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welcome back to basic electronics in the last
lecture we obtained analytic expressions for
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the capacitor of voltage and current for a
series r c circuit with the step input voltage
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we considered both charging and discharging
transients in this lecture we will look at
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the graphs for these quantities and learn
a lot more about these transient we will then
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look at how to handle an r c or r l circuit
with the piece wise constant source and follow
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that up with an example ok
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let us start these are the plots the input
voltage is given by this ah step v form the
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light blue curve and the capacitor of voltage
is given by this curve here the dark blue
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curve and let us check whether this does correspond
to the expression for the capacitor voltage
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that we have over here what is v of t at t
equal to zero at t equal to zero this term
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is one so therefore we have v zero times one
minus one that is zero so at t equal to zero
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we expect v of t to be equal to zero what
about t equal to infinity at t equal to infinity
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this term is zero and therefore we get v equal
to v zero and that is what we see over here
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and how long does the transient take as we
are seen earlier the transient takes five
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ten constant
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let us check whether that is happening we
have r equal to one k c equal to one micro
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farad so the time constant is one k times
one micro that is one milliseconds so we expect
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the transient to last for five times the time
constant that is five milliseconds five milliseconds
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is over here and we observed that the transient
does vanish at about that time and there after
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the circuit teaches a steady state and there
is no perceptual variation in either the capacitor
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voltage or the capacitor current beyond that
point what about the current the current is
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given by these expression at t equal to zero
plus this term is one and therefore the current
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v zero by r our we zero is five volts r is
one k so this number is five volts divided
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by one k or five millions and that is what
we observe over here
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at zero minus the current is zero at zero
plus it is five millions as t to needs to
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infinity this term goes to zero and therefore
the current is expected to go to zero and
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that is indeed what we observe over here all
right
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let us know make a five observation first
the capacitor of voltage is continuous and
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that is of course expected as we explained
earlier because if there was a discontinuity
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in the capacitor voltage that would need to
very large currents which would not satisfy
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our circuit equations on the other hand there
is no subscription on the capacitor current
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and we do observe discontinuity at t equal
to zero second observation the capacitor current
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is always positive as is also obvious from
this equation this term e raised to minus
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t by tau e is always positive and v zero and
r are positive constants and therefore the
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capacitor current is always positive
now since the capacitor current is c d v by
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d t a positive capacitor current means a positive
d v d t and therefore we expect the capacitor
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voltage to always keep rising like that eventually
of course the capacitor current becomes zero
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and therefore the capacitor voltage becomes
constant if the current is zero that means
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d v d t is zero that means v becomes constant
as you know over here in this graph since
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the current is large in the beginning we have
a larger slope d v d t here now the current
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goes on decreasing so this slope also goes
on decreasing and eventually of course the
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current becomes zero and the slope becomes
zero that means v becomes constant
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finally notice that the shape of v of t or
i of t is one of the two shapes we saw earlier
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in the graph of f of t equal to a e raised
to minus t by tau plus b if you dont remember
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that you can go back a few slides and make
sure that these two shapes indeed follow what
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we predicted in that slide over there ok
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let us now look at the discharging transient
this is the circuit v s goes from five volts
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to zero volts at t equal to zero and the expressions
for v of t and i of t which we derived earlier
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are reproduced over here v of t is v zero
e raised to minus t by tau i of t is minus
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v zero by r e raised to minus t by tau here
are the plots this is our source voltage v
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s it starts at five volts it has been five
volts for a long time and then at t equal
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to zero it abruptly changes to zero and then
remains zero thereafter this dark blue curve
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is the capacitor voltage as a function of
time and this graph here is the capacitor
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current notice that the capacitor current
is negative
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let us check whether these results the plots
here correspond to the equations that we have
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derived
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let us look at the capacitor voltage first
at t equal to zero this term is one and therefore
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v of t is v zero v zero is five volts and
therefore the capacitor voltage is also five
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volts as t tends to infinity this term becomes
zero and therefore the capacitor voltage becomes
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zero as in this graph and how long does this
take we expect that to take five time constants
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r is one k here c is one micro farad as in
the previous case and therefore the time constant
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is one millisecond five tau is five milliseconds
therefore at five milliseconds we expect the
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transient to vanish and that is what we observe
over here v becomes constant after about five
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milliseconds
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let us look at the current plot now here is
the expression for the current and remember
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that this is valid for t greater than zero
now at t equal to zero plus this exponential
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factor becomes one and we have i of t equal
to minus v zero by r v zero is five volts
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r is one k and therefore we have i of t equal
to minus five volts by one k that is minus
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five millions and that is indeed what we observe
in this graph here as t increases this exponential
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factor decreases and therefore we expect the
capacitor current to decrease in magnitude
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it will of course remain negative because
of this minus sign eventually as t tends to
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infinity this factor will become zero and
therefore the current will become zero and
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that is what we observe in this graph as t
increases the current decreases in magnitude
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and eventually it becomes zero as we would
expect the capacitor voltage is continuous
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but the current is discontinuous there is
a discontinuity at t equal to zero and this
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observation is similar to what we saw in the
charging case
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now the capacitor current in this case is
always negative eventually of course it becomes
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zero and what does that mean since the capacitor
current i is c d v d t negative current means
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a negative d v d t that means the slope of
the capacitor voltage versus time graph would
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always be negative and that is what we observed
over here
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so therefore the capacitor voltage keeps decreasing
eventually of course the current becomes zero
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d v d t becomes zero therefore the voltage
becomes constant so there is a lot to learn
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from graphs and you would have heard that
a picture is worth a thousand words and that
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is also true about graphs a graph is packed
with information we only need to look for
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it and there are several features that we
should look for and we should make sure that
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they are compatible with the analytical understanding
or the equations that we have
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let us now look at the effect of the time
constant on the capacitor voltage transient
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first let us consider the charging case in
which the source voltage changes from zero
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volts to five volts abruptly at t equal to
zero and as a result the capacitor voltage
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also changes it follows this equation which
we are derived earlier now with r equal to
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one k the time constant is one k times one
micro that is one millisecond and in about
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five time constants that is in five milliseconds
we expect the capacitor voltage transient
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to vanish and a steady state to be established
and that is what we observe in this plot this
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sky blue curve corresponds to r equal to one
k in the beginning the capacitor voltage has
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been zero volts for a long time at t equal
to zero it starts rising towards five volts
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and in about five milliseconds this time point
here it reaches almost five volts and does
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not change after that all right
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now let us look at the second case namely
r equal to one hundred ohms that is point
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one k what is the time constant in this case
it is point one k times one micro that is
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point one milliseconds now in this case we
expect the transient to last for five times
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point one milliseconds that is point five
milliseconds and that is indeed what we observe
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in this plot this is one millisecond point
five millisecond is somewhere here and we
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observe that for r equal to hundred ohms we
reach the steady state in about point five
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milliseconds and after that point the capacitor
voltage does not change
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let us now consider the discharging case so
v s changes from five volts to zero volts
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at t equal to zero and as a result the capacitor
voltage also changes and it follows this equation
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which we have derived earlier now with r equal
to one k the time constant is one millisecond
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and in about five time constants the transient
vanishes as seen in this graph this is five
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volts and ah t equal to five milliseconds
we reach steady state and the capacitor voltage
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does not change thereafter for r equal to
hundred ohms the time constant is point one
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millisecond and in five tau that is in point
five milliseconds we reach steady state here
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and the capacitor voltage becomes constant
so far we have considered r c and r l circuits
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with the constant source a voltage source
or a current source
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let us now look at r c and r l circuits with
a piece wise constant source here is an example
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there is a voltage source here which varies
with time like that in this interval t less
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than t one this voltage is zero between t
one and t two it is non zero lets say five
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volts and then for t greater than t two it
is zero again let us see how we can handle
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such a situation clearly this is not a d c
source or a constant source but it is piece
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wise constant we have one piece here which
is constant we have another piece here which
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is constant and we have a third piece here
which is constant
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and in each of these intervals we can use
the expressions that we have derived that
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is for any current or voltage x of t in the
circuit we can write a general expression
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such as x of t equal to a one e raise to minus
t by tau plus v one for the first piece that
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is t less than t one similarly for the second
interval from t one to t two we can write
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x of t as a two e raised to minus t by tau
plus b two now this a two and b two are also
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constants and generally they would be different
from a one and b one and finally in this third
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interval that is t greater than t two we can
write x of t as a three e raised to minus
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t by tau plus b three in order to get a complete
picture what we need to do next is to find
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this tau the circuit time constant and also
these constants a one b one a two b two etcetera
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now the time constant is given by r times
c for an r c circuit where r is the thevenin
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resistance seen by the capacitor for an r
l circuit tau is given by l divided by r where
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r is the thevenin resistance seen by the inductor
we have already seen these formulas earlier
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how do we obtain these constants a one b one
a two b two and so on what we need to do is
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to work out suitable conditions on x of t
which could be a current or a voltage at specific
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time points for example in this case we might
look at x of t at t one t two minus infinity
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and plus infinity
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and in order to do that in order to find these
conditions we can use the following considerations
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a if the source voltage or current has not
changed for a long time long compared to the
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circuit time constant tau then we know that
all derivatives must be zero because all quantities
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in the circuit all voltages and currents must
have become constant the circuit must be in
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a steady state for example take this piece
if v s has been zero for a long time then
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we know that when we come to t one minus that
is just a little bit before t one then the
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circuit is in steady state and all voltages
and currents must be constant and once we
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know that we know that the capacitor current
which is given by c d v c d t must be equal
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to zero
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so the capacitor can be treated as an open
circuit and the inductor voltage which is
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given by l d i l d t must also be zero because
i l would have become constant that means
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the inductor can be replaced with a short
circuit since the voltage across the inductor
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is zero b when a source voltage or current
changes say at t equal to t zero in this example
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t zero would be t one or t two then the capacitor
voltage or the inductor current cannot change
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abruptly as we have seen earlier that means
we see at t zero plus the capacitor voltage
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at t zero plus must be equal to v c at t zero
minus and in an r l circuit the inductor current
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i l at t zero plus must be equal to the inductor
current at t zero minus
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so using these considerations we can work
out suitable conditions on the current or
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voltage at specific time points and then use
those to compute these constants a one b one
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a two b two and a three b three in this example
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so once we have tau and once we have these
constants we have a complete description for
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the quantity of interest let us now apply
the ideas that we have just learnt in the
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last slide to this particular example we have
an r l circuit here with two registers r one
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and r two and a single inductor r one is ten
ohms r two is forty ohms and the inductance
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is point eight henry the input voltage is
a pulse given by this graph here t zero is
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zero and t one is point one second up to t
zero the source voltage v s is zero volts
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between t zero and t one it is ten volts and
after t one it is again zero volts and we
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are interested in finding this ah current
through the inductor as a function of time
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step number one is to identify intervals in
which the input voltage is constant and from
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this graph you can see that there are three
such intervals interval one t less than t
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zero where v s is zero volts interval two
t zero less than t less than t one in which
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v s is ten volts and interval three that is
t greater than t one in which v s is zero
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volts like we mentioned in the last slide
what we need to do next is to consider each
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of these intervals identify the circuit time
constant in each interval write a general
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expression for i of t in each of these intervals
and then find suitable conditions on i of
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t at suitable time points which will enable
us to calculate the coefficients the constants
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involved in those expressions so let us begin
with the circuit time constant
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now we figure that in all of these intervals
the time constant does not change because
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the circuit topology is the same in all three
cases so let us proceed with that calculation
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now what is the circuit time constant tau
it is l by r t h where r t h is the thevenin
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resistance as seen by the inductor to make
this calculation a little easier let us redraw
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the circuit as shown here what we have done
is we have taken this inductor on the other
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side that of course does not change the circuit
and now we look at the rest of the circuit
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this circuit here from the inductor and note
that the circuit topology is the same in all
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three intervals the only thing that is changing
is the value of v s in the first interval
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v s is zero in the second interval it is ten
and in the third interval it is zero
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so the thevenin resistance seen by the inductor
is the same in all three cases and how do
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we find that we deactivate the voltage source
that means we short this voltage source and
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then what happens is r one and r two come
in parallel
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so therefore r t h is simply r one parallel
r two so that turns out to be ten parallel
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forty or eight ohms then the time constant
is given by l divided by r t h over l is point
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eight n d r t h is eight ohms so therefore
the time constant is point one second next
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let us find suitable conditions on i of t
which will eventually enable us to find the
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constants involved in the expressions for
i of t in these three intervals let us consider
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t zero minus that means just before t zero
what is the situation at t equal to t zero
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minus the source voltage has been zero for
a long time we have a steady state all currents
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and voltages have become constants and therefore
this v which is l d i by d t is zero
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if v is zero the voltage across r two is zero
that means no current can flow through r two
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so the current paths that we have is like
this and since this voltage drop is zero the
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current i must be v s divided by r one now
v s is zero in this interval up to t zero
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minus and therefore the current i at t zero
minus must be zero like that now since this
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i is an inductor current we know that it must
be continuous that means i at t zero plus
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must be equal to i at t zero minus and therefore
we get i at t zero plus equal to zero ampere
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all right
let us now obtain another condition on the
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current and we will do that by assuming that
this v s is did not change in real life of
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course v s has changed at t one but let us
pretend that it has remained constant
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the question is does this make sense how can
we assume that v s is ten volts in this interval
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when it is actually zero volts the answer
to that question is that it does make sense
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as long as we are in this region up to t one
so what we will do is we will assume this
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condition but we will consider the solution
to be valid only up to this point t one to
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put it a bit informally the inductor or the
circuit does not really know that a change
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is going to take place add this point and
therefore the response of the circuit would
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be the same in this interval
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whether or not v s state constant or it went
back to zero at this point ok with this situation
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let us now find this current as t tends to
infinity what do we have at t equal to infinity
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the circuit would have reached steady state
all currents and voltages would have become
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constant therefore this v which is l d i d
t would have become zero so we have a short
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circuit here and the current i would then
be v s divided by r one the current through
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r two of course would be zero because this
voltage drop is zero at t equal to infinity
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v s would be ten volts so therefore the current
i would be ten divided by r one that is ten
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volts divided by ten ohms that is one ampere
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let us now use these two conditions namely
i at t zero plus and i at infinity to obtain
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the solution for i of t in this interval and
we will do that in the next slide
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in conclusion we have started looking at r
c and r l circuits with a piece wise constant
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source in the next lecture we will continue
with the r l circuit example and obtain the
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complete solution until then goodbye