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welcome back to basic electronics we will
come across a few circuits in this course
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for which we will need to find the time taken
by a capacitor to charge or discharge from
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an initial voltage v one to a final voltage
v two in this lecture we will make a beginning
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in that direction we will first look at the
general solution for n r c or r l circuit
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see how to plot the solution as a function
of time in subsequent lectures we will consider
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some specific examples so let us begin
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let us now start our discussion of r c circuits
with d c sources and this situation as we
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will find out is very common in electronic
circuits and once we have this background
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we will be able to calculate for example pulse
width or oscillation frequency in a circuit
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all right here is the circuit a linear circuit
which consists of resistors independent voltage
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sources which are d c independent d c current
sources and then dependent sources such as
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current controlled voltage source current
controlled current source etc to this circuit
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we connect ah capacitor and now let us see
how this capacitor voltage or current evolves
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with time
since this circuit is a linear circuit we
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can represent it with this thevenin equivalent
consisting of a voltage source v t h in series
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with a resistance r t h and this capacitor
of course appears as it was earlier over here
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ok let us now make a few important points
if all sources are d c or constant then v
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t h is a constant and its important to bear
in mind this assumption if for example we
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have a sinusoidal voltage source here or a
triangular current source here then our analysis
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is not going to be valid
let us begin by writing the k v l equation
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for this loop and what is that that says that
this v t h must be equal to this voltage drop
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r t h times i plus v like that now this current
i that is flowing through this resistor is
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also the capacitor current and therefore i
is equal to c d v d t so we substitute for
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i c d v d t and then we get v t h equal to
r t h c d v d t plus v we can rewrite this
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equation as t v d t plus v divided by r t
h times c equal to v t h divided by r t h
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times c all we have done here is divided both
sides by this vector here
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notice that our right hand side here is a
constant because our sources are d c sources
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and therefore v t h is a constant all right
now how do we solve this ah ordinary differential
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equation we first obtained the homogenous
solution and we do that by replacing this
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right hand side the forcing function with
zero and then we get this equation here we
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have defined tau which is called the time
constant as r t h times c over here this time
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constant tau has units of time and let us
quickly verify that what are the units of
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r t h its a resistance so the units are ohms
ohms is volts by ampere and ampere is coulombs
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per second what about c c has units of charge
by voltage that is coulombs per volts so this
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volts cancels out this coulombs cancels out
leaving behind seconds that is the unit of
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time
let us now get the homogeneous solution we
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rewrite this equation as d v by v equal to
minus d t by tau now we can integrate both
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sides the integral of this left hand side
will give us log of v this log of course is
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to the base e and the integral of this part
the right hand side will give us minus t by
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tau and there is an integration constant here
which is k zero so this equation can be rewritten
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as v equal to e raised to minus t by tau plus
k zero and then we can define e raised to
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k zero as another constant k and therefore
we have v the homogenous part indicated here
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with this superscript h equal to k times exponential
minus t by tau
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the second part of the solution is the particular
solution what is the particular solution it
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is a specific function that satisfies the
differential equation in this case this is
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our differential equation so we are looking
for some solution which satisfies this equation
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and in r c circuits also r l circuits it is
very convenient to think of the steady state
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and that can be used as a particular solution
and we do that as follows we know that all
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time derivatives vanish as t tends to infinity
that makes the current equal to zero for a
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capacitor because the current is i equal to
c d v d t if v becomes constant then i becomes
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zero and then we get v particular indicated
here with this superscript equal to v t h
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for this specific case
how do we get this this current has become
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zero so therefore no voltage dropped here
and therefore we have v equal to v t h so
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that is how we have got this particular solution
and let us check that this solution does indeed
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satisfy the differential equation let us substitute
v equal to v t h in this equation here since
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v t h is a constant d v t h d t would be zero
and then we have on the left hand side v t
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h divided by r t h times c and that is precisely
the same as the right hand side so therefore
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this particular solution does satisfy our
ordinary differential equation all right
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now finally how about net solution is given
by the addition of the homogenous part which
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is here and the particular part which is here
so we have v equal to k exponential minus
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t by tau plus v t h and in general v of t
can be written as some constant times c raised
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to minus t by tau plus some other constant
where a and b can be obtained from known conditions
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on v and we will very soon see how that can
be done
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so here is the summary of what we studied
in the last slide if all sources all independent
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sources these are d c or constant then we
have v of t equal to a e raised to minus t
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by tau plus b where a and b are constants
and tau is called a time constant and its
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given by r thevenin times c where r thevenin
is the thevenin resistance as seen from the
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capacitor all right what about the current
the current can be computed once we know the
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voltage because we know that i if c d v d
t so all we need to do now is to differentiate
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this part and then multiply that by c so then
that gives us c times a a raised to minus
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t by tau times minus one by tau so that can
be rewritten as some other constant a prime
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times e raised to minus t by tau
one observation that we can make from this
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equation all of t equal to a prime e raised
to minus t by tau is the following as t times
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to infinity this e raised to minus t by tau
becomes zero and therefore the current becomes
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zero and this observation also agrees with
our intuition because as t tends to infinity
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we reach a steady state and in steady state
all quantities tend to be constant so therefore
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this v will also become a constant and since
the current here is c d v d t if v is constant
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that current is going to become zero
so therefore the capacitor behaves like an
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open circuit in steady state as t tends to
infinity what we have looked at so far is
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the voltage across the capacitor and the current
through over capacitor but because this circuit
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is a linear circuit we can actually comment
on all variables inside the circuit that is
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all voltages and all currents since the circuit
in the black box each linear any variable
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that is any voltage or any current in the
circuit can be expressed as x of t equal to
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k one a raised to minus t by tau plus k two
where k one and k two can be obtained from
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suitable conditions on that particular variable
x of t and as we look at some examples this
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procedure will become very clear
let us now look at this function f of t equal
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to e raised to minus t by tau which appears
in our solution as we saw in the last slide
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here is a table which shows how this function
varies with time the first column here is
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t by tau so this is t equal to zero this is
t equal to one times tau that is t equal to
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tau this is t equal to two tau this is t equal
to three tau and so on this column here is
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our f of t that is e raised to minus t by
tau and this column here is one minus f of
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t for t equal to zero e raised to minus t
by tau is one for t equal to tau it is point
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three seven for two tau it is point one three
for three tau point zero five four tau point
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zero one eight and five times tau it is point
zero zero six seven
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so we seen that as t increases this function
f of t gets closer and closer to zero what
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about one minus f of t to begin with it is
one minus one equal to zero at t equal to
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zero and then as f of t goes on decreasing
one minus f of t goes on increasing and eventually
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of course as e raised to minus t by tau approaches
zero one minus e raised to minus t by tau
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approaches one
so by the time we come to five times the time
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constant this function has almost become equal
to one so for t by tau equal to five e raised
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to minus t by tau is nearly equal to zero
this number here and one minus e raised to
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minus t by tau is nearly equal to one this
number here and therefore we can say that
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the transient lasts for about five time constants
after that things become constant here is
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the figure showing e raised to minus t by
tau as a function of t by tau this curve here
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and one minus e raised to minus t by tau as
a function of t by tau the blue curve here
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and as we have seen in this table at t equal
to zero we have e raised to minus t by tau
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equal to one and as t increases this function
goes to zero and after about five time constants
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that is after t by tau equal to five we see
that this function becomes approximately equal
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to zero and will stay at zero thereafter
similarly one minus f of t that is one minus
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e raised to minus t by tau starts off at zero
and eventually it becomes equal to one after
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about five time constants
let us now take a look at the plot of f of
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t equal to a e raised to minus t by tau plus
b where a and b are constants and why do you
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want to do that that is because as we have
seen in the earlier slides this is the general
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form for our solution and this will apply
to any quantity any current or any voltage
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in an r c circuit or r l circuit with d c
sources all right so let us first figure out
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some constraints on this function and then
we will look at the plot what about t equal
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to zero if we substitute t equal to zero here
e raised to minus t by tau becomes one and
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00:14:05,890 --> 00:14:15,050
therefore we have f equal to a plus b what
about t equal to infinity as t tends to infinity
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e raised to minus t by tau becomes zero and
therefore we have f tending to b
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from these two observations we conclude that
the graph of f of t will always lie between
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a plus b and b if a is positive then a plus
b is going to be greater than b and otherwise
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it is going to be less than b ok let us now
look at the derivative of f what is t f d
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t we need to differentiate this part that
is a e raised to minus t by tau times minus
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one by tau and the derivative of b of course
is zero because that is a constant so therefore
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we get t f d t equal to a e raised to minus
t by tau times minus one by tau at t equal
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to zero e raise to minus t by tau is one and
therefore we get d f d t equal to minus a
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over tau
now if a is positive then this derivative
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at t equal to zero is going to be negative
because of this minus sign here and what is
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the derivative this derivative is nothing
but the slope of the function at t equal to
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zero if a is less than zero then this derivative
is going to be positive what about t tending
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to infinity as t tends to infinity this term
becomes zero and therefore d f d t will tend
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to zero that means f becomes constant and
what is the constant we have already seen
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that that constant is b all right so as t
tends to infinity d of d t tends to zero irrespective
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of the sign of a because a simply does not
come into the picture it gets multiplied by
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this zero here
let us now look at these two cases namely
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a greater than zero a positive and a negative
let us start with a less than zero in this
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case a plus b is less than b so our initial
value which is a plus b is going to be smaller
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than our final value which is b what about
the slope if a is less than zero then the
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slope at t equal to zero is positive like
that and as t tends to infinity the slope
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anyway goes to zero irrespective of the sign
of a and we notice that also over here and
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notice also that things stop wearing after
five times tau as we have already seen in
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the last slide so after five tau this function
f of t becomes approximately constant does
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not change thereafter all right
now let us look at the second case that is
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a positive if a is positive a plus b is greater
than b that means our initial value is greater
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than our final value and that is what we see
over here this is the initial value that is
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the final value what about the slope at t
equal to zero if a is positive then the derivative
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at t equal to zero is negative and that is
indeed what we observe over here there is
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a negative slope here at t equal to zero
and after five time constants the function
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becomes constant and that is equal to d as
we expect and of course t f d t becomes equal
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to zero so these are the only two possibilities
we have if a is negative then f of t starts
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of with a positive slope at t equal to zero
there is some variation in f of t up to five
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tau and then it approaches the constant that
is b and the slope of course becomes zero
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in the other case when a is positive f of
t starts with a negative slope at t equal
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to zero then there is some variation up to
five tau and after that f of t becomes constant
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that means the derivative becomes zero
let us notice also that these functions are
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continuous the derivatives are also continuous
and therefore we dont see any abrupt changes
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00:19:10,309 --> 00:19:18,399
anywhere either in this case or in this case
and all of these are very important points
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and they will help us when we want to plot
a current as a function of time or a voltage
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as a function of time in r c or r l circuits
with b c sources now very soon we are going
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00:19:30,559 --> 00:19:37,350
to work out some examples and the plots that
we have going to obtain lets say for a current
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or a voltage would be either of this type
or of this type so whenever you see a plot
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you should really come back to this slide
and confirm that what we see over there is
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one of these forms
let us now look at r l circuits with d c sources
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so here is a linear circuit once again consisting
of resistors independent voltage sources independent
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00:20:07,510 --> 00:20:15,020
current sources and dependent sources to this
circuit we connect an inductor and our interest
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00:20:15,020 --> 00:20:21,980
is to find the voltage across the inductor
as a function of time and the current through
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00:20:21,980 --> 00:20:29,159
the inductor as a function of time step number
one we replace this original circuit with
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00:20:29,159 --> 00:20:37,090
its thevenin equivalent consisting of a voltage
source v t h in series with the resistance
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00:20:37,090 --> 00:20:49,850
r t h and the inductor of course appears as
it is as in the r c circuit case we will assume
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that all our sources these independent voltage
sources and independent current sources are
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00:20:56,850 --> 00:21:05,399
d c that means they are constant and therefore
this v t h is going to be a constant ok
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00:21:05,399 --> 00:21:14,179
let us now write the k v l equation for this
loop and what is that this v t h must be equal
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to this voltage drop plus this voltage drop
now this voltage drop across the resistance
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00:21:20,600 --> 00:21:33,240
is r t h times i and the voltage drop across
the inductor is given by l d i by d t to solve
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00:21:33,240 --> 00:21:41,279
this differential equation we first obtain
the homogenous solution and we do that by
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replacing this forcing function with zero
like that and then we divide both sides by
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l so we get d i d t plus one over tau times
i equal to zero where tau is l by r t h this
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00:21:59,100 --> 00:22:04,749
tau has units of time and that is the circuit
time constant
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00:22:04,749 --> 00:22:11,950
the homogenous solution indicated here with
the superscript h is then given by k times
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exponential minus t by tau where this k is
a constant to obtain a particular solution
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00:22:21,690 --> 00:22:26,350
once again we will use information from the
steady state condition like we did in the
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r c circuit case that is as t tends to infinity
we expect all quantities to become constant
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00:22:34,769 --> 00:22:42,190
therefore this current will also become constant
and v which is l times d i d t well then become
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00:22:42,190 --> 00:22:52,679
zero when v become zero this current is equal
to v t h divided by r t h so that is a particular
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00:22:52,679 --> 00:23:00,840
solution that we can use and finally we can
add the homogenous and particular solutions
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00:23:00,840 --> 00:23:08,299
to obtain the net current as a function of
time so i is i h plus i p i h is given by
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00:23:08,299 --> 00:23:14,509
k e raised to minus t by tau i p is given
by v t h by r t h so therefore our net or
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00:23:14,509 --> 00:23:23,399
total current is k e raised to minus t by
tau plus v t h by r t h
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00:23:23,399 --> 00:23:28,289
and we dont need to really go through this
derivation every time the moment we see a
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00:23:28,289 --> 00:23:38,110
linear circuit with a single inductor connected
to it and when the sources are d c sources
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00:23:38,110 --> 00:23:44,700
we can write the current as a e raised to
minus t by tau plus b where a and b are constants
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00:23:44,700 --> 00:23:53,690
to be obtained from known conditions on the
current ok to summarize the discussion from
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our last slide if we have a linear circuit
with d c sources then we can write this inductor
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current as a e raise to minus t by tau plus
b where a and b are constants and tau the
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time constant is given by l divided by r t
h
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what about the voltage across the inductor
once we know the current we can obtain the
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voltage as l d i d t so all we need to do
now is to differentiate this current and then
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multiply that by l and that gives us l times
a e raised to minus t by tau times minus one
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over tau
now all of this l times a times minus one
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over tau can be clubbed together into another
constant called a prime so therefore the voltage
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is given by a prime e raised to minus t by
tau what happens as t tends to infinity as
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t tends to infinity this e raised to minus
t by tau tends to zero and therefore the voltage
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across the inductor becomes zero and what
is the meaning of that when the voltage here
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is zero that means it is just like a wire
with no voltage drop across it so the inductor
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behaves like a short circuit all right
now we have already looked at the voltage
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across the inductor and the current through
the inductor but we can make a much more general
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comment because this circuit is a linear circuit
and that is since the circuit in the black
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box is linear any variable that is any current
or any voltage inside the circuit can be expressed
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as x of t equal to k one e raised to minus
t by tau plus k two where k one and k two
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are constants and they can be obtained using
suitable conditions on x of t to summarize
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if we have a linear circuit with d c voltage
sources or d c current sources connected to
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a capacitor or an inductor then any quantity
in the circuit that is any current or any
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voltage is given by this expression here x
of t equal to k one e raised to minus t by
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tau plus k two
if we have a capacitor connected to this linear
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circuit then this time constant tau is given
by r t h times c where r t h is the thevenin
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resistance seen by the capacitor if we have
an inductor then this tau is given by l by
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r t h where r t h is the thevenin resistance
seen by the inductor
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in summary we have looked at the general form
of the solution in n r c or r l circuit with
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d c sources we have also seen how to plot
the solution as a function of time these observations
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will be very useful when we look at some examples
in the next lecture that is all for now
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see you next time