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welcome back to basic electronics in the last
lecture we have introduced phasors and seen
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how to interpret them in the sinusoidal steady
state in this lecture we want to apply phasors
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to r l c circuits to figure out the steady
state currents and voltages in addition we
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will look at the maximum power transfer theorem
for r l c circuits in the sinusoidal steady
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state so let us start
let us now look at how we can use phasors
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in circuit analysis the time domain k c l
and k v l equations sigma i k equal to zero
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and sigma v k equal to zero can be written
as phasor equations sigma i k equal to zero
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where i k are current phasors now and sigma
v k equal to zero where v k are voltage phasors
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in the frequency domain
resistors capacitors and inductors can be
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described by v equal to z times i in the frequency
domain which is similar to v equal to r i
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in d c conditions the only difference is that
we are now dealing with complex numbers when
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we talk about this equation an independent
sinusoidal source in the frequency domain
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behaves like a d c source for example for
a voltage source a sinusoidal voltage source
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we can say that the phasor v s is the constant
which is the complex number for dependent
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sources a time domain relationship such as
i of t equal to beta times i c of t translates
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to the phasor relationship phasor i equal
to beta times phasor i c in the frequency
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domain so the equation looks very similar
except we have complex numbers here
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so from all of these remarks we conclude that
circuit analysis in the sinusoidal steady
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state using phasors is very similar to d c
circuits with independent and dependent sources
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and registers and therefore all the results
that we derived for d c circuits are valid
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also for sinusoidal steady state analysis
therefore series parallel formulas for resistors
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nodal analysis mesh analysis thevenin and
nortons theorems can be directly applied to
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circuits in the sinusoidal steady state totally
difference of course is we are now dealing
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with complex numbers let us now consider an
r l circuit in the sinusoidal steady state
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where r and l in series the impedance of the
register is r as we have seen before and the
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impedance of the inductor is j omega l
this is our sinusoidal source voltage v m
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angle zero that is v m cos omega t in the
time domain and we are interested in this
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current when we use phasors this becomes an
extremely simple calculation as we will see
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let us imagine that we have a d c circuit
with a d c source here v s a resistor r one
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here and a register r two here what would
the current b in that case the d c current
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it would be simply v s divided by r one plus
r two so we can use exactly the same equation
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except instead of v s we have a phasors source
now that is v m angle zero and instead of
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r one and r two they have r and j omega l
the impedances of the resistor and the inductor
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all right so then we can write i as v m angle
zero this source divided by r plus j omega
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l as simple as that
we can write this phasor current as i m angle
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minus theta where i m is the magnitude of
this expression and that is simply v m divided
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by the magnitude of the denominator which
is square root r squared plus omega squared
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l squared the angle of the denominator is
tan inverse of omega l by r and the angle
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of the numerator is zero so the net angle
of i is zero minus tan inverse of omega l
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by r that is all we get this minus sign over
here so theta here is tan inverse of omega
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l by r
in the time domain we have i of t equal to
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i m cos omega t minus theta i m is given by
this expression and theta is this one all
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right now let us take some component values
say r equal to one ohm l equal to one point
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six millihenry and f equal to fifty hertz
then theta turns out to be twenty six point
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six degrees and here are the plots for the
source voltage and the current this is our
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source voltage v m is one so it is one times
cos omega t and this is our current the dark
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one and let us now try to understand this
current plot in terms of this equation
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our i m value turns out to be about point
nine ampere and therefore we see that the
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current goes from ah minus point nine amperes
to plus point nine amperes all right and as
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we have said earlier theta is twenty six point
six degrees now how do we relate this value
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to this plot this is an angle and this is
time over here so what we can do is to convert
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this angle to time we know that three sixty
degrees corresponds to one period and that
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in this case is twenty milliseconds because
one over f is twenty milliseconds and therefore
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we can use that fact to find the time that
corresponds to this angle that turns out to
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be one point four eight milliseconds
ok now lets get back to this equation i of
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t equal to i m cos omega t minus theta and
ask the question when does this i of t go
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through is speak that is when is omega t minus
theta equal to zero and the answer is that
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happens when t is equal to theta divided by
omega and that turns out to be exactly one
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point four eight milliseconds all right so
what it means is that the current does not
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go through the peak here when the voltage
goes through each speak but a little later
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that is at one point four eight or nearly
one point five milliseconds and that is why
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we say that it lags the source voltage
notice how easy this entire calculation was
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we did not write down any differential equation
we simply used this expression which is like
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two resistors in series and then we managed
to get all the information that we required
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so that is how phasors really help in analyzing
circuits in the sinusoidal steady state all
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right so here is the circuit file you can
simulate the circuit and maybe change some
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component values and look at the results
let us now represent the k v l equation that
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is v s equal to v r plus v l in a graphical
form in the complex plane using a phasor diagram
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and to do that we require v s which is v m
angle zero and v r v l let us look at v r
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first what is v r v r is i times the impedance
of the register which is r so that is r times
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i m angle minus theta because our i is i m
angle minus theta what about v l v l is i
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times the impedance of the inductor which
is j omega l now this j is nothing but angle
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pi by two and i is i m angle minus theta so
therefore this quantity is omega i m times
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l angle minus theta plus pi by two
so this is our phasor diagram this vector
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is v s and v s as angle zero therefore it
is a longer x axis that is the real part of
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v all right this is our v r and v r has angle
minus theta that means theta in the clockwise
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direction and this is our v l what is the
angle of v l it is minus theta plus pi by
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two that means we go clockwise by theta and
then we go anti clockwise by pi by two that
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brings us to this angle all right and now
this equation v s equal to v r plus v l is
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essentially a vector equation if we add v
l and v r then we get v s
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next let us take an r c circuit a series r
c circuit and this can be analyzed in exactly
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the same manner as the r l circuit which we
just saw what do we do in this case we replace
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the register with its impedance which is r
the capacitor with its impedance which is
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one over j omega c and then we can obtain
this current as v m angle zero divided by
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r plus one over j omega c that we write as
i m angle theta where i m its given by this
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quantity you should really verify this and
theta is pi by two minus tan inverse omega
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r c now this tan inverse omega r c can vary
from zero two pi by two and therefore this
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angle is basically a positive angle
in the time domain we write the current as
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i m cos omega t plus theta and now let us
calculate theta for some component values
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say r equal to one ohm c equal to five point
three millifarad and f equal to fifty hertz
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for this combination theta turns out to be
thirty one degrees and that corresponds to
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a time of one point seven two milliseconds
all right now let us ask this question when
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does i of t go through its peak and that happens
when omega t plus theta becomes equal to zero
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that means t is equal to minus theta divided
by omega and that time is exactly one point
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seven two milliseconds
ok let us look at the plots law this light
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curve is the source voltage one angle zero
and it goes through its peak at t equal to
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zero because there is the cos function this
dark curve is the current and that goes through
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its peak when t is equal to minus one point
seven two milliseconds as we just discussed
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in other words the current goes through its
peak before the source voltage goes through
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its peak and that is why we say that the current
leads the source voltage since the peak of
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i of t occurs theta by omega seconds before
that of the source voltage
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let us draw a phasor diagram to represent
the k v l equation in this case that is v
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is equal to v r plus v c what is v r i times
r where i is i m angle theta so therefore
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v r is r times i m angle theta what about
v c v c is i times the impedance of the capacitor
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which is one over j omega c now one over j
is the same as minus j that is an angle of
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minus pi by two and therefore we get v c equal
to i m by omega c that is the magnitude and
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for the angle we get theta which comes from
i and then we have this minus pi by two coming
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from this one over j
and as we mentioned earlier our theta is a
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positive angle between zero and pi by two
all right with that information we can now
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draw the phasor diagram this is our v s and
that is along the x axis because it has an
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angle of zero what about v r v r has got an
angle theta which is positive and between
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zero and pi by two so that is what v r looks
like what about v c v c has an angle of theta
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minus pi by two so we go anti clockwise by
theta and then become clockwise by pi by two
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that brings us to this angle so that is our
v c and now we can see that the vector equation
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v s equal to v r plus v c each satisfied this
is our v s so v r plus v c brings us to v
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s so this is the phasor diagram corresponding
to this k v l equation in this case
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let us consider a little more complex circuit
now the one shown here so we have a sinusoidal
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voltage source ten angle zero frequency fifty
hertz and we want to find these currents i
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s i c and i l step number one we convert all
the components to their impedances so this
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two ohms of course remains two ohms ten ohms
remains ten ohms to millifarads becomes z
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three where it z three is one over j omega
z that turns out to be minus j one point six
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ohms fifteen milli entry becomes z four where
is z four is j omega l omega is two pi times
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fifty that is l so that turns out to be j
four point seven ohms and now next step is
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we can calculate the equivalent impedance
of this combination and this is exactly like
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series parallel register combination we have
z two and z four in series that combination
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in parallel with z three and the whole thing
in series with z one so that is what z equivalent
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is z one plus z three parallel z two plus
z four
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you are definitely encouraged to go through
all of these steps and arrive at this final
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result for z equivalent but also look up your
calculator and it is possible that you will
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be able to do this calculation in a smaller
number of steps depending on what your calculator
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allows all right now let us go ahead so we
have come up to this step you have found said
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equivalent we already have v s so now we can
calculate i s so i s is v s divided by z equivalent
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v s is ten angle zero and z equivalent from
the last slide is this number here so that
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turns out to be three point five eight angle
thirty six point eight degrees ampere
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once we get i s we can now get i c by using
the current division formula that is i c equal
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to z two plus z four divided by z two plus
z four plus z three times i s like that and
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that turns out to be this number what about
i l you can get i l in two ways one i s minus
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i c gives us i l by k c l or we can use the
current division formula i l is equal to z
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three by z three plus z two plus z four times
i s either way you should get this number
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for i l and now let us draw the phasor diagram
which describes the k c l equation at this
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node here is the phasor diagram and notice
that we have used the same scale for the x
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and y axis that is this distance which represents
one unit on the x axis also represents one
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unit on the y axis and we do that so as to
represent angles correctly that means a forty
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five degree angle would you indeed look like
a forty five degree angle if we follow this
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practice
all right let us now verify whether the k
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c l equation at this node is satisfied what
is the equation we have i s equal to i c plus
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i l our i s is three point five eight angle
thirty six point eight degrees that is this
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vector this magnitude is three point five
eight and this angle is thirty six point eight
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degrees i c has a magnitude of three point
seven nine so little bit larger than i s and
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an angle of forty four point six degrees that
is i c this angle is forty four point six
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degrees i l its much smaller in magnitude
point five four six and it has a negative
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angle minus seventy point six degrees so that
is our i l and now we see that i c plus i
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l is indeed equal to i s so that means k c
l is verified
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we have looked at the maximum power transfer
theorem for linear d c circuits now let us
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look at maximum power transfer in the sinusoidal
steady state so let us consider ah circuit
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whose devin an equivalent is given by this
circuit here namely a voltage source v t h
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in series with an impedance z t h both of
these of course are complex numbers now we
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connect a load impedance to the circuit and
as a result a current is going to flow and
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that current is denoted by this spacer i
we want to find the condition for which the
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power transfer from this circuit two z l is
maximum all right let us begin with z l equal
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to r l plus j x l where this is the real part
of z l and that is the imaginary part of z
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l and similarly let z t h b r t h plus j x
t h let i b i m angle phi i m is the magnitude
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of this phasor i here and phi is its angle
now the power absorbed by z l is given by
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p equal to half i m squared r l where i m
is the magnitude of this phasor i and r l
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is the real part of z l what is i m it is
the magnitude of i and what is i is simply
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v t h divided by z t h plus z l so we get
p equal to this expression over here let us
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rewrite this expression as half mod v t h
squared divided by r t h plus r l squared
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plus x t h plus x l squared and that multiplied
by r l now this v t h squared is independent
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of z l and as far as we are concerned that
is just a constant so what we need to do now
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is to find conditions on z l for which this
whole expression is maximum
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now for p to be maximum clearly this denominator
must be ah minimum and that will happen if
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x t h plus x l squared is zero because there
is a square here the smallest value that this
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term can take is zero and therefore that gives
us x l equal to minus x t h and with x l equal
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to minus x t h this second term disappears
and we get p equal to half mod v t h squared
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divided by r t h plus r l squared times r
l so for maximum power transfer we need to
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maximize this expression now
how do we do it we differentiate p with respect
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to r l and we equate d p d r l to zero and
then find that p is going to be maximum when
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r l is equal to r t h so we have two conditions
one the imaginary part of z l that is x l
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must be equal to negative of the imaginary
part of z t h that is x t h and the real part
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of z l must be equal to the real part of z
t h
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to summarize for maximum power transfer to
the load z l we need r l equal to r t h and
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x l equal to minus x t h that means over load
impedance must be the complex conjugate of
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the thevenin equivalent impedance that is
z t h so z l must be equal to z t h star
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let us now look at an application of the maximum
power transfer theorem for sinusoidal steady
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state here it is an audio amplifier driven
by an audio input signal so the frequencies
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here would be in the range twenty hertz to
say sixteen kilo hertz or so this audio amplifier
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is followed by ah transformer and we will
soon comment on why this transformer is required
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and then finally we have this speaker this
speaker has a complex impedance which varies
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with frequency but in the audio range its
resistance is more or less constant typically
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eight ohms and its imaginary part can be ignored
so the equivalent circuit represent take this
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entire situation is given by this circuit
here this source here represents the input
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signal amplified by the gain of the audio
amplifier and then this one k resistance here
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is the output resistance of the audio amplifier
and then we have the transformer with a turns
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ratio n one two and two and finally the speaker
which is represented by this a eight ohms
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resistance here
our objective of course is to maximize the
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power transfer from this circuit to the speaker
that is how we will hear the loudest sound
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that is possible with the given input signal
all right so let us simplify the circuit we
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can transfer this resistance to the other
side of the transformer and then it becomes
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n one by n two squared times eight ohms
here is our actual problem statement we look
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at this circuit and calculate the turns ratio
to provide maximum power transfer of the audio
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signal what is the maximum power transfer
theorem it says that z l must be equal to
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z t h star and in this case since the imaginary
parts of z t h and z l are zero all it means
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is that the real parts must be equal that
means we must have n one by n two squared
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times eight ohms equal to one k like that
and we can now solve this equation for n one
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by n two and that gives us n one by n two
of about eleven so if we pick a transformer
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with this translation then it is current it
that the audio signal will deliver maximum
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power to the speaker and we will hear the
loudest possible sound that is possible with
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this input signal
in conclusion we have seen how to use phasors
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to analyze r l c circuits in the sinusoidal
steady state this background is going to be
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very useful when be look at filter circuits
a little later we also looked at the maximum
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power transfer theorem for r l c circuits
in the sinusoidal steady state we considered
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an example which is very important in practice
namely how to obtain maximum audio power from
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a speaker that is all for now
see you next time