charge

6.2415e18 electrons = 1 coulomb = 1 C

current
ampere = A = coulomb/second
for DC
I = ∆q / ∆t
for variable current
i(t) = dq(t) / dt
current is the net charge flow through a cross sectional area of a material
positive flow of current is in the opposite direction of the flow of electrons
labeling two-terminal devices with variables does not assign actual values of + or –
once the + or – is known through analysis, the variable’s value will show that
ac current is rectified – current is always absolute value of true ac

sinusoidal waveform
Asin(ωt + φ) + B
A = amplitude, ω = frequency, t = time, φ = horizontal phase angle, B = vertical shift
peak value = 2*Amp, period = T = 1/f = 2π/ω

potential = voltage = V = J/C

V = IR
vR(t) = iR(t)R(t)
current flows from higher potential to lower potential, + end to – end
Vab is the voltage drop from point a to b, if Vab = +, then point a is at a higher potential than b
symbols on a circuit diagram – +’s, –’s, subscripts are all relative and are used as references

power = P = Watt = W
rate of energy = W = J/s = QV = dW(t)/dt
power absorbed = P = IV
P = IABVAB = I2 R = V2 / R
the power delivered by the source is the power dissipated by the rest of the circuit
sum(Pcircuit ) = 0
when P of active element is + , the device dissipates power, when P is – , it delivers power

energy = joule = J = work = W

energy absorbed = W = VIT = PT
work = integral(P)dt

ideal voltage source, independent voltage source, constant voltage regardless of circuit, Rint = 0
non- ideal is usually mapped as an ideal with a resistor in series
ideal current source, independent current source, constant current regardless of circuit, Rint = ∞
non- ideal is usually mapped as an ideal with a resistance in parallel
dependent voltage source, voltage depends on another element in the circuit
dependent current source, current depends on another element in the circuit
controlled sources allow the possibility of negative resistances

prepared by michael c. pappas – michaelcpappas.com

 voltage controlled voltage source
VCVS, voltage depends on the voltage, v(t) = µvx . µ = voltage gain

current controlled voltage source

CCVS, voltage depends on the current, v(t) = rmvx . rm = Ω = transfer resistance

voltage controlled current source

VCCS, current depends on the voltage, i(t) = gmvx . gm = S = transfer conductance

current controlled current source

CCCS, current depends on the current, i(t) = βix . β = current gain

resistance, ohm = Ω
ohm’s law VAB = IABR, v- i characteristic, ideal linear resistor, v = f(i) or i = f(v)
resistors always absorb power

when does ohm’s law apply

the physical dimensions of the circuit element must be small relative to the wavelength of the
sinusoidal frequencies expected to drive the device, called a lumped circuit elements, the
opposite of lumped circuits is called a distributed circuit

resistivity = R = ρL/A, Ohm- meters

conductance, siemens = S, mho = Ω −1
short circuit = zero resistance, infinite conductance, modeled as ideal wire
open circuit = infinite resistance, zero conductance, modeled as open switch
memoryless device – v = f(i) or i = f(v)
memory device – voltage and current relationships satisfy differential equation

nodal and loop analysis are needed when there is not a parallel-series circuit

kirchhoff’s current law (KCL), nodal analysis

maximum number of equations to n nodes and b braches is (n – 1) equations
for lumped circuits, the algebraic sum of the currents entering a node is zero
for lumped circuits, the algebraic sum of the currents leaving a node is zero
current sources cannot be in series
make supernodes for floating voltage sources, neither node is connected to the reference
KCL applies for gaussian curves drawn around elements
nodal analysis usually involves less unknowns in larger circuits than loop analysis
nodal analysis is easy when the circuit contains only resistances, interdependent current
sources, and VCCS’s or R-Is-gm networks

kirchhoff’s voltage law, loop analysis

maximum number of equations to n nodes and b braches is (b – n + 1) equations
for lumped circuits, the algebraic sum of the voltages around a closed path is zero
for lumped circuits, the algebraic sum of the voltages around a closed node sequence is zero

prepared by michael c. pappas – michaelcpappas.com

 the sum of the node voltages in a closed path is zero
the voltage of vAB = vAB + … + vYZ
vab = va – vb
make supermeshes when current sources are common to two more loops
modified loop analysis is easier actually
voltages sources are never connected in parallel
make a reference node 0 V to be ground
nodal analysis is usually better to choose when circuits get bigger

in series in parallel

i1 = i2 = ... v1 = v2 = ...
veq = v1 + v2 + ... ieq = i1 + i2 + ...
1
Req = R1 + R2 + ... Req =
1/ R1 + 1/ R2 + ...
1
Leq = L1 + L2 + ... Leq =
1/ L1 + 1/ L2 + ...
R1 1/ R1
v1 = vin i1 = iin
R1 + R2 + ... 1/ R1 + 1/ R2 + ...
1
Ceq = Ceq = C1 + C2 + ...
1/ C1 + 1/ C2 + ...

when there are no dependent sources present in the circuit, the matrix of the node equation is
always symmetric in the natural order

when there are no dependent sources, the node equations can be written by inspection, the i-th
entry is simply the sum of the independent source currents injected into the i-th node where KCL
is applied

nonideal sources can be modeled as an ideal independent voltage source (with zero resistance) in
series with a resistor or an ideal independent current source (with infinite resistance) in parallel
with a resistor

to measure the voltage of an battery, supply a typical current flow through the battery so that the
internal resistance will be accounted for. internal resistance is not seen by very small current
supplied by a voltmeter

modified loop analysis

add auxiliary voltage variables for current sour ces common to two or more loops
apply kvl
write constraint equations determined by the dependent current sources
solve in matrix form

prepared by michael c. pappas – michaelcpappas.com

modified nodal analysis

the unknown are the usual nodal voltages plus auxiliary currents
auxiliary currents:
current through independent voltage sources
current through dependent voltage sources
current through short-circuit elements (0 V independent)
controlling currents of dependent sources
currents declared as output quantities
for every element x whose current source has been chosen as an auxiliary current:
temporarily replace that element by an independent current source having the value Ix
write an equation that relates the elements of the original to the new setup
write matrix the matrix

writing loop equations for non-planar circuits

just keep track of the wires, when a wire hops over another, your loop equation should too

if the voltage source is not connected to ground, then enclose it in a gaussian surface, if a
supernode is created, one of the variables in it will need to be directed related to the other
known one.

if an independent or dependent current source is common in two current loops, then you will
right one current in terms of another using the source, you might also need to create auxiliary
voltage variables

microprocessors used in everything

microprocessors interact with analog
analog produces continuous output and input signals for objectives
microprocessors process numbers
analog to digital, A/D converter, digital to analog, D/A converter
operational amplifier, used in A/D and D/A converters, it is an active element, it supplies energy
voltage gain = A = Vout / Vin
vd = v+ – v–

ideal op amp

usually modeled with a VCVS, Rout being a wire, and Rin being open
does not load the circuit
Rin = ∞ Rout = 0
the full output voltage appears across any circuit connected to the output part
v+ = v– A= ∞
no current runs through
i+ = 0 i– = 0
an ideal op amp supplies voltage to a resistance without letting the resistance load the circuit

prepared by michael c. pappas – michaelcpappas.com

 nodal analysis is used very often, label two terminals with the same nodal voltage, make sure
you do not imply that current runs through the op amp while setting up nodal analysis, you
will eventual solve for the two terminal nodal voltage in the two equations the op amp
makes, then set those two equations equal to each other, do not assume too many zeros and
infinities with op amp without doing the analysis

nonideal op amp
vout = Avd

inverting op amp
input and output voltages are always positive

noninverting
does not reveres polarity

voltage follower op amp

used to prevent the load, RL, from drawing current directly from the source by vin
it supplies current directly to the load

difference op amp
used to compare voltages, or to amplify the difference
magnitude scaling, scale the ratio of the resistance and achieve the same gain
useful to make the resistance values more practical

summing amplifiers, multi- input single-output amplifiers

general scale of multi- input single-output

Vout = −( a1Va1 + a2Va 2 + ... + anVan ) + (b1Vb1 + b2Vb2 + ... + bmVbm )

feedback resistor is generalize to 1 Ω, then the resistances would be scaled to more practical
values without changing the gain

prototype design
set Gf = 1S
Gai = ai S for i = 1,…n
Gbi = bi S for i = 1,…m
total conductance incident on the inverting terminal will need to be the same as the total
conductance incident on the no ninverting terminal by a proper choice of ∆G and Gground

compute Gground and ∆G so the total conductance on the two terminals are equal
define δ = (1 + a1 + ... + an )− (b1 + ... + bm )
if δ > 0, then set Gground = δ and ∆G = 0
if δ ≤ 0, then set Gground = some value so that Gground = 1 S and ∆G = |δ| + Gground

prepared by michael c. pappas – michaelcpappas.com

scaling to achieve practical element values

multiply all the conductance incidences at the inverting and non- inverting terminals by a
constant Ka and Kb respectively, good if Ka = Kb

saturation and the active region of the op amp

when input increases or decreases beyond a certain range, the output will clamp at minimum
or maximum voltage levels, in between that is the active region

the critical threshold voltages of vd, difference between the terminals, where saturation
occurs is ± Vsat /A

comparator
used to compare input voltage with reference voltage, it utilizes the saturation levels

a/d converter
converts signals to a discrete quantization

d/a converter
converts a binary number
[bn−1 ,..., b2 , b1b0 ] , where bi is either 1 or 0
to an output voltage
vout = [bn−1 2 n−1 + ...b2n 2 + b1 n 1 + b0 n0 ]E0 , E0 is some constant
the circuit that allows the voltage to be converted is a R-2R ladder network
voltage division is applied and switches are used to direct current through different branches

linearity theorem, for linear circuits

output voltage or current is linearly related to all of the independent sources by the equation:
y = a1u1 + a2u2 + ... + amum
where y is output, a’s are dimensionless quantities and u’s are the independent sources

superposition, for linear circuits

finding the contribution from one independent source by setting all other to zero
this is like setting up a matrix table

if there are fixed sources, then all of them are represented by one constant, the fixed sources
contribute a fixed amount to the circuit, the constant
vout = Ai s + B
superposition does not apply to power calculations
when setting values to zero, treat voltage sources as wires and current sources as open
leave dependent sources alone and calculate the contribution in terms of those values

prepared by michael c. pappas – michaelcpappas.com

proportionality property
when any independent sources are acting alone, the output is proportional to that single input
if a linear circuit is dependent only upon one sources, then when that source is multiplied by
a constant, the current and voltage of the circuit will also be multiplied by that constant

source transformation
when a voltage source in series with a resistor is converted to a current in parallel with that
same resistor value, or the other way around, this is the equivalent two-terminal network, it
applies to both dependent and independent voltage and current sources

when doing source transformation on dependent sources, it is important to keep the

controlling source unaffected, not unchanged, while transforming the circuit

make little square circuits when the sources are deep in the network

modified superposition analysis

when there are dependent sources, the analysis contains more variables, this modified
analysis shows how to deal with this

replace the dependent source with an independent source and label it

use superposition to solve for the unknown in terms of the different sources alone
also solve for the controlling variables of the replaced dependent source
restore the constraint replacement

this method can also be used to replace R with a dependent voltage source Ri, and G with a
dependent current source Gv where the original circuit current’s direction is kept the same

then you will replace and restore the dependent source that you created

bilinear form theorem

 b0 
output b0 + b1 x
= , i.e. [1 R A ]  b1  = RAa1 , where A = Vout / Vin
input a0 + a1 x
 a0 
x is a resistance or ant dependent source parameter
this ratio can be (vout / vin , vout / iin , iout / iin , iout / vin , Rin or Gin )
for the relationship between some varying resistance R and gain A, the matrix above

thevenin theorem for passive networks, voc = Rth isc

for a linear circuit containing resistances and independent sources, there exists an equivalent
two-terminal network consisting of an equivalent thevenin resistance Rth , when all the
independent voltage sources are deactivated, in series with an open-circuit voltage voc, when
no other network is attached

prepared by michael c. pappas – michaelcpappas.com

 find Rth by deactivating all independent sources
voltage sources become short and current sources become open
if the circuit is a series-parallel, use inspection to reduce the circuit
if it is not a series-parallel, use KCL or KVL, whatever makes finding the unkown easier

norton theorem for passive networks, voc = Rth isc

for a linear circuit containing resistances and independent sources, there exists an equivalent
two-terminal network consisting of an equivalent thevenin resistance Rth , when all the
independent voltage sources are deactivated in parallel with an independent short circuit
current source isc

proceed to reduce the circuit like the thevenin theorem procedures

when a network contains no independent sources, voc = isc = 0 with a single resistance Rth
to find voc, use thevenin or norton equivalent
to find isc, connect a wire across terminals

thevenin and norton equivalents for active networks, containing dependent sources and op amps

if it also has any dependent sources, deactivate them to find the network equivalent
apply a voltage or current source to the two terminals from which you want to find the
equivalent resistance
use Rth = vs / is to find either vs or is
use vs or is to find the unknown part of the dependent source or whatever else

remember that Rth is the equivalent resistance depending on what two terminals you pick
and depending on what state the circuit is at that instant, the state will especially change
when dealing with switches

when you replace a circuit with a thevenin, you have vAB = Rth is + voc, where Rth is is the voltage
created by a fictitious current source, and voc is the voltage created by the independent sources in
the circuit across the two terminals. these two things will allow you to find vAB, the total voltage
across the two terminals opening and closing

when you replace a circuit with a norton, you have iA = vs / Rth – isc, where vs / Rth is the current
created by a fictitious voltage source, and isc is the current created by the independent sources in
the circuit across the two terminals. these two things will allow you to find iA, the total current
across the two terminals
thevenin and norton equivalents from measured data

you can make two measurements in the laboratory of vab to make two equations and two
unknowns

multiple thevenin equivalents can be made to model nonlinear circuits

prepared by michael c. pappas – michaelcpappas.com

other theoretical considerations

any circuit that has a horizontal v- i characteristic in the v- i plane has a Norton equivalent but
not a Thevenin

any circuit that has a vertical v- i characteristic in the v-i plane has a Thevenin equivalent but
not a Norton

norton and thevenin equivalents can not be made with v- i characteristics described as points

(there is only one voltage and current for all voltages appearing across the terminals) or as
planes (any i is possible for each v)

an independent current source does not have a thevenin equivalent

an independent voltage source does not have a norton equivalent

to find out whether there is equivalent, if the determinant of the matrix of vectors of
unknown voltages and currents containing, v1 , equaling vectors of independent source values
containing i1 or zero values is not equal to zero, then the ne twork has a thevenin equivalent
 M   v1   i1 
 ...  ...  =  ...
    
or, if the determinant of the coefficients matrix in the modified loop equation is not equal to
zero, then there is a Norton

if Rth is not equal to 0, then there is a norton equivalent

if 1 / Rth is not equal to zero, there is a thevenin

maximum power
maximum power is transferred to the load when RL = Rth
this is important when trying to match loud speaker “resistances” to output “resistances” of a
stereo amplifier or when trying to get as much power as possible out of an antenna and into a
receiver, but it is not important for power transmission networks

the condition where maximum power transfer is delivered to a load network will occur when
v = 0.5voc

where v is the terminal voltage, the i through v will thus be

i = 0.5voc / Rth
and the pL, max will be voc2 / 4Rth

inductance, a coiled conductor

measures the magnitude of the voltage induced by a change in the current through an
inductor
energy storage occurs, the induced voltage is proportional to the derivative of the current, the
inductance of the coil which is a constant of proportionality is L

prepared by michael c. pappas – michaelcpappas.com

 − Rth t
(t −t 0 ) di (t ) 1 0
iL (t ) = i L (t0 ) e
L
vL (t ) = L L iL ( t ) = ∫ vL (τ ) dτ p L (t ) = vL ( t )i L (t )
dt L −∞
energy stored in a inductor
iL (t1 )
1 2 1 2 1 2
WL (t 0 , t1) = L ∫ iL diL =
2
LiL (t1 ) − LiL (t 0 ) , WL (t ) = Li L (t )
2 2
iL ( t0 )

stored energy is independent of voltage waveform in between ti and tf, only initial an final
if the voltage across an indictor is bounded, the current through the inductor is continuous
the inductor is a lossless device, all energy stored is returned
inductors initially behave like an open in the circuit
inductors behave like a short or an zero resistance wire after a very long time

capacitor, separated conductors

current is proportionality to the time rate of change of its voltage, the ability to produce a
current from the changes in the voltages across it
q (t ) = CvC (t )
C1v1 (t ) + C2v2 ( t ) + ... = 0 , junction of more than one capacitors
−1
( t −t 0 ) t
dvC ( t ) 1
vC ( t ) = vC (t 0 )e iC ( t ) = C vC ( t ) = ∫ iC (τ ) dτ pC (t ) = vC (t )iC (t )
Rth C

dt C −∞
energy stored in a capacitor
t1
1 2 1 2 1 2
WC (t 0 , t1) = C ∫ pC (τ ) dτ = CvC ( t1 ) − CvC ( t0 ) , WC (t ) = CvC (t )
t0
2 2 2
stored energy is independent of voltage waveform in between ti and tf, only initial an final
if the current through a capacitor is bounded, the voltage across the capacitor is continuous,
unless two are connected on parallel or when some capacitors and some voltage sources form
a loop, but it really doesn’t break the rule, it’s just fast
the capacitor is a lossless device, all energy stored is returned
initially capacitors behave like a short or a zero resistance wire
after a long time capacitors behave like an open in the circuit

dual circuits
two circuits are dual if KCL, KVL, and v- i branch equations in one circuit become KVL,
KCL, and v- i branch equations upon interchanging ik <>vk * and vk <>ik *. any solution in N
leads to a solution to N* and relationship among variables in N lead to relationship among
variables in N*. N* will only exist if N is planar. N and N* must have the same number of
branches, but not nodes. a dual circuit elements guide will help build dual circuits.
there are problems making dual circuits when:
the circuit is not series-parallel
the circuit contains controlled sources
the reference direction of the circuit elements play an important role in the solution
when this happens: follow procedures below

prepared by michael c. pappas – michaelcpappas.com

procedures for determining the connection of dual network branches

draw the given planar circuit N without branch crossings, and identify the meshes (regions)
place a node of N* inside each mesh and one in the outside, infinite region

for each branch b in N that is on the boundary of two regions, draw the dual branch b*
joining the nodes placed in each of these regions. if the branch arrow in N is clockwise
around the node of N*, the arrow for the corresponding branch in N* is away from the node
of N*

the step function

1 t ≥ t0
u (t − t 0 ) = 
 0 t < t0

first-order RL and RC circuits

circuits have behaviors modeled by differential equations

the solution to a first-order differential equation is a waveform, or a signal or a response
the solution works for all continuous and piecewise continuous time functions
a solution to a differential equation means that the waveform satisfies the given differential
equation with the proper initial condition

source free or zero input response

Rth
diL (t ) R − ( t −t 0 )
= − i L (t ) iL ( t ) = i L (t 0 )e L
dt L
1
− ( t − t0 )
dvC (t ) 1
=− vC (t ) vC ( t ) = vC (t 0 ) e Rth C
dt RC
τ = L / R = RC , time it takes the source-free circuit response to drop to e-1 of the initial value

dc or step response of first-order circuits

for a linear resistive circuit with constant sources

V − v (t )
vL (t ) = Voc − Rthi L (t ) iC ( t ) = oc c
Rth
diL ( t ) R R diL ( t ) R 1
= − th iL (t ) + th I sc = − th iL (t ) + Voc
dt L L dt L L
dvC (t ) 1 1 dvC (t ) 1 1
=− vC (t ) + Voc =− vC (t ) + I sc
dt RthC RthC dt RthC C
Voc implies in series with Rth
Isc implies parallel with Rth

prepared by michael c. pappas – michaelcpappas.com

 ( t −t 0 )
−
general solution to the equations above, x (t ) = Fτ + [ x( t0+ ) − Fτ ]e τ
x(t) = iL Fτ = Voc / Rth = Isc
x(t) = vC Fτ = Voc
when [x(t) = constant] satisfies a differential equation, it is called the equilibrium state
Fτ is the equilibrium state

when τ > 0, or when Rth > 0, C > 0, and L > 0, or when the circuit is passive
Rth
− (t −t 0 )
+
iL (t ) = i L (∞ ) + [iL (t ) − iL (∞)]e
0
L

1
− ( t − t0 )
+
vC (t ) = vC (∞ ) + [vC (t ) − vC ( ∞)]e
0
RthC

these equations are easier to use if you can use them because only the initial and final
values of iL and vC are needed

when looking at a first-order RC or RL circuit

does have dependent sources?
does it not have independent sources?
if it also has any dependent sources, deactivate them to find the network equivalent
apply a voltage or current source to the two terminals from which you want to find
the equivalent resistance
use Rth = vs / is to find either vs or isc
use vs or isc to find the unknown part of the dependent source or whatever else
now that you have Rth
does it have the initial inductor current or capacitor voltage given?
is it not driven by a dependent source, AT the specific time?
1
−
Rth
( t −t 0 ) − ( t − t0 )
use iL ( t ) = i L (t 0 )e or vC ( t ) = vC (t 0 ) e
L Rth C

is it driven by a dependent source, AT the specific time?

1
−
Rth
(t −t 0 ) − ( t − t0 )
+ +
use iL (t ) = i L (∞ ) + [iL (t ) − iL (∞)]e
0
L
or vC (t ) = vC (∞ ) + [vC (t ) − vC ( ∞)]e
0
RthC

superposition and linearity

again, like before, one sets all the independent sources to zero and computes the response due
to each initial condition with all other initial conditions set to zero. the sum of all of the
responses to each of the independent sources plus the individual initial condition responses
yields the complete circuit response, by the principle of superposition

1 t  1 t 
vC ( t ) = a1  ∫ iC 1(τ )dτ  + a2  ∫ iC 1(τ ) dτ  = a1vC 1(t ) + a2vC 2 (t )
 C −∞   C −∞ 

for example, if there two independent sources and an initial inductor current, set the sources
to zero, find the response due to the initial condition of initial current. set the initial current

prepared by michael c. pappas – michaelcpappas.com

 and one of the sources to zero, and find the contribution from the other source, then find do
the same thing for the other source

superposition is good when you want to change one of the sources or just the initial condition
because you do not have to recalculate the equivalent circuit network

voltage and current in these circuits may not always behave continuously, this may be the
case when capacitors, inductors, and switches turn into shorts and opens

when vin (t ) = −18u ( −t) + 9u( t ) , v = –18 V when t < 0 and v = 9 V when t > 0

the general form of an RC and RL circuit has the form

( t − t0 )
−
+
x (t ) = Xe + [ x (t ) − X e ]e
0
τ
, where X is the voltage or current level, and where
 X − x (∞) 
t 2 − t1 = τ ln  1  , is the elapsed time formula, the subscripts match up, X and t
 X 2 − x( ∞ ) 

first-order RC op amp circuits

ideal differentiator op amp ideal integrator op amp

t
dv ( t ) 1
RC ∫0 in
vout ( t ) = − RC in vout (t ) = − v (τ ) dτ
dt
LC circuits
 1 
vC ( t ) = K cos(ωt + θ ) vC ( t ) = V0 cos  t
 LC 
C  t 
iL ( t ) = K cos(ωt + θ ) iL ( t ) = V0 sin  
L  LC 
source-free LC circuits are undamped

series RLC parallel RLC

R 1 1 1
v′′C + v′C + vC = F iL′′ + iL′ + iL = F
L LC RC LC

to find initial conditions, replace each capacitor by an independent current source with value
vc(0+) and each inductor with an independent current source with value iL (0+)

solutions to RLC,

−b ± b2 − 4 c b b 2 − 4c
s1, s2 = σ = ωd =
2 2 2

prepared by michael c. pappas – michaelcpappas.com

real and distinct, overdamped
x (t ) = K1es1 t + K 2e s2 t + X F XF = final value of either voltage or current
x (0 + ) = K1 + K2 + X F x′(0 + ) = s1K1 + s2 K 2

complex and distict, underdamped

x (t ) = e−σ t [ K1 cos(ω dt ) + K 2 sin(ωd t ) ] + X F
x (0 + ) = K1 + X F x′(0+ ) = −σ K`1 + ωd B
s1 = −σ + jωd s2 = −σ − jω d

real and equal, critically damped

x (t ) = ( K1 + K 2t ) est + X F
x (0 + ) = K1 + X F x′(0+ ) = sK1 + K 2

when dealing with the initial conditions of an RLC circuit, you can turn capacitors and inductors
into independent voltage and current sources

complex number review

a + jb = ρ∠θ = ρ e jθ ρ = a2 + b2 cos(θ ) = a / ρ sin(θ ) = b / ρ

(a + jb) ρ
( a + jb)(c + jd ) = foil = ρ1ρ2∠ (θ 1 + θ 2 ) = rationalize = 1 ∠ (θ1 − θ 2 )
(c + jd ) ρ2
Re[ρ e jθ ] = ρ cos(θ ) −1/ j = j sin(θ ) = cos(θ − 90°)
c + jd −1  d  −1  b 
Ve jθ = θ = tan   − tan   V = Vm∠φ = Vm cos(ω t + φ )
a + jb c  a

impedance
1
ZR = R ZC = Z L = jω L R = Re[ Z ] Re[V/Z] = i(t) Re[IZ] = v(t)
jωC

impendance magnitude is max, or admitatnce min, and values are purely real when Im[Z] = 0
when Im[Z] > 0, you have an inductor
capacitor voltage and inductor current are always continous
1 T V2
p (t ) = v (t )i (t ) Pav = ∫ v( t) i( t)dt Pa v, R = m
T 0 2R
Pav = Veff I eff cos(θ v − θi ) (θ v − θi ) = the angle of the impendance of the two terminal element

effective value = rms

Fm F
sinusoidal Feff = triangular Feff = m square Feff = Fm Ssource = S1 + S2 + S3
2 3
complex power absorbed = S = Veff I *eff I *eff = the complex conjugate

prepared by michael c. pappas – michaelcpappas.com

average power - Re[S] = P
reactive power - Im[S] = jQ
apparent power - S = P 2 + Q2
P P
power factor = pf = av = cos(θ v − θi ) < power factor angle pf =
S P +Q
2 2

pf lagging 0 < (θ v − θi ) < 180 pf leading 0 < (θ i − θ v ) < 180

Q > 0 pf lagging Q < 0 pf leading
low pf, large Q, large loss high pf, low Q, small loss

there is maximum power transfer when

VS2,eff
ZL = Z *
S RL = RS XL = −X S Pmax,av =
4R S

prepared by michael c. pappas – michaelcpappas.com