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welcome back to basic electronics so for we
have looked at circuits with dc sources there
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are many situations in which there is a sinusoidal
voltage source in the circuit and in particular
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we are interested in the solution in the so
called sinusoidal steady state in this lecture
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we will first look at the meaning of the term
sinusoidal steady state we will then look
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at a convenient way to represent voltages
and currents in that situation using a new
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concept called phasor we will also look at
how phasors can be used to represent r l and
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c in the sinusoidal steady state ok
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so let us begin let us try to understand the
meaning of this term sinusoidal steady state
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with help of this example its an r c circuit
with a sinusoidal input voltage there is a
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switch here which closes at t equal to zero
and initially the capacitor is uncharged that
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means v c is zero all right let us begin with
the circuit equation this equation one here
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what does it say it says that this voltage
drop plus this voltage drop must be the same
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as the source voltage this voltage drop is
r times the current and the current is c b
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b c d t so that is what this says here r times
c v c prime plus v c must be equal to v m
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cos omega t for t greater than zero that is
when the switch is closed
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the solution v c of t is made up of two components
a homogeneous component indicated with this
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superscript h here and a particular component
indicated with this superscript p the homogeneous
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component we c h t satisfies the homogeneous
differential equation r c v c prime plus v
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c equal to zero so we drop the source term
and that is all we get this equation and this
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equation has this solution a e raise to minus
t by tau with tau equal to r times c let us
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now look at the second part the particular
solution since the forcing function is v m
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cos omega t we can try v c p of t equal to
c one cos omega t plus c two sin omega t as
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a candidate solution and when we substitute
this in equation one we get this equation
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here and the constants c one and c two can
be found by equating the coefficients of sin
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omega t and cos omega t on the left and right
sides
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all right here is a specific example with
v m equal to one volts f is equal to one kilo
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hertz two k here point five micro here for
the capacitance and then we get this v c shown
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in this figure the complete solution is a
e raise to minus t by tau from the previous
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slide that is the homogeneous part plus c
one cos omega t plus c two sin omega t this
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is the particular part of the solution all
right now as t tends to infinity the exponential
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term becomes zero this goes to zero and we
are left with v c of t equal to c one cos
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omega t plus c two sin omega t this part here
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and we can observe that in this plot as well
this is our t equal to zero that is the time
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when the switch closes and in the beginning
there is some exponential transient and then
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finally the transient vanishes and we have
the sinusoidal steady state so this is known
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as the sinusoidal steady state response since
all quantities currents and voltages in the
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circuit are sinusoidal in nature and this
turns out to be generally true for any circuit
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containing resistors capacitors inductors
sinusoidal voltage and current sources dependent
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sources such as c c c s c c v s etcetera so
any circuit containing these components behaves
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in a similar manner that is each current and
voltage in the circuit becomes purely sinusoidal
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as t tends to infinity so each quantity each
current and each voltage in the circuit would
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have this form in the sinusoidal steady state
ok
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let us continue and now we want to introduce
phasors in the sinusoidal steady state phasors
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can be used to represent currents and voltages
so let us see what a phasor is a phasor is
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a complex number thats why it is written in
boldface here it has got a magnitude of x
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m and an angle of theta so that is the phasor
x and we can rewrite this as x m here is to
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j theta and it has the following interpretation
in the time domain corresponding to the phasor
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x we have a time domain quantity denoted by
x of t and the interpretation is like this
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x of t is d l part of this number that is
phasor x multiplied by e raised to j omega
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t
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so let us see what that turns are to be our
phasor is nothing but x m times e raised to
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j theta that gets multiplied by e raised to
j omega t now we can combine these two terms
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to get e raised to j omega t plus theta what
is e raised to j omega t plus theta it is
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cos of omega t plus theta plus j sin of omega
t plus theta and we are only taking the real
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part of that expression so therefore we get
x m cos omega t plus theta so that is what
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a phasor x corresponds to in the time domain
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let us make some more comments use of phasors
substantially simplifies analysis of circuits
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in the sinusoidal steady state and we will
look at some examples of that note that a
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phasor can be written in the polar form or
rectangular form this is the polar form x
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m angle theta also called the magnitude angle
form that is the same as x m e raise to j
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theta and that can be written as x m cos theta
plus j x m sin theta this is called the rectangular
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form this figure shows how the polar and rectangular
forms of x can be represented this axis is
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the real part of x which is the complex number
this axis is the imaginary part of x this
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is our phasor it has got a magnitude of x
m and angle of theta
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alternatively we can also write this component
which is the real component of x and this
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component which is the imaginary component
of x and that gives us the rectangular form
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all right now one final remark very important
the term omega t is always implicit so when
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we write a phasor for example x here we dont
write omega t over there but it is understood
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that omega t is always implicit and that is
how it comes in this time domain expression
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here let us now take a few examples to see
how something can be represented in the time
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domain and in the frequency domain here is
a voltage v one of t three point two cos omega
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t plus thirty degrees volts we can write this
in the phasor form the magnitude is three
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point two and the angle is thirty degrees
so that is what the phasor looks like three
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point two angle thirty degrees
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we can also write that as three point two
e raise to j pi by six because thirty degrees
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is the same as pi by six radians another example
i of t is minus one point five cos omega t
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plus sixty degrees now it is not quite in
the form that we would like because we have
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this minus sin over here so let us rewrite
this expression as one point five cos omega
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t plus pi by three that is sixty degrees minus
pi this minus pi accounts for this minus sin
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here and that follows form integrametric identities
all right now we can combine these these two
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terms and get omega t minus two pi by three
and now we can write the corresponding phasor
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like that now we took one point five and angle
minus two pi by three next example v two of
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t equal to minus point one cos omega t once
again we dont want this minus sin here so
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we rewrite this expression as point one cos
omega t plus pi this plus pi accounts for
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this minus sin over here and note that we
can either write plus pi here or we can write
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minus pi like we did in the last example here
ok
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now the corresponding phasor is magnitude
point one and angle equal to pi next we have
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i two of t equal to point one eight sin omega
t now we require cos over here not sin so
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therefore let us rewrite this expression as
point one eight cos omega t minus pi by two
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and now we can write the corresponding phasor
like that i two equal to point one eight that
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is the magnitude point one eight angle minus
pi by two so that is the angle let us take
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an example of the reverse transformation now
the frequency domain description is given
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phasor i three that is one plus j one in amperes
now we want to convert that into the time
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domain form now this is in the rectangular
form we first convert that into the polar
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form like that so they are square root two
angle forty five degrees and this we can write
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in the time domain directly like that
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so i three of t is square root two cos omega
t plus forty five degrees and remember always
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that this omega t is implicit so we dont write
that in the phasor but we understand that
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it is always there let us now talk about addition
of phasors first let us consider adding two
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sinusoidal time domain quantities like this
v of t equal to v one of t plus v two of t
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v one of t is given by v m one cos omega t
plus theta one and v two of t is given by
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v m two cos omega t plus theta two all right
now consider the addition of the corresponding
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phasors this phasor corresponds to v one of
t this phasor corresponds to v two of t so
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v one therefore is v m one e raise to j theta
one and it would v m one and angle theta one
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v two similarly is v m two e raise to j theta
two in the time domain this phasor v which
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is v one plus v two corresponds to v tilde
t with v tilde t equal to real part of phasor
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v times e raised to j omega t as we have seen
earlier
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this is how a phasor can be written in the
time domain all right so now we substitute
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for v that is v m one e raise to j theta one
plus v m two e raise to j theta two like that
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and now we can combine this e raise to j theta
one and e raise to j omega t to obtain e raise
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to j omega t plus theta one and similarly
e raised to j omega t plus theta two here
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all right now we need to take the real part
of this entire expression and that turns out
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to be v m one cos omega t plus theta one plus
v m two cos omega t plus theta two this in
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fact is the same as v of t right there all
right now what is the conclusion the conclusion
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is that if we have two sinusoidal varying
quantities such as v one of t and v two of
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t here we can obtain their sum by either directly
adding them or we can add the corresponding
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phasors like here and then from the resulting
phasor v here we can obtain the time domain
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quantity like that
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so from the last slide we can say that addition
of sinusoidal quantities in the time domain
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can be replaced by addition of the corresponding
phasors in the sinusoidal steady state and
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this is very important for us because of the
following point the k c l and k v l equations
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that is sigma i k of t equal to zero at a
node that is all currents at a given node
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add up to zero and sigma v k of t equal to
zero in a loop that is all voltages as we
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go along a loop add up to zero these equations
amount to addition of sinusoidal quantities
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and can therefore be replaced by the corresponding
phasor equations what are the phasor equations
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sigma of the phasor currents equal to zero
at a node and sigma of the phasor branch voltages
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is equal to zero in a loop let us now introduce
the concept of an impedance starting with
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impedance of a register first here is a register
in the time domain that is the register current
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i of t and v of t is the voltage across the
resistor
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now in the frequency domain we replace this
current with phasor i and the voltage with
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phasor v and now we can write phasor v equal
to phasor i times something called z where
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z is called the impedance and let us actually
show that we can do that let i of t v i m
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cos omega t plus theta then we have v of t
equal to r times i of t that is valid at all
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times and that is equal to r times i m cos
omega t plus theta after substituting for
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i of t from there and we might define that
as v m cos omega t plus theta where v m is
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r times i m all right now the phasors corresponding
to i of t and v of t are respectively for
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i we have i m angle theta i m is the magnitude
and theta is the angle for v we have r i m
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angle theta r i m is the magnitude and theta
is the angle
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we have therefore the following relationship
between phasor v and phasor i that is v is
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equal to r times i as simple as that and we
can see that from here v is r i m angle theta
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i is i m angle theta so therefore v is simply
r times i and therefore the impedance of a
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register which is defined as z equal to v
by i is z equal to r plus j zero from here
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z is equal to r and r is a real number so
we write that real number as r plus j zero
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so the impedance of a resistor is the resistance
itself let us follow the same process now
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for a capacitor and find its impedance marked
as z over here again ok
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so let this voltage across the capacitor v
of t be equal to v m cos omega t plus theta
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what is the current current in that direction
its c d v d t and when we differentiate this
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quantity we get minus c omega v m sin omega
t plus theta and now we can use the identity
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cos phi plus pi by two is minus sin pi and
we get i of t equal to c omega v m these three
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terms here times cos of omega t plus theta
plus pi by two like that in terms of phasors
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phasor v is v m angle theta v m is the magnitude
and theta is the angle and phasor i is omega
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c v m angle theta plus pi by two omega c v
m is the magnitude and theta plus pi by two
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is the angle all right
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now we can rewrite i as i equal to omega c
v m e raise to j theta plus pi by two that
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is omega c v m times e raise to j theta times
e raised to j pi by two and this quantity
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e raised to j pi by two is nothing but j so
we get then j omega c times v m e raise to
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j theta now this quantity is nothing but the
phasor v and therefore we can write i as j
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omega c times phasor v in short phasor i is
equal to j omega c times phasor v and therefore
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the impedance of a capacitor z is equal to
v by i is z is equal to one over j omega c
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and the admittance which is defined as y equal
to i by v is y equal to j omega c notice that
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we have omega here now that means at low frequencies
the impedance of a capacitor is large and
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at high frequencies the impedance is small
let us now repeat the process for an inductor
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let i of t b i m cos omega t plus theta then
we get v of t as l d i d t we can differentiate
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i of t and get v of t equal to minus l omega
i m times sin omega t plus theta
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now we can use the identity cos phi plus pi
by two is equal to minus sin phi and get v
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of t has l omega i m times cos omega t plus
theta plus pi by two so we have replaced this
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minus sin omega t plus theta by cos omega
t plus theta plus pi by two all right now
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in terms of phasors we have i equal to i m
angle theta i m is the magnitude and theta
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is the angle and v equal to omega l i m omega
l pi m is the magnitude angle theta plus pi
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by two that is the angle and v can be rewritten
as v equal to omega l i m times e raised to
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j theta plus pi by two that is omega l i m
e raised to j theta times e raised to j pi
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by two this is nothing but j so that becomes
j omega l times i m e raised to j theta and
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this quantity is nothing but our phasor i
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so finally we get phasor v equal to j omega
l times phasor i and from there the impedance
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of an inductor z equal to v by i is z equal
to j omega l and the admittance of an inductor
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y equal to i by v is y equal to one over j
omega l notice that the inductor impedance
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also depends on the frequency but it is directly
proportional to the frequency so at low frequencies
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the impedance is small and at high frequencies
the impedance is large
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let us now look at sources sinusoidal current
source and sinusoidal voltage source and independent
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sinusoidal current source i s of t equal to
i m cos omega t plus theta can be represented
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by the phasor i m angle theta magnitude i
m angle theta that is simply a constant complex
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number so we have this source in the time
domain and the corresponding phasor in the
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frequency domain and that i s is a constant
all right and similarly an independent sinusoidal
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voltage source v s of t is equal to v n cos
omega t plus theta can be represented by the
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phasor v m angle theta that is the constant
complex number like that what about dependent
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sources dependent sources can be treated in
the sinusoidal steady state in the same manner
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as we treated a register that is by the corresponding
phasor relationship
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for example let us consider a c c v s that
is current controlled voltage source in the
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time domain we have v of t that is the voltage
across the c c v s equal to r a constant times
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i c of t where i c of t is the controlling
current now in the frequency domain this relationship
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becomes v phasor equal to r times i c phasor
as simple as that and similarly we can write
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relationships for the other dependent sources
such as c c c s or v c v s etcetera
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in summary we have seen what phases are and
how they can be used to represent currents
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and voltages in the sinusoidal steady state
we have also seen how r l and c can be represented
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in the sinusoidal steady state using phasors
in the next lecture we will extend these ideas
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to circuits until then goodbye