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welcome back to basic electronics in this
lecture we will look at a series r c circuit
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with a single resistor a single capacitor
and a step input voltage it is a simple circuit
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and the solution is well known we will not
simply state the solution but derive it using
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what we already know and when we do that we
will learn some useful ideas which would certainly
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help in the long run so lets begin
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we have looked at the general solution for
a current or a voltage in an r c circuit with
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d c sources let us now answer this important
question given an r c circuit we want to find
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out if the capacitor voltage can change suddenly
or abruptly and we will answer this question
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with the help of this example here what do
we have here we have a series r c circuit
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r is one k c is one micro farad and this r
c circuit is given by a step voltage source
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up to t equal to zero the voltage is zero
volts and at t equal to zero it changes from
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zero volts to five volts
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the capacitor voltage is given to be zero
volts at t equal to zero so we see zero up
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to this point and now as a result of ah v
s changing from zero volts to five volts we
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expect this ah capacitor voltage to change
so this is our ah question v s changes from
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zero volts at t equal to zero minus to five
volts at t equal to zero plus that is just
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after t equal to zero and as a result of this
change v c will rise the question is how fast
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can v c change
so for example what would happen if v c changes
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by one volt in one microsecond and ah let
us assume that this is happening at a constant
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rate that is one volt divided by one microsecond
or ten raise to six volts per second because
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of this d v c d t a capacitor current will
flow which is given by i equal to c d v c
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d t c is one micro farad and d v c d t is
ten raise to six volts per second so this
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product turns out to be one ampere with i
equal to one ampere let us calculate the voltage
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drop across this resistance down this r is
one k or one thousand ohms and r times i is
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the voltage drop so that would be one thousand
ohms times one ampere
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so that is one thousand volts thats a very
large drop and ah this would simply not be
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allowed by k v l because we have only five
volts here we cannot have one thousand volts
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over here and therefore we conclude that v
c at t equal to zero plus must be equal to
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v c at t equal to zero minus if this were
not equal that would lead to a very large
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capacitor current and ah the k v l equation
would not be satisfied
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so therefore a capacitor does not allow abrupt
changes in v c if there is a finite resistance
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in the circuit as long as there is a finite
resistance in the circuit which is usually
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the case we must have the capacitor voltage
to be continuous all right now in an r l circuit
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we must have continuity of the inductor current
because any abrupt change in inductor current
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would lead to a very large l d i l d t that
is a very large voltage across the inductor
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and that again would not be allowed by k v
l
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so the take home point from this slide is
that the capacitor voltage must be continuous
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and [noises] the inductor current also must
be continuous as long as there is a finite
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resistance in the circuit as our first example
let us consider this simple r c series circuit
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so we have r and c in series and this combination
is driven by a voltage source with a step
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voltage so this v s is zero volts up to t
equal to zero and then it abruptly changes
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to v zero and it remains constant after that
point and we are interested in finding this
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ah capacitor voltage and the capacitor current
as a function of time
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let us begin by writing v of t the capacitor
voltage as a e raised to minus t by tau plus
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b where a and b are constants to be determined
now this expression if you recall is valid
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only as long as the independent sources in
the circuit are constant now in our case we
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should use this equation only for t greater
than zero because our voltage is constant
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only for that region we cannot use this same
equation to describe this region for example
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what is this time constant in this ah equation
it is simply the product of c and ah thevenin
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resistance as seen by the capacitor
now this circuit is already in the terminate
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form and r t h as seen by the capacitor is
simply r so therefore this time constant is
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r times c let us now obtain these constants
a and b and ah first let us look at this circuit
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at t equal to zero minus what is the situation
that we have at t equal to zero minus the
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source voltage has been zero volts for a long
time and ah that means that we have a steady
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state the voltages and currents have settled
down to some constant values
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now if this v is constant then this current
i which is c d v d t must be zero and if that
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is the case there is no voltage drop across
this resistance and v is then equal to v s
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and v s is zero at t equal to zero minus therefore
v is also zero so we have v at zero minus
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equal to zero volts now as we have seen earlier
the capacitor voltage cannot change abruptly
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and therefore v at zero plus must be the same
as v at zero minus therefore v at zero plus
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is also equal to zero volts
now why are we concerned about v at zero plus
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that is because this equation applies only
for t greater than zero it does not apply
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at zero minus so therefore we can use this
condition to get a and b but not this condition
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ok let us now look at the circuit at t equal
to infinity what is the situation at t equal
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to infinity we have v s equal to v zero and
ah sufficient time has passed to reach steady
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state and therefore v and i and all other
variables in the circuit would have become
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constant and ah once again this i which is
c d v d t would be zero because v is a constant
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and then we have no voltage drop across this
resistance and the capacitor voltage v is
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the same as v s at t equal to infinity and
v s at t equal to infinity is the same as
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v zero and therefore we have v at infinity
equal to v s at infinity equal to v zero
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so that is our second condition we can now
use these two conditions v at zero plus equal
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to zero volts and v at infinity equal to v
zero to obtain these constants a and b so
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at zero plus we have v equal to zero so we
put zero over here this e raised to minus
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t by tau is one so therefore we have zero
equal to a plus b like that at t equal to
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infinity we have v equal to v zero so we put
v zero over here e raised to minus t by tau
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a zero as t tends to infinity therefore we
have v zero equal to b like that so our b
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is v zero and therefore a must be minus b
zero like that
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we now substitute these values for a and b
back into this equation and then that gives
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us the final result for v of t so v of t is
v zero times one minus e raised to minus t
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by tau so this equation describes the charging
transient let us now look at the discharging
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transient we now have v s going from v zero
to zero at t equal to zero and we are interested
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in finding v and i as a function of time as
a result of this change
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let us begin with v of t equal to a e raised
to minus t by tau plus b where a and b are
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constants and tau is the time constant given
by r times c because the thevenin resistance
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seen by the capacitor is once again r as in
this circuit here now this equation is valid
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for t greater than zero for t greater than
zero we have only a d c source in the circuit
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namely v s equal to zero and therefore this
equation is valid it is not valid in this
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region where v s is changing all right
now like we did in the charging transient
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let us find the conditions on v of t and then
calculate a and b we need two conditions one
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condition is v at zero plus and the other
condition is v at infinity so let us first
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get this condition and to do that let us look
at the circuit at t equal to zero minus what
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is the situation at t equal to zero minus
v s has been v zero for a long time that means
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all the variables in the circuit have become
constants this v has become constant and therefore
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i which is c d v by d t is zero at t equal
to zero minus if i is zero there is no voltage
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drop across r and therefore this v is the
same as v s that is v zero so that tells us
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that we at zero minus must be equal to v zero
and because the capacitor voltage must be
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continuous we must have v at zero plus equal
to v at zero minus therefore we get v at zero
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plus equal to v zero all right
what about t equal to infinity at t equal
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to infinity again the various quantities in
the circuit would have become constants because
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we have steady state therefore this v is a
constant i which is c d v by d t is zero this
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voltage drop is zero and this v is equal to
v s now v s at t equal to infinity is simply
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zero and therefore we have v at infinity equal
to zero volts
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now using these two conditions we can calculate
a and b let us see how to do that at t equal
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to zero plus we have v zero plus equal to
v zero therefore we substitute for v v zero
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and then for t zero here so e raised to minus
t by tau is one for t equal to zero therefore
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we get v zero equal to a plus b this equation
here for t equal to infinity we have v equal
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to zero so we substitute v equal to zero here
t equal to infinity this term now becomes
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zero so therefore we have zero equal to b
so our constant b is zero and therefore a
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is equal to v zero like that
now putting this back over here we get the
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final expression for v of t for the discharging
transient so that is v of t equal to v zero
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e raised to minus t by tau at this point it
is a good idea to stop the video and derive
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these equations for the charging transient
and the discharging transient it is of course
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always possible to memorize these equations
but that is not such a good idea because if
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we memorize things they dont stay with us
for a long time they tend to evaporate after
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some time whereas if we derive the same equations
they will stay with us for a much longer time
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so that is always advisable whenever possible
let us now find the current as a function
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of time for t greater than zero for the charging
transient there are two ways of doing this
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one we already know how this voltage varies
with time so what we can do is simply calculate
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i as c d v d t that is c d d t of v zero times
one minus e raised to minus t by tau so we
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differentiate this entire expression this
one of course is a constant so that drops
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out then we get c v zero by tau that tau coming
from this term here times e raised to minus
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t by tau that is v zero by r e raised to minus
t by tau because this tau is equal to r times
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c and that c will cancel out and we get v
zero over r here the other way to find i of
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t is to start from scratch that is we pretend
that we dont really know what v of t looks
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like we begin with an expression for i of
t and then find the constants involved in
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that expression
so let i of t be a prime e raised to minus
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t by tau plus b prime for t greater than zero
for which region we have d c conditions because
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our v s is simply a constant equal to v zero
now we can do that because as we have said
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earlier any current or any voltage in the
circuit has this form and therefore we can
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write i of t like this what is the next step
the next step is to find conditions on i of
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t which will enable us to find a prime and
b prime
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so let us do that now as we have seen earlier
at t equal to zero plus the voltage across
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the capacitor is zero and that tells us what
the current should be if we have v s equal
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to v zero at ah zero plus v equal to zero
the voltage difference across this resistance
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is v zero minus zero and therefore the current
is v zero minus zero divided by r that is
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v zero by r what about t equal to infinity
as t tends to infinity we have steady state
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and all quantities including this voltage
become constant and therefore i which is c
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d v by d t becomes zero so at t equal to infinity
we have i equal to zero using these conditions
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we obtain these constants now we substitute
t equal to zero here zero plus i equal to
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v zero by r here that gives us an equation
for a prime plus b prime then we substitute
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t equal to infinity here so this term drops
out i equal to zero here so that tells us
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that b prime must be zero so that gives us
a prime equal to v zero by r and b prime equal
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to zero and finally putting these back over
here we get i of t equal to v zero by r e
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raised to minus t by tau and that of course
is exactly the same as what we got in the
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first case that of course is expected
let us consider the discharge in case now
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and find i of t for t greater than zero there
are two ways to do this again one we already
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know v of t for t greater than zero that we
calculated in the last slide since v of t
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is known the current is given by c d v d t
so therefore i is c d d t of v zero e raised
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to minus t over tau that is v of t as we obtained
in the last slide when we differentiate this
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quantity we get minus c v zero by tau times
e raised to minus t by tau and since tau is
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equal to r c we get minus v zero over r e
raised to minus t by tau
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in the second method we start from scratch
assume some form for i of t and then find
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the constants involved in that expression
using conditions found for i of t so let us
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do that let i of t be a prime e raised to
minus t by tau plus b prime and this constants
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a prime and b prime need to be determined
using conditions on i of t now this expression
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holds for t greater than zero for which we
have d c conditions namely the source voltage
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is a constant that is zero volts all right
at t equal to zero plus as we found in the
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last slide we are v equal to v zero across
the capacitor and that gives us i at zero
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plus what is i at zero plus it is this voltage
minus this voltage divided by r this voltage
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for zero plus is zero this voltage is v zero
so the current is zero minus v zero divided
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by r that is minus v zero over r
next let us consider t equal to infinity what
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is the situation as t tends to infinity we
have steady state all currents and voltages
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in the circuit have become constant therefore
this voltage has also become constant and
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i which is c d v d t is zero so for t equal
to infinity we have i equal to zero now using
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these two conditions i at zero plus equal
to minus v zero over r and i at infinity equal
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to zero we can obtain these constants a prime
and b prime a prime turns out to be minus
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v zero by r v prime turns out to be zero and
that gives us i of t equal to minus v zero
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over r times e raised to minus t over tau
let us now compare the charging and discharging
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transients in particular let us look at the
currents in the charging case the current
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is given by v zero by r e raised to minus
t by tau and in the discharging case the current
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is given by minus v zero by r e raised to
minus t by tau
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so in the charging case this current is always
positive for t greater than zero and in the
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discharging case the current is always negative
for t greater than zero and ah that is exactly
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what we would expect if the current is positive
that means this capacitor voltage will go
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on increasing and that corresponds to the
charging process for the capacitor on the
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other hand when the current is negative that
is when this i is negative the actual current
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is flowing in that direction and that will
decrease the capacitor voltage so therefore
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the capacitor is actually discharging ok
let us now look at the plots for v of t as
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well as i of t both in the charging case and
the discharging case here is the charging
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case the source voltage is going from zero
volts to five volts at t equal to zero and
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these are the expressions which we already
obtained earlier for the voltage and the current
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this is the capacitor voltage and that is
the current through the capacitor so v of
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t is v zero one minus e raised to minus t
by tau and i of t is plus v zero by r e raised
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to minus t over tau
to summarize we have derived analytic expressions
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for the capacitor voltage and current for
a series r c circuit when a step input voltage
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is applied in the next lecture we will look
at graphs showing these quantities as a function
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of time until then goodbye