = \frac{I _{2}}{I _{1}}_{V_{2} = 0}, h_{12} h_{22} = \frac{V _{1}}{V _{2}}_{I_{1} = 0} = \frac{I _{2}}{V _{2}}_{I_{1} = 0}

$h_{11}$ is the short-circuit input impedance (ohms), $h_{12}$ the open-circuit reverse voltage gain, $h_{21}$ the short-circuit forward current gain, and $h_{22}$ the open-circuit output admittance (siemens). Port 2 is shorted for $h_{11}$ and $h_{21}$ ; port 1 is open for $h_{12}$ and $h_{22}$ .

G (INVERSE HYBRID) PARAMETERS

[I_{1} V_{2}] = [g_{11} g_{21} g_{12} g_{22}] [V_{1} I_{2}], g_{11} g_{21} = \frac{I _{1}}{V _{1}}_{I_{2} = 0}, = \frac{V _{2}}{V _{1}}_{I_{2} = 0}, g_{12} g_{22} = \frac{I _{1}}{I _{2}}_{V_{1} = 0} = \frac{V _{2}}{I _{2}}_{V_{1} = 0}

$[g] = [h]^{- 1}$ whenever both exist. $g_{11}$ is the open-circuit input admittance (siemens), $g_{12}$ the short-circuit reverse current gain, $g_{21}$ the open-circuit forward voltage gain, and $g_{22}$ the short-circuit output impedance (ohms).

ABCD (TRANSMISSION) PARAMETERS

[V_{1} I_{1}] = [A C B D] [V_{2} - I_{2}], A C = \frac{V _{1}}{V _{2}}_{I_{2} = 0}, = \frac{I _{1}}{V _{2}}_{I_{2} = 0}, B D = - \frac{V _{1}}{I _{2}}_{V_{2} = 0} = - \frac{I _{1}}{I _{2}}_{V_{2} = 0}

Written out, $V_{1} = A V_{2} - B I_{2}$ and $I_{1} = C V_{2} - D I_{2}$ . $A$ and $D$ are dimensionless, $B$ is in ohms and $C$ in siemens. $A$ and $C$ are measured with port 2 open, $B$ and $D$ with port 2 shorted. The minus sign is there because $- I_{2}$ is the current that leaves port 2 and enters whatever is connected to it.

CONDITION FOR RECIPROCITY

z_{12} = z_{21} ⟺ y_{12} = y_{21} ⟺ h_{12} = - h_{21} ⟺ g_{12} = - g_{21} ⟺ A D - B C = 1 ⟺ A^{'} D^{'} - B^{'} C^{'} = 1

Any one of these implies all the others (for the sets that exist). A reciprocal two-port has only three independent parameters.

CONDITION FOR SYMMETRY

z_{11} = z_{22} ⟺ y_{11} = y_{22} ⟺ A = D ⟺ Δ_{h} = h_{11} h_{22} - h_{12} h_{21} = 1 ⟺ Δ_{g} = 1

These express electrical symmetry of the two ports, and they hold whether or not the network is reciprocal. For a network that is also reciprocal, only two parameters are independent (for example $z_{11}$ and $z_{12}$ ). Ports that can be swapped with no change at all need both $z_{11} = z_{22}$ and $z_{12} = z_{21}$ .

DETERMINANTS AND QUICK IDENTITIES

Δ_{z} = z_{11} z_{22} - z_{12} z_{21}, Δ_{y} = \frac{1}{Δ _{z}}, Δ_{T} = A D - B C = \frac{z _{12}}{z _{21}} = \frac{y _{12}}{y _{21}} = - \frac{h _{12}}{h _{21}}

Use $Δ_{T}$ for a quick reciprocity check from any set. The determinant of a product is the product of the determinants, so after multiplying the ABCD matrices of reciprocal sections (built from $R$ , $L$ , $C$ , coupled coils and ideal transformers), $Δ_{T}$ must still equal 1. That catches arithmetic slips.

Z AND ABCD

A = \frac{z _{11}}{z _{21}}, B = \frac{Δ _{z}}{z _{21}}, C = \frac{1}{z _{21}}, D = \frac{z _{22}}{z _{21}} z_{11} = \frac{A}{C}, z_{12} = \frac{Δ _{T}}{C}, z_{21} = \frac{1}{C}, z_{22} = \frac{D}{C}

$[z]$ exists only if $C \neq = 0$ ; $[T]$ exists only if $z_{21} \neq = 0$ .

Y AND ABCD

A = - \frac{y _{22}}{y _{21}}, B = - \frac{1}{y _{21}}, C = - \frac{Δ _{y}}{y _{21}}, D = - \frac{y _{11}}{y _{21}} y_{11} = \frac{D}{B}, y_{12} = - \frac{Δ _{T}}{B}, y_{21} = - \frac{1}{B}, y_{22} = \frac{A}{B}

$[y]$ exists only if $B \neq = 0$ ; $[T]$ exists only if $y_{21} \neq = 0$ .

H FROM Z AND FROM Y

h_{11} = \frac{Δ _{z}}{z _{22}} = \frac{1}{y _{11}}, h_{12} = \frac{z _{12}}{z _{22}} = - \frac{y _{12}}{y _{11}}, h_{21} = - \frac{z _{21}}{z _{22}} = \frac{y _{21}}{y _{11}}, h_{22} = \frac{1}{z _{22}} = \frac{Δ _{y}}{y _{11}}

Note the pairs that are true reciprocals, each pair measured under the same termination: $h_{11} = 1/ y_{11}$ (port 2 shorted), $h_{22} = 1/ z_{22}$ (port 1 open), $g_{11} = 1/ z_{11}$ (port 2 open) and $g_{22} = 1/ y_{22}$ (port 1 shorted).

G FROM Z, FROM Y AND FROM ABCD

g_{11} g_{21} = \frac{1}{z _{11}} = \frac{Δ _{y}}{y _{22}} = \frac{C}{A}, = \frac{z _{21}}{z _{11}} = - \frac{y _{21}}{y _{22}} = \frac{1}{A}, g_{12} g_{22} = - \frac{z _{12}}{z _{11}} = \frac{y _{12}}{y _{22}} = - \frac{Δ _{T}}{A} = \frac{Δ _{z}}{z _{11}} = \frac{1}{y _{22}} = \frac{B}{A}

$g_{21} = 1/ A$ is the open-circuit forward voltage gain. $[g]$ exists only if $z_{11} \neq = 0$ (equivalently $A \neq = 0$ ).

T NETWORK AND T EQUIVALENT

[z] = [Z_{a} + Z_{c} Z_{c} Z_{c} Z_{b} + Z_{c}] Z_{a} = z_{11} - z_{12}, Z_{b} = z_{22} - z_{12}, Z_{c} = z_{12} = z_{21}

$Z_{a}$ is the series arm at port 1, $Z_{b}$ the series arm at port 2, and $Z_{c}$ the shunt arm to the common terminal. Every reciprocal two-port that has a $[z]$ has this T equivalent, though an arm may come out negative (not realisable with positive elements). The T joins the two lower terminals, so it is equivalent only in its port voltages and currents.

PI NETWORK AND PI EQUIVALENT

[y] = [Y_{a} + Y_{c} - Y_{c} - Y_{c} Y_{b} + Y_{c}] Y_{a} = y_{11} + y_{12}, Y_{b} = y_{22} + y_{12}, Y_{c} = - y_{12} = - y_{21}

$Y_{a}$ is the shunt arm at port 1, $Y_{b}$ the shunt arm at port 2, and $Y_{c}$ the series arm between the upper terminals. Every reciprocal two-port that has a $[y]$ has this $π$ equivalent, with the same caveats as the T.

COUPLED COILS

[z] = [j ω L_{1} j ω M j ω M j ω L_{2}], T equivalent: L_{1} - M, L_{2} - M (series arms), M (shunt arm)

Coil $L_{1}$ across port 1 and coil $L_{2}$ across port 2, mutual inductance $M$ , with both dots at the upper terminals, so both port currents enter dotted terminals. Moving one dot to its lower terminal changes $M$ to $- M$ everywhere. In the $s$ -domain replace $j ω$ by $s$ . The T equivalent (which may need a negative inductance) holds only if the two lower terminals may be joined.

SYMMETRIC LATTICE

z_{11} = z_{22} = \frac{Z _{b} + Z _{a}}{2}, z_{12} = z_{21} = \frac{Z _{b} - Z _{a}}{2}

Series arms $Z_{a}$ join upper terminal to upper terminal and lower terminal to lower terminal; cross arms $Z_{b}$ join each upper terminal to the lower terminal of the other port. Its transfer impedance changes sign when $Z_{a} > Z_{b}$ .

PARAMETER ADDITION

series: [z] = [z_{a}] + [z_{b}], parallel: [y] = [y_{a}] + [y_{b}], series-parallel: [h] = [h_{a}] + [h_{b}], parallel-series: [g] = [g_{a}] + [g_{b}]

Valid only when each sub-network still satisfies the port condition after the connection (see the validity note below).

CASCADE CONNECTION

[A C B D] = [A_{a} C_{a} B_{a} D_{a}] [A_{b} C_{b} B_{b} D_{b}], n identical sections: [T]^{n}

$N_{a}$ is the section nearer the source; keep the order. Always valid, because the current $- I_{2}$ leaving $N_{a}$ is exactly the current entering $N_{b}$ . That is why the minus sign is built into the ABCD definition.

INPUT IMPEDANCE OF A TERMINATED TWO-PORT

Z_{in} = \frac{V _{1}}{I _{1}} = z_{11} - \frac{z _{12} z _{21}}{z _{22} + Z _{L}} = \frac{A Z _{L} + B}{C Z _{L} + D} = h_{11} - \frac{h _{12} h _{21} Z _{L}}{1 + h _{22} Z _{L}}, Y_{in} = y_{11} - \frac{y _{12} y _{21}}{y _{22} + Y _{L}}

Port 2 terminated in $Z_{L}$ (so $V_{2} = - I_{2} Z_{L}$ ). As $Z_{L} \to \infty$ this becomes $z_{11} = A / C$ , the open-circuit input impedance; at $Z_{L} = 0$ it becomes $1/ y_{11} = h_{11} = B / D$ , the short-circuit input impedance.

VOLTAGE AND CURRENT GAIN WITH A LOAD

\frac{V _{2}}{V _{1}} = \frac{z _{21} Z _{L}}{z _{11} Z _{L} + Δ _{z}} = \frac{Z _{L}}{A Z _{L} + B} = \frac{- y _{21}}{y _{22} + Y _{L}}, \frac{I _{2}}{I _{1}} = - \frac{z _{21}}{z _{22} + Z _{L}} = \frac{- 1}{C Z _{L} + D} = \frac{h _{21}}{1 + h _{22} Z _{L}}

Same termination, with port 1 driven directly by $V_{1}$ . The current delivered to the load is $- I_{2}$ , so the load-current gain is $- I_{2} / I_{1}$ . With the output open, $V_{2} / V_{1} = 1/ A = z_{21} / z_{11}$ .

THEVENIN EQUIVALENT AT PORT 2

V_{t h} = \frac{z _{21} V _{s}}{z _{11} + Z _{s}} = \frac{V _{s}}{A + C Z _{s}}, Z_{t h} = z_{22} - \frac{z _{12} z _{21}}{z _{11} + Z _{s}} = \frac{D Z _{s} + B}{C Z _{s} + A}

Port 1 driven by $V_{s}$ in series with $Z_{s}$ . For $Z_{t h}$ , set $V_{s}$ to zero, which leaves $Z_{s}$ across port 1. With $V_{s}$ and $Z_{s}$ fixed, the load receives maximum power when $Z_{L} = Z_{t h}^{*}$ .

OPEN- AND SHORT-CIRCUIT IMPEDANCES

Z_{oc 1} = z_{11} = \frac{A}{C}, Z_{sc 1} = \frac{1}{y _{11}} = \frac{B}{D}, Z_{oc 2} = z_{22} = \frac{D}{C}, Z_{sc 2} = \frac{1}{y _{22}} = \frac{B}{A}

Each is measured at one port with the other port open (oc) or shorted (sc). For any two-port, $Z_{oc 1} / Z_{sc 1} = Z_{oc 2} / Z_{sc 2} = A D / (B C)$ . So only three of the four are independent.

IMAGE IMPEDANCES

Z_{i 1} = Z_{oc 1} Z_{sc 1} = \frac{A B}{C D}, Z_{i 2} = Z_{oc 2} Z_{sc 2} = \frac{B D}{A C}, symmetric (A = D) : Z_{0} = \frac{B}{C}

The image impedances are the pair for which terminating port 2 in $Z_{i 2}$ makes the input impedance at port 1 equal to $Z_{i 1}$ , and terminating port 1 in $Z_{i 1}$ makes the impedance at port 2 equal to $Z_{i 2}$ . For a symmetric network both equal the characteristic impedance $Z_{0}$ : a load $Z_{0}$ is reproduced at the input. For a reciprocal network the image transfer constant $θ$ satisfies $cosh θ = A D$ , which becomes $cosh θ = A$ when the network is also symmetric.

Balanced three-phase circuits

Read the notes

BALANCED POSITIVE-SEQUENCE (ABC) SET

V_{an} = V_{p} ∠ 0^{\circ}, V_{bn} = V_{p} ∠ - 12 0^{\circ}, V_{c n} = V_{p} ∠ + 12 0^{\circ}

Phase b lags phase a by $12 0^{\circ}$ and phase c lags phase b by $12 0^{\circ}$ . For negative sequence (acb) the angles of $V_{bn}$ and $V_{c n}$ are interchanged: $V_{bn} = V_{p} ∠ + 12 0^{\circ}$ , $V_{c n} = V_{p} ∠ - 12 0^{\circ}$ .

OPERATOR A AND THE ZERO SUM OF A BALANCED SET

a = 1∠12 0^{\circ} = - \frac{1}{2} + j \frac{3}{2}, 1 + a + a^{2} = 0

For abc: $V_{bn} = a^{2} V_{an}$ and $V_{c n} = a V_{an}$ . Hence any balanced set sums to zero, $V_{an} + V_{bn} + V_{c n} = 0$ , and the same holds for a balanced set of currents.

STAR: LINE AND PHASE RELATIONS (ABC)

V_{ab} = 3 V_{an} ∠ + 3 0^{\circ}, V_{L} = 3 V_{p h}, I_{L} = I_{p h}

Each line voltage leads the phase voltage of its first subscript by $3 0^{\circ}$ ( $V_{b c}$ leads $V_{bn}$ , $V_{c a}$ leads $V_{c n}$ ). For negative sequence the magnitudes are unchanged but $V_{ab} = 3 V_{an} ∠ - 3 0^{\circ}$ .

DELTA: LINE AND PHASE RELATIONS (ABC)

I_{a} = 3 I_{ab} ∠ - 3 0^{\circ}, I_{L} = 3 I_{p h}, V_{L} = V_{p h}

Phase currents $I_{ab}$ , $I_{b c}$ , $I_{c a}$ are taken in the direction of their subscripts; each line current lags, by $3 0^{\circ}$ , the phase current leaving that line's corner ( $I_{b}$ lags $I_{b c}$ , $I_{c}$ lags $I_{c a}$ ). For negative sequence, $I_{a} = 3 I_{ab} ∠ + 3 0^{\circ}$ .

BALANCED STAR-DELTA CONVERSION

Z_{Y} = \frac{Z _{Δ}}{3}, Z_{Δ} = 3 Z_{Y}

Identical impedances only. The division is by a real number, so the impedance angle $ϕ$ , and hence the power factor, is unchanged.

THREE-PHASE ACTIVE POWER

P = 3 V_{p h} I_{p h} cos ϕ = 3 V_{L} I_{L} cos ϕ = 3 I_{p h}^{2} R_{p h}

Balanced load, star or delta; rms values; $ϕ$ = angle of the per-phase impedance. Pair the delta resistance with the delta phase current, or the star-equivalent resistance with the line current.

THREE-PHASE REACTIVE POWER

Q = 3 V_{p h} I_{p h} sin ϕ = 3 V_{L} I_{L} sin ϕ = 3 I_{p h}^{2} X_{p h}

$Q > 0$ (absorbed) for an inductive, lagging load; $Q < 0$ for a capacitive, leading load.

COMPLEX AND APPARENT POWER, POWER FACTOR

S = 3 V_{p h} I_{p h}^{*} = P + j Q, ∣ S ∣ = 3 V_{L} I_{L} = P^{2} + Q^{2}, pf = cos ϕ = \frac{P}{∣ S ∣}

$V_{p h}$ and $I_{p h}$ are the phasors of the same phase and $^{*}$ is the complex conjugate. $∣ S ∣$ is in VA (or kVA); a three-phase kVA figure means this total for all three phases, not the per-phase value.

INSTANTANEOUS POWER OF A BALANCED LOAD

p (t) = v_{a} i_{a} + v_{b} i_{b} + v_{c} i_{c} = 3 V_{p h} I_{p h} cos ϕ (constant)

Balanced sinusoidal voltages and a balanced load; $V_{p h}$ , $I_{p h}$ rms. The total does not vary with time and equals the average power $P$ .

TWO-WATTMETER SUM (ANY THREE-WIRE LOAD)

P = W_{1} + W_{2}

Readings are taken with their signs. Valid for balanced and unbalanced three-wire loads.

WATTMETER READINGS FOR A BALANCED LOAD

W_{1} = V_{L} I_{L} cos (3 0^{\circ} - ϕ), W_{2} = V_{L} I_{L} cos (3 0^{\circ} + ϕ)

Balanced sinusoidal system. The current coils, in lines a and c, carry the line currents towards the load; the pressure coils measure $V_{ab}$ and $V_{c b}$ (each coil's $\pm$ terminal on its own current-coil line, the other end on line b). For abc sequence the meter in line c reads $W_{1}$ and the meter in line a reads $W_{2}$ . The formulas hold for either sign of $ϕ$ ( $ϕ > 0$ lagging, $ϕ < 0$ leading), so a leading load, or a reversed sequence, swaps which meter reads more.

READINGS IN TERMS OF P AND Q (BALANCED LOAD)

W_{1} = \frac{P}{2} + \frac{Q}{2 3}, W_{2} = \frac{P}{2} - \frac{Q}{2 3}

The same pair written with $P = 3 V_{L} I_{L} cos ϕ$ and $Q = 3 V_{L} I_{L} sin ϕ$ ; handy for predicting the readings when $P$ and $Q$ (or $P$ and the power factor) are given. $W_{2} < 0$ exactly when $Q > 3 P$ , i.e. $ϕ > 6 0^{\circ}$ .

POWER-FACTOR ANGLE AND REACTIVE POWER FROM THE READINGS

tan ϕ = \frac{3 ( W _{1} - W _{2} )}{W _{1} + W _{2}}, Q = 3 (W_{1} - W_{2})

Balanced load only. Follows from $W_{1} + W_{2} = 3 V_{L} I_{L} cos ϕ$ and $W_{1} - W_{2} = V_{L} I_{L} sin ϕ$ . The two readings alone fix $∣ ϕ ∣$ ; whether the load lags or leads needs either the nature of the load or knowledge of which physical meter reads $W_{1}$ , which depends on the connection and the phase sequence.

POWER FACTOR DIRECTLY FROM THE READINGS

cos ϕ = \frac{W _{1} + W _{2}}{2 W _{1}^{2} - W _{1} W _{2} + W _{2}^{2}}

Balanced load; the same content as the $tan ϕ$ form. Insert a negative reading with its sign.

THREE-WATTMETER AND ONE-WATTMETER METHODS

P = W_{a} + W_{b} + W_{c} (four-wire, any load), P = 3 W (balanced load, one meter)

Three-wattmeter method: one meter per line, each pressure coil from its line to the neutral, so each meter reads one phase power; valid for balanced and unbalanced loads. One-wattmeter method: for a balanced load, a single meter with its current coil in one line and its pressure coil from that line to the neutral (or to an artificial neutral formed by three equal impedances, one of them the pressure coil) reads one-third of the total.

CAPACITOR KVAR REQUIRED

Q_{C} = P (tan ϕ_{1} - tan ϕ_{2})

Total for all three phases; $ϕ_{1}$ is the original angle and $ϕ_{2}$ the target angle, both lagging. For a unity target, $Q_{C}$ equals the original $Q$ .

CAPACITANCE PER PHASE OF THE BANK

C_{Y} = \frac{Q _{C}}{ω V _{L}^{2}}, C_{Δ} = \frac{Q _{C}}{3 ω V _{L}^{2}} = \frac{C _{Y}}{3}

$V_{L}$ = line voltage (rms), $ω = 2 π f$ . Each star capacitor sees $V_{L} / 3$ and supplies $Q_{C} /3 = ω C_{Y} V_{L}^{2} /3$ ; each delta capacitor sees the full $V_{L}$ and supplies $Q_{C} /3 = ω C_{Δ} V_{L}^{2}$ .

NEUTRAL-SHIFT VOLTAGE (UNBALANCED STAR, NO NEUTRAL)

V_{N^{'} N} = \frac{Y _{a} V _{an} + Y _{b} V _{bn} + Y _{c} V _{c n}}{Y _{a} + Y _{b} + Y _{c}}, Y_{k} = \frac{1}{Z _{k}}

Nodal equation at N', with the source phase voltages measured from N; any line impedance is included in $Z_{k}$ . It is zero when $Y_{a} = Y_{b} = Y_{c}$ , which recovers the balanced case. If N' and N are joined through a neutral impedance $Z_{n}$ , add $Y_{n} = 1/ Z_{n}$ to the denominator.