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welcome back to basic electronics in this
lecture we will continue with the r l circuit
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example from the previous lecture we will
then look at an r c circuit with a switch
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and learn how to obtain currents and voltages
in the circuit before and after the switch
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position changes we will then look at an r
c circuit in the periodic steady state with
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a square wave input ok let's get started
so this is the solution that we obtained using
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the conditions that we found for i of t at
zero plus and infinity at zero plus i is zero
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and as t tends to infinity i is one ampere
all right now this of course is only part
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of the story why because in reality the source
voltage v s changes at t equal to t one so
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therefore this solution is really valid up
to t one that is up to point one second and
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we need to work out the rest of the solution
that is the solution for t greater than t
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one this interval here and how do we go about
that we know that as we go from this interval
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t zero to t one to this next interval t one
to infinity the inductor current must remain
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continuous that means the value of the inductor
current i at t one minus that is just before
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t one must be the same as the value of i at
t one plus that is just after t one
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for this interval t zero less than t less
than t one this interval here i of t turns
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out to be one minus e raised to minus t by
tau ampere and this expression describes this
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part of the solution and ah you are of course
encouraged to derive this expression how do
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you do that let i of t be k one e raised to
minus t by tau plus k two where k one and
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k two are constants to be determined and now
use the conditions that we derived in the
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last slide namely i at zero plus equal to
zero and i at infinity equal to one ampere
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in those conditions you can find k one and
k two and then you will end up with this expression
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here and once we have this expression we can
find i at t one minus that means i at point
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one second all we need to do is replace this
t with point one second this tau as we saw
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in the last slide is also point once again
so we end up with i of t equal to one minus
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e raised to minus point one by point one that
is e raised to minus one
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so that turns out to be point six three two
amperes and ah i at t one plus as we just
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argued is the same as i at t one minus so
therefore i at t one plus is point six three
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two amperes so that is one condition that
we have for i of t in this last interval here
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what is the second condition the second condition
is i at infinity so let us see what that should
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be what is the situation at t equal to infinity
v s is zero and it has been zero for a long
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time the circuit is in steady state all quantities
voltages and currents would have become constant
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this current would have become constant so
therefore this v which is l d i d t would
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be zero this voltage is therefore zero no
current flows through r two and i is simply
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v s divided by r one since v s is zero at
t equal to infinity the current is zero amperes
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like that all right
so now we have two conditions i at t one plus
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and i at infinity and now we can get the complete
solution for this last interval here let i
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of t be equal to a e raised to minus t by
tau plus b in this interval that is t greater
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than t one and now let us find a and b using
these two conditions it is convenient to rewrite
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i of t as a prime e raised to minus t minus
t one by tau plus b we are not really changing
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this equation all we are doing is shifting
the origin from t equal to zero to t equal
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to t one all right
let us now use these two conditions i at t
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one plus is point six three two amperes and
i at infinity equal to zero ampere at t one
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plus we have i equal to point six three two
amperes so we put point six three two here
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t equal to t one plus here so this argument
becomes zero and e raised to zero is equal
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to one so therefore we get point six three
two equal to a prime plus b at infinity i
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is zero so we put zero here and e raised to
minus infinity is zero therefore this first
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term goes away and we get zero equal to b
and putting these together we get our final
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expression for i of t in this last interval
t greater than t one and that is i of t equal
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to point six three two e raised to minus t
minus t one divided by tau amperes and this
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is what the solution looks like in the final
interval t greater than t one at t one i is
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point six three two amperes and at t equal
to infinity it is zero ampere
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so now we have different pieces of the solution
available to us one piece in this interval
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and one piece and this interval all we need
to do now is to put these together like that
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so this is our complete solution for i of
t the sequel file for this particular simulation
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is available to you given here and you can
play with it for example you can change r
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two from forty ohms to twenty ohms and ah
predict first of all what should happen and
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then run the simulation and check that ah
your prediction is correct
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here is our next example it is an r c circuit
with a switch that opens at t equal to zero
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so the switch has been closed for a long time
and opens at t equal to zero and ah we are
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interested in finding this current i as a
function of time now this situation is very
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different than what we have been looking at
so far because when the switch opens our circuit
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changes before t equal to zero we have one
circuit and after t equal to zero we have
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another circuit so let us see how to handle
that
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here is the situation for t less than zero
that is when the switch is closed when the
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switch is closed we have a short circuit here
assuming the switch is ideal and therefore
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this circuit data applies after the switch
opens that is when t is greater than zero
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this switch is an open circuit and therefore
this entire branch is not in the circuit anymore
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and the circuit reduces to this circuit shown
over here
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now this situation for t less than zero has
been there for a long time so therefore this
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circuit is in steady state you can use that
fact to find v c over here and that capacitor
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voltage value serves as the bridge between
these two situations because as we know the
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capacitor voltage must be continuous so we
see at zero plus must be equal to v c at zero
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minus
to begin with let us look at the situation
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at t equal to zero minus that means we are
looking at this circuit and ah at zero minus
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the switch has been closed for a long time
so this circuit is in steady state what is
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the meaning of steady state that means all
currents and voltages have become constant
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so therefore this voltage we see has also
become constant and therefore i c which is
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c d v c d t is equal to zero so this is an
open circuit and ah the current path is given
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by this path here and that tells us what this
current i should be it is simply six volts
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divided by the total resistance in the circuit
that is one k plus five k so six volts divided
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by six k that is one milliamp
so we have i at zero minus equal to one milliamp
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what about v c at zero minus we have seen
that this current is zero at zero minus and
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therefore there is no voltage drop across
this ah resistance and then this voltage v
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c is the same as this voltage here that is
i multiplied by five k and i is one milliamp
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so therefore we have one milliamp times five
k or five volts so v c at zero minus is five
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volts and because the capacitor voltage must
be continuous v c at zero plus in this circuit
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must also be five volts
so therefore we have v c zero minus equal
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to five volts and v c zero plus also equal
to five volts and that tells us what i at
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zero plus should be let us look at this circuit
because now we are talking about t greater
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than zero this voltage is five volts and i
therefore is five volts divided by five k
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plus five k that is point five milliamps all
right
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now let us find the current i as a function
of time for t greater than zero and let us
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begin by assuming that i of t has this form
a e raised to minus t by tau plus b so we
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need to determine three things tau the time
constant and then these constants a and t
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what is the time constant for the circuit
it is r t h times c where r t h is the terminal
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resistance as seen by the capacitor and ah
in this case the terminal resistance is simply
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these two resistors in series so therefore
five k plus five k or ten k so the time constant
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then is ten k times five micro farad that
is fifty milliseconds all right
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what about a and b to find a and b we need
two conditions on i we already have one condition
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here i at zero plus equal to point five milliamp
let us now find i at infinity and that gives
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us the second condition to find i at infinity
let us look at this circuit what is i here
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i is v c divided by ten k now as this ah current
flows the capacitor gets discharged so that
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will keep happening until the entire charge
on the capacitor there is exhausted so finally
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we have no charge on the capacitor v c becomes
equal to zero and therefore i becomes equal
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to zero
another way to find i at infinity is to use
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the fact that the circuit would be in steady
state as t tends to infinity and all quantities
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currents and voltages would have become constant
at that time so therefore this v c would have
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become constant and the current i c which
is c d v d t would be zero if i c is zero
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then of course i is also equal to zero
so now we have two conditions i at zero plus
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equal to point five milliamp and i at infinity
equal to zero and using these two conditions
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we can find a and b and we get this expression
finally i of t equal to point five e raised
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to minus t by tau milliamps
here is a plot of i as a function of time
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and the current i is in milliamps here this
is zero and that is one milliamp at zero minus
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i is equal to one milliamp and in fact it
has been one milliamp for a long time because
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the switch has been closed for a long time
so therefore the current i has been constant
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when the switch opens the current changes
and at zero plus that means just after t equal
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to zero where i is equal to point five milliamp
point five milliamp is write here
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so therefore at t equal to zero there is a
discontinuity before zero we have one milliamp
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and after zero we have point five milliamp
and subsequently i of t is given by point
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five e raised to minus t by tau milliamps
and that part of the curve is shown over here
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eventually of course as t tends to infinity
this term becomes zero and i tends to zero
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as we see here
now how long does the transient last we expected
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to last for five time constants what is the
time constant the time constant is fifty milliseconds
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five times that is two fifty milliseconds
that is point two five seconds here is zero
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here is point one second point two second
and so on so point two five seconds is here
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and ah we do observe that after point two
five seconds the current does not change it
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becomes constant
here is the capacitor current as a function
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of time and that is also plotted in milliamps
this is zero and this is minus point five
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milliamps what is the situation at zero minus
at zero minus we have this circuit and it
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is in steady state v c is constant and therefore
the capacitor current is zero and that is
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what we see over here and it has been zero
for a long time because the switch has been
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closed for a long time so that explains this
part of the plot what about zero plus after
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t equal to zero this circuit comes into picture
and then we have i c equal to minus i
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and since i starts off at point five milliamps
and goes to zero eventually i c starts off
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at minus point five milliamps and then goes
to zero finally and notice that there is a
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discontinuity at t equal to zero all right
let us now look at the capacitor voltage what
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is v c at zero minus we already looked at
that v c at zero minus is five volts and it
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has been five volts for a long time and that
explains this part of the plot what about
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t greater than zero for t greater than zero
we have this circuit and v c now is equal
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to i multiplied by five k plus five k that
is ten k multiplied by i now i is point five
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milliamp at zero plus and then eventually
it goes to zero so therefore v c would go
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from point five milliamps times ten k that
is five to zero like that
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and as we would expect the capacitor voltage
is of course continuous there is no discontinuity
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at t equal to zero the sequel file for this
particular simulation is given over here and
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you can run this simulation and look at all
of these variables and also some other variables
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that you may be interested here here is our
next example it is a series r c circuit with
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r equal to five k and c equal to one micro
farad and the input voltage is a square wave
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in this interval the input voltage is high
that is ten volts and in this interval it
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is low that is zero volts and then it repeats
it is periodic
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let us say that our capacitor is initially
uncharged that means v c is equal to zero
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volts at t equal to zero as shown over here
and ah at t equal to zero the input voltage
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goes high to ten volts as a result of that
the capacitor is going to start charging like
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that toward ten volts but that does not happen
of course because at this point the input
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voltage goes back to zero volts and now the
capacitor starts discharging towards zero
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and that discharging process is also not completed
because the input voltage once again becomes
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high so therefore the capacitor starts charging
again and this process repeats until finally
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we have a steady state situation
this kind of steady state is called periodic
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steady state as suppose to the d c steady
state that we have looked at earlier what
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happens in the periodic steady state let us
take this v c as an example v c starts off
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at this point it does change within one period
but then after one period is over it comes
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back to the same value where it started so
that is the meaning of periodic steady state
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let us now look at the current waveform the
current is denoted by i c here the capacitor
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current in this first interval when the input
voltage is high the current is positive and
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in fact that is responsible for this increase
in the capacitor voltage in the second interval
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the input voltage has gone back to zero and
the current becomes negative as shown over
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here and ah this negative current is responsible
for this decrease in v c so this keeps happening
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the current is positive here negative here
positive again negative again and so on and
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finally the capacitor current waveform also
reaches the periodic steady state that means
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within one period there is a change in i c
but after one period is over the i c value
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comes back to where it started all right and
our problem has to do with finding v c of
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t and i c of t in the periodic steady state
let us look at the problem statement now this
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is our input voltage waveform it's a square
wave wave from zero to v zero and then back
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to zero t one is the interval in which the
input voltage is high that is v zero and t
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two is the interval in which the input voltage
is low that is zero volts all right so the
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problem is to find expressions for v c of
t that is the capacitor voltage and i c of
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t the capacitor current in the periodic steady
state and these expressions will be in terms
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of r c v zero this value t one and t two
here is a schematic diagram showing the capacitor
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voltage and the capacitor current as a function
of time in the periodic steady state the light
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blue graph is the input voltage v s during
this interval marked by t one the input voltage
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is high equal to v zero and during this interval
marked by t two here the input voltage is
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low that is zero during the t one phase the
capacitor voltage rises from v one to v two
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and during the t two phase the capacitor voltage
decreases from v two to v one so within one
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period the capacitor voltage comes back to
where it started off namely v one
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in the t one phase when the capacitor voltage
is increasing we have a positive capacitor
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current and in the t two phase when the capacitor
voltage is decreasing that means when the
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capacitor is discharging we have a negative
capacitor current and ah the capacitor current
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is also periodic
now the maximum value of i c is denoted by
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i one over here and the minimum value is denoted
by minus i two now this minus sign is introduced
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to make our algebra easier so note that this
i two is a positive number so minus i two
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is negative our objective is to find expressions
for v c of t and i c of t let us begin with
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v c and ah we should note first that v c will
be represented by two expressions one expression
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in this t one phase and another expression
in this t two phase all right so let us begin
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with the t one phase and in that phase let
v c be a e raised to minus t by tau plus b
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00:24:57,220 --> 00:25:08,410
and we will call that as v c one where the
superscript one indicates this t one phase
195
00:25:08,410 --> 00:25:15,650
so we now need to find a and b the time constant
of course in this case is simply r times c
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to summarize we learned how to treat an r
c circuit with a switch we also looked at
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the salient features of the current and voltage
plots for that circuit we then started with
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00:25:31,090 --> 00:25:37,600
an r c circuit with a square wave input we
looked at the meaning of the term periodic
199
00:25:37,600 --> 00:25:43,919
steady state and started analyzing the circuit
in that condition in the next class we will
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00:25:43,919 --> 00:25:49,929
complete the analysis and also compare the
results with plots obtained by circuit simulation
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00:25:49,929 --> 00:25:50,940
until then goodbye