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Transient Analysis - First order R C and R L Circuits

WEBVTT
Kind: captions
Language: en

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Hey friends, welcome to the YouTube channel
ALL ABOUT ElECTRONICS.
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So, in the last video, we had seen that what
is transient analysis and what is the importance
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of this transient analysis.
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And then after we had seen the behaviour the
basic components during the transients.
00:00:32.239 --> 00:00:37.860
So, in this video, we will see the transient
analysis for the first order RC and RL circuits.
00:00:37.860 --> 00:00:42.360
analysis for the first order RC circuit and
00:00:42.360 --> 00:00:47.720
first order RL circuits and for this circuits,
we will see the two kinds of responses.
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The first is the source free response and
second is the forced response.
00:00:51.460 --> 00:00:55.730
So, the source free response is the response
of the circuit when there is no source connected
00:00:55.730 --> 00:00:56.860
into the circuit.
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While this forced response is the response
of the circuit when there is some kind of
00:01:01.920 --> 00:01:04.449
forced input excitation is given to the circuit.
00:01:04.449 --> 00:01:10.780
So, in this video, we will provide forced
DC excitation to the RC and RL circuit and
00:01:13.900 --> 00:01:17.260
we will see how this circuit behaves for this
forced excitation.
00:01:17.260 --> 00:01:17.799
So, let's start with source free response
for the RC circuits.
00:01:17.799 --> 00:01:21.429
So, in this circuit, the switch is at position
number 1 since a long time.
00:01:21.429 --> 00:01:26.619
And at time t=0, this switch has been moved
from position number 1 to the position number
00:01:26.619 --> 00:01:27.619
2.
00:01:27.619 --> 00:01:32.280
So, if we see the equivalent circuit at time
t=0+, which is the just after the switch has
00:01:32.280 --> 00:01:36.289
been moved from position number 1 to position
number 2, then the equivalent circuit will
00:01:36.289 --> 00:01:37.340
look like this.
00:01:37.340 --> 00:01:41.590
Now, let's say the voltage across the capacitor
is Vc.
00:01:41.590 --> 00:01:49.609
Now, before this time t=0, let's say time
t=0-, since this voltage Vis applied continuously
00:01:49.609 --> 00:01:55.109
across this capacitor, so the voltage across
this capacitor will be V volt.
00:01:55.109 --> 00:02:00.999
Now, at time t=0+, this switch will be at
position number 2.
00:02:00.999 --> 00:02:04.759
And we know that the capacitor opposes the
instantaneous change of voltage.
00:02:04.759 --> 00:02:11.390
So, at time =0+ also, the voltage across the
capacitor will remain V.
00:02:11.390 --> 00:02:13.780
So, this will be our initial condition.
00:02:13.780 --> 00:02:19.790
So, now intuitively if you see, the charge
across the capacitor will get discharge through
00:02:19.790 --> 00:02:20.790
this resistor.
00:02:20.790 --> 00:02:23.980
So, initially, the voltage across the capacitor
will be V Volt.
00:02:23.980 --> 00:02:30.040
And with time this voltage gets reduced and
at time t is equal to infinity or after sufficient
00:02:30.040 --> 00:02:33.980
time the voltage across this capacitor will
go to the zero.
00:02:33.980 --> 00:02:37.769
Now, we don't know, how this voltage is getting
reduced.
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The voltage can go from voltage V to the 0,
either linearly or exponentially or any other
00:02:43.879 --> 00:02:44.879
way.
00:02:44.879 --> 00:02:48.560
So, let's find out how this capacitor is getting
discharged through this resistor.
00:02:48.560 --> 00:02:53.100
So, for that let's apply Kirchhoff's Voltage
Law in this circuit.
00:02:53.100 --> 00:02:58.890
So, applying KVL we can write
VR+VC=0
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Where VCis the voltage across this capacitor
and VRis the voltage across this resistor.
00:03:05.230 --> 00:03:12.140
Thar means I*R+VC=0
Where I is the current that is flowing through
00:03:12.140 --> 00:03:13.140
this circuit.
00:03:13.140 --> 00:03:17.280
Now, the same current I will also flow through
this capacitor.
00:03:17.280 --> 00:03:20.360
And we know that the capacitor current can
be given as
00:03:20.360 --> 00:03:26.489
I=C*dVc/dt
So, we can write
00:03:26.489 --> 00:03:35.170
R*[C*(dVc/dt)]+VC=0
So, if we further simplify it then we can
00:03:35.170 --> 00:03:42.690
write
dVc/dt + (1/RC)*VC=0
00:03:42.690 --> 00:03:48.939
So, if you see this equation, this is first
order linear differential equation.
00:03:48.939 --> 00:03:52.270
And there is no forced excitation in this
equation.
00:03:52.270 --> 00:03:54.590
That means this Q is Zero.
00:03:54.590 --> 00:04:00.810
And this Q term is zero, so the solution of
this equation will only contain source free
00:04:00.810 --> 00:04:03.840
response or complementary function only.
00:04:03.840 --> 00:04:07.540
And P.I term, that is particular Integral
will be zero.
00:04:07.540 --> 00:04:12.879
So, the complementary function or source free
response for the first order differential
00:04:12.879 --> 00:04:15.420
equation can be given as
Ae^(-P*t)
00:04:15.420 --> 00:04:26.670
So, similarly, for this equation, we can write
the solution VC as A*e^(-t/RC)
00:04:26.670 --> 00:04:35.640
So, now to find this coefficient A, let's
apply initial condition into this equation.
00:04:35.640 --> 00:04:41.670
So, at time t=0+, we can write this equation
as
00:04:41.670 --> 00:04:47.110
A*e^(-0)
That means VC=A
00:04:47.110 --> 00:04:55.920
So, now we know that at time t=0+, the voltage
across this capacitor will be V volt.
00:04:55.920 --> 00:05:01.170
So, we can say that the value of A will be
V Volt.
00:05:01.170 --> 00:05:13.850
And if you put this value into this equation
then we will get Vc(t)= V*e^(-t/RC)
00:05:13.850 --> 00:05:17.100
This is how the capacitor will get discharged
through this resistor R.
00:05:17.100 --> 00:05:21.200
So, now here this RC is known as the time
constant of the circuit.
00:05:21.200 --> 00:05:29.080
And we can denote this RC as T.
So, now we can represent this equation graphically
00:05:29.080 --> 00:05:30.080
as follows.
00:05:30.080 --> 00:05:35.200
So, at time t=0, the voltage across the capacitor
will be V volt.
00:05:35.200 --> 00:05:41.090
So, now in this equation, if we put the value
of t is equal to 1RC or 1-time constant then
00:05:41.090 --> 00:05:44.880
the capacitor voltage will be 0.37 Volt.
00:05:44.880 --> 00:05:52.500
So, in one time constant the capacitor voltage
will be only 37% of the original value.
00:05:52.500 --> 00:05:56.010
Similarly, in this equation, if we put the
value of t as
00:05:56.010 --> 00:06:03.470
2T, 3T, 4T and 5T then you will get corresponding
voltage across this capacitor.
00:06:03.470 --> 00:06:08.830
And as you can see here after 5 time constant,
the capacitor will get almost fully discharged.
00:06:08.830 --> 00:06:14.060
So, we can say that capacitor will get fully
discharged approximately in 5 time constant.
00:06:14.060 --> 00:06:18.490
So, suppose in your circuit, if you want that
your capacitor should get discharged in some
00:06:18.490 --> 00:06:23.320
particular time then by choosing the value
of R and C, you can decide this discharging
00:06:23.320 --> 00:06:24.320
time.
00:06:24.320 --> 00:06:29.260
Let's say in your circuit you want that capacitor
should get discharged in 5 microseconds.
00:06:29.260 --> 00:06:34.590
Now, we know that the capacitor will get discharged
in 5 RC time constant.
00:06:34.590 --> 00:06:39.060
So, this RC time constant will be 1 microsecond.
00:06:39.060 --> 00:06:47.320
Now, suppose if you have chosen value of R
as 1 Kilo Ohm, then value of C will be 1 microsecond
00:06:47.320 --> 00:06:50.850
/ (1 Kilo Ohm)
So, the value of C will come out as 10^-9
00:06:50.850 --> 00:06:55.950
F.
So, in this way, you can choose your value
00:06:55.950 --> 00:06:59.740
of R and C to get desired discharging time.
00:06:59.740 --> 00:07:05.060
So similarly, let's see now Sorce free Response
for the RL circuit.
00:07:05.060 --> 00:07:10.250
number 1 since a long time and the current
00:07:10.250 --> 00:07:13.070
I is flowing through this RL circuit.
00:07:13.070 --> 00:07:17.780
And at time t=0, the switch has been moved
from this position number 1 to the position
00:07:17.780 --> 00:07:18.810
number 2.
00:07:18.810 --> 00:07:23.210
So, at time t=0+, the equivalent circuit will
00:07:23.210 --> 00:07:30.750
Now, at time t=0-, let's say current that
is flowing through this inductor is IL.
00:07:30.750 --> 00:07:35.260
And this IL can be given as I.
00:07:35.260 --> 00:07:39.110
That means the current that is flowing through
this inductor will be I.
00:07:39.110 --> 00:07:43.120
Now, we know that the inductor opposes the
instantaneous change of current.
00:07:43.120 --> 00:07:50.140
So, at time t=0+ also, the value of this inductor
current will be I.
00:07:50.140 --> 00:07:55.300
So, now let's find out how this inductor current
will change during this transient.
00:07:55.300 --> 00:07:59.830
Law inside this circuit.
00:07:59.830 --> 00:08:04.570
Let's assume that the current I is flowing
inside this circuit.
00:08:04.570 --> 00:08:09.110
And VR is the voltage across this resistor
and VL is the voltage across this inductor.
00:08:09.110 --> 00:08:14.450
So, applying KVL we can write,
VR+VL=0
00:08:14.450 --> 00:08:22.660
Now, this VR can be given as I*R.
And VL can be given as L*di/dt
00:08:22.660 --> 00:08:27.640
write it as
00:08:27.640 --> 00:08:39.850
di/dt + (R/L)*i =0
Now, if you compare this equation with di/dt+Pi=
00:08:39.850 --> 00:08:45.520
Q
Then you can see here, this Q term is zero.
00:08:45.520 --> 00:08:50.440
So, the response of the circuit will be only
source free response, as there is no forced
00:08:50.440 --> 00:08:53.030
excitation into this circuit.
00:08:53.030 --> 00:08:54.860
And the response of the circuit can be given
00:08:54.860 --> 00:09:03.150
i(t)= A*e^[(-R/L)*t]
So, now to find the value of this coefficient
00:09:03.150 --> 00:09:05.340
A, let's apply initial conditions.
00:09:05.340 --> 00:09:11.640
So, as we have already found out earlier,
at time t=0, current that is flowing through
00:09:11.640 --> 00:09:13.050
00:09:13.050 --> 00:09:24.010
So, we can say that I(0+) = A*e^(-0)
So, we can say that I(0+) =A.
00:09:24.010 --> 00:09:29.410
And the value of this current I(t) at time
t=0+ is I Ampere.
00:09:29.410 --> 00:09:32.260
So, the value of A will be I.
00:09:32.260 --> 00:09:37.470
So, if we put this value of A, inside this
equation, then we will get
00:09:37.470 --> 00:09:44.800
i(t)= I*e^(-R*t/L)
which is
00:09:44.800 --> 00:09:46.950
the equation for the inductor current.
00:09:46.950 --> 00:09:51.630
And graphically if you see this inductor current,
the plot for this inductor current will be
00:09:51.630 --> 00:09:54.510
similar to the discharging curve for the capacitor.
00:09:54.510 --> 00:10:00.680
So, at time t=0, the value of this inductor
current will be I, and as the time progresses,
00:10:00.680 --> 00:10:05.050
the value of this inductor current will reduce
and at time t is equal to infinity or in steady
00:10:05.050 --> 00:10:08.660
state condition, the value of current I will
be zero.
00:10:08.660 --> 00:10:14.390
Now, here the ratio of L/R is known as the
time constant and it can be denoted as T.
00:10:14.390 --> 00:10:23.230
So, we can write this equation as I*e^(-t/T)
where t is nothing but L/R.
00:10:23.230 --> 00:10:29.480
So, now suppose if you want to find the value
of inductor voltage then this inductor voltage
00:10:29.480 --> 00:10:31.360
VL can be given as
L*di/dt
00:10:31.360 --> 00:10:37.680
So, now if you put the value of inductor current
IL inside this equation, and solve it then
00:10:37.680 --> 00:10:40.670
we will get the value of inductor voltage
VL as -I*R*e^(-R*t/L).
00:10:40.670 --> 00:10:47.730
So, in this way you can find the inductor
voltage as well.
00:10:47.730 --> 00:10:54.649
So, for this inductor voltage at time t=0,
the value of inductor voltage will -I*R.
00:10:54.649 --> 00:10:59.810
And as the time progresses, inductor voltage
gets exponentially decreased.
00:10:59.810 --> 00:11:03.710
And at time t is equal to infinity, inductor
voltage goes to the zero.
00:11:03.710 --> 00:11:08.050
So, so far we have seen the source free response
for the RC and the RL circuits.
00:11:08.050 --> 00:11:11.690
So, now let's see forced response for this
RC and RL circuits.
00:11:11.690 --> 00:11:16.620
So, we will provide some DC excitation to
this RC and RL circuit and we will see that
00:11:16.620 --> 00:11:19.430
how this circuit behaves during this transients.
00:11:19.430 --> 00:11:25.520
So, let's say in this circuit at time t=0,
the switch is getting closed.
00:11:25.520 --> 00:11:28.800
And this voltage V is getting applied across
00:11:28.800 --> 00:11:35.740
Now, before this time t=0, here we assume
that there is no charge across this capacitor.
00:11:35.740 --> 00:11:42.900
And at time t=0, once the switch is closed,
the equivalent circuit will look loke this.
00:11:42.900 --> 00:11:50.750
So, at time t=0-, since there is no charge
across this capacitor, so the value of capacitor
00:11:50.750 --> 00:11:52.420
voltage VC will be zero.
00:11:52.420 --> 00:11:57.790
Now, we know that the capacitor opposes the
instantaneous change of voltage, so at time
00:11:57.790 --> 00:12:02.250
t=0+ also, the capacitor voltage will be zero
as well.
00:12:02.250 --> 00:12:09.200
So, intuitively is you see, as this voltage,
V is applied across this RC circuit, the capacitor
00:12:09.200 --> 00:12:10.760
will start charging.
00:12:10.760 --> 00:12:15.830
So, at time t=0, there will not be any charge
across this capacitor.
00:12:15.830 --> 00:12:19.900
And as the time progresses, the capacitor
00:12:19.900 --> 00:12:22.830
And it will charge up to the voltage V.
00:12:22.830 --> 00:12:27.850
So, now let's find out how this capacitor
is getting charged up to this voltage V.
00:12:27.850 --> 00:12:32.990
It could be either linear, or it could be
exponential or it could be of any other form.
00:12:32.990 --> 00:12:37.720
charged through this voltage V.
00:12:37.720 --> 00:12:42.020
00:12:42.020 --> 00:12:48.650
V= R*i + Vc
00:12:48.650 --> 00:12:53.060
Where Vc is the voltage across this capacitor.
00:12:53.060 --> 00:12:57.160
Now, the same current I is also flowing through
00:12:57.160 --> 00:13:01.350
So, we can write this current I as capacitor
current.
00:13:01.350 --> 00:13:14.589
And capacitor current can be given as C*dVc/dt
So, we will get V= R*C*dVc/dt + Vc
00:13:14.589 --> 00:13:24.980
And if we further simplify this equation,
then we will get dVc/dt+ (1/RC)*Vc = V/RC
00:13:24.980 --> 00:13:39.190
So, now if you compare this equation with
di/dt + Pi=Q, then here P is nothing but 1/RC
00:13:39.190 --> 00:13:45.310
and Q is V/RC
So, now if you see here we also have this
00:13:45.310 --> 00:13:47.380
forced excitation into this circuit.
00:13:47.380 --> 00:13:52.150
So, now the solution of this equation will
contain two terms.
00:13:52.150 --> 00:13:55.180
Particular Integral and the Complementary
Function.
00:13:55.180 --> 00:13:59.300
So, the complementary function part will remain
same as we have found earlier.
00:13:59.300 --> 00:14:03.080
Which will be the source free response for
the circuit.
00:14:03.080 --> 00:14:08.600
And the Particular Integral is the solution
of the equation at time t is equal to infinity.
00:14:08.600 --> 00:14:13.700
So, we can say that for the given circuit,
particular integral will nothing but the voltage
00:14:13.700 --> 00:14:17.290
across this capacitor Vc at time t is equal
to infinity.
00:14:17.290 --> 00:14:20.650
So, let's, first of all, find this particular
integral.
00:14:20.650 --> 00:14:25.690
That is Vc at time t is equal to infinity.
00:14:25.690 --> 00:14:31.340
So, now we know that at time t is equal to
infinity or in steady state condition, this
00:14:31.340 --> 00:14:33.850
capacitor acts as an open circuit.
00:14:33.850 --> 00:14:38.950
So, the value of Vc or capacitor voltage at
time t is equal to infinity will be nothing
00:14:38.950 --> 00:14:40.250
but value V.
00:14:40.250 --> 00:14:45.089
So, our particular integral will be nothing
but V volt.
00:14:45.089 --> 00:14:48.240
So, now let's find out the complementary function.
00:14:48.240 --> 00:14:57.240
response can be given as A*e^(P*t).
00:14:57.240 --> 00:15:04.140
And here the value of Pis nothing but 1/RC
So, the source e free response for the given
00:15:04.140 --> 00:15:11.820
equation will be A*e^(-t/RC)
So, the overall response for the capacitor
00:15:11.820 --> 00:15:18.210
voltage can be given as
Vc(t) = V + A*e^(-t/RC)
00:15:18.210 --> 00:15:24.830
So, now let's find out this coefficient A.
And to find this coefficient A, let's apply
00:15:24.830 --> 00:15:26.270
the intial condition.
00:15:26.270 --> 00:15:30.970
So, at time t=0+, the capacitor voltage can
00:15:30.970 --> 00:15:46.750
Vc(0+)= V+A*e^(-0)
That is nothing but Vc(0+)= V+ A
00:15:46.750 --> 00:15:53.350
And at time t=0+, as there is no charge across
this capacitor, so the value of Vc will be
00:15:53.350 --> 00:15:54.350
0.
00:15:54.350 --> 00:16:00.510
So, we will get V+A= 0
Or in another term, we can say that A is equal
00:16:00.510 --> 00:16:02.510
to nothing but -V.
00:16:02.510 --> 00:16:06.750
So, the overall response Vc(t) can be given
00:16:06.750 --> 00:16:12.899
Vc(t)= V-V*e^(-t/RC)
And if we this equation graphically, then
00:16:12.899 --> 00:16:14.480
it will look like this.
00:16:14.480 --> 00:16:20.400
across the capacitor.
00:16:20.400 --> 00:16:23.140
And the value of capacitor voltage will be
00:16:23.140 --> 00:16:28.020
will charge exponentially and it will reach
00:16:28.020 --> 00:16:32.800
maximum value up to the voltage V at time
t is equal to infinity or in steady state
00:16:32.800 --> 00:16:33.800
condition.
00:16:33.800 --> 00:16:38.560
So, in this equation, first of all, let's
put the value of t as 1RC or 1 Time Constant.
00:16:38.560 --> 00:16:43.149
So, we will get the value of capacitor voltage
as 0.63 volts.
00:16:43.149 --> 00:16:51.000
So, we can say that in one time constant capacitor
used to get charge up to the 63% of its final
00:16:51.000 --> 00:16:52.000
value.
00:16:52.000 --> 00:16:57.820
And similarly, if we put a value of T as 2-time
constant, 3 time constant, 4 time constant
00:16:57.820 --> 00:17:03.620
and 5-time constant, then you will see that
capacitor will reach to its final value or
00:17:03.620 --> 00:17:06.659
approximately to its final value in 5 time
constant.
00:17:06.659 --> 00:17:11.730
So, in your design suppose if you want to
charge your capacitor very fast then by choosing
00:17:11.730 --> 00:17:15.769
the value of R and C you can decide the charging
time for the capacitor.
00:17:15.769 --> 00:17:19.480
So, this is all about forced excitation for
the RC Circuit.
00:17:19.480 --> 00:17:23.429
So, now let's see the forced responce for
the RL Circuit.
00:17:23.429 --> 00:17:29.230
So, in this circuit, we are providing the
DC excitation to this RL circuit at time t=0.
00:17:29.230 --> 00:17:35.899
So, before this time t=0, we are assuming
that there is no energy inside this inductor.
00:17:35.899 --> 00:17:39.929
And as soon as time t=0, the switch is getting
closed.
00:17:39.929 --> 00:17:42.879
then this voltage will get applied across
this inductor.
00:17:42.879 --> 00:17:45.110
And current will start flowing through this
circuit.
00:17:45.110 --> 00:17:49.259
00:17:49.259 --> 00:17:56.389
Now, we know that at time t=0-, there will
not be any current flowing through this inductor.
00:17:56.389 --> 00:17:59.659
Let's say IL is the current that is flowing
through this inductor.
00:17:59.659 --> 00:18:06.150
So, this IL can be given as 0 Ampere, as there
is a connection with this voltage source.
00:18:06.150 --> 00:18:09.850
Ad we know that capacitor opposes the instantaneous
change of current.
00:18:09.850 --> 00:18:15.720
current will be zero.
00:18:15.720 --> 00:18:20.649
So, now for the forced free response, let's
find out the equation for the inductor current.
00:18:20.649 --> 00:18:25.070
00:18:25.070 --> 00:18:33.440
So, applying KVL we can write, V= VR+VL
Where VLis the voltage across this inductor
00:18:33.440 --> 00:18:35.629
and VR is the voltage across this resistor.
00:18:35.629 --> 00:18:44.299
So, we can write it as, V= iR+L*di/dt
And if we further simplify it then we can
00:18:44.299 --> 00:18:54.600
di/dt + (R/L)*i = V/L
00:18:54.600 --> 00:18:59.659
di/dt + P*i = Q
00:18:59.659 --> 00:19:04.400
Then the value of P will be R/L and value
of Q will be V/L
00:19:04.400 --> 00:19:11.220
So, this V/L is nothing but forced excitation
for this given circuit.
00:19:11.220 --> 00:19:15.700
And as the circuit contains forced excitation
as well, so the solution of the equation will
00:19:15.700 --> 00:19:17.799
be having two terms.
00:19:17.799 --> 00:19:20.480
particular Integral and Complementary function.
00:19:20.480 --> 00:19:26.390
So, this Particular Integral in nothing but
the solution of the circuit at time t is equal
00:19:26.390 --> 00:19:27.940
00:19:27.940 --> 00:19:33.730
And this complementary function is the solution
of the circuit or equation for the source
00:19:33.730 --> 00:19:34.730
free response.
00:19:34.730 --> 00:19:37.970
That means when there is no source connected
00:19:37.970 --> 00:19:42.360
So, first of all, let's find out this Particular
Integral.
00:19:42.360 --> 00:19:47.610
So, this particular integral is nothing but
the inductor current at time t is equal to
00:19:47.610 --> 00:19:48.610
infinity.
00:19:48.610 --> 00:19:54.020
So, at time t is equal to infinity or in steady
state condition, this inductor will act as
00:19:54.020 --> 00:19:55.929
a short circuit.
00:19:55.929 --> 00:20:02.890
So, the current that is flowing through this
inductor IL can be given as V/R.
00:20:02.890 --> 00:20:09.090
So, the value of Particular Integral for our
case will be nothing but V/R.
00:20:09.090 --> 00:20:17.090
Now, the complementry function or source free
response for the circuit can be given as A*e^(-Pt)
00:20:17.090 --> 00:20:23.820
And here the value of P is nothing but R/L
So, the solution or complementary function
00:20:23.820 --> 00:20:27.460
solution, for the circuit will be nothing
but A*e^(-Rt/L).
00:20:27.460 --> 00:20:36.950
Now, if we combine this particular integral
and complementary function, then the overall
00:20:36.950 --> 00:20:46.690
solution for the circuit will be nothing but
V/R + A*e^(-Rt/L)
00:20:46.690 --> 00:20:54.590
00:20:54.590 --> 00:21:02.529
as V/R + A.
00:21:02.529 --> 00:21:07.070
And we know that at time t=0+, the value of
inductor current is 0.
00:21:07.070 --> 00:21:16.220
So, we will get V/R + A = 0
Or we can say that A is nothing but -V/R.
00:21:16.220 --> 00:21:26.169
So, we will get inductor current i(t) as,
V/R- (V/R)*e^(-R/t/L)
00:21:26.169 --> 00:21:30.029
And graphically if you see this equation,
the plot will be similar to the charging for
00:21:30.029 --> 00:21:31.029
00:21:31.029 --> 00:21:35.400
So, at time t=0, the inductor current IL will
00:21:35.400 --> 00:21:39.889
And as the time progresses, the inductor current
will increase exponentially and it will reach
00:21:39.889 --> 00:21:44.660
maximum value up to V/R.
So, this all about the forced response for
00:21:44.660 --> 00:21:45.660
the RL circuit.
00:21:45.660 --> 00:21:49.289
that means the circuit has been given some
DCexcitation at time t=0.
00:21:49.289 --> 00:21:55.309
So, in this way we got equations for the capacitor
voltage and inductor current for the RC and
00:21:55.309 --> 00:21:59.419
RL circuits for the forced excitation as well
as the source free responses.
00:21:59.419 --> 00:22:03.619
So, now let's see the shortcut method using
which we can easily find this equation.
00:22:03.619 --> 00:22:08.899
So, using this shortcut method, the capacitor
voltage or inductor current can be given as
00:22:08.899 --> 00:22:18.090
Final Value + (Initial Value -Final Value)*e^(-t/T)
So, let's take the case of this forced excitation
00:22:18.090 --> 00:22:19.539
for the capacitor.
00:22:19.539 --> 00:22:25.609
So, for the forced excitation, the final value
of the voltage across the capacitor at time
00:22:25.609 --> 00:22:31.809
t is equal to infinity is nothing but V.
And we know that initially, the voltage across
00:22:31.809 --> 00:22:38.669
the capacitor is nothing but 0 and the final
value of the voltage for the capacitor is
00:22:38.669 --> 00:22:40.090
nothing but V.
00:22:40.090 --> 00:22:44.529
So, initial value - final value will come
out as -V.
00:22:44.529 --> 00:22:52.759
And that will be multiplied by the e^(-t/T)
Now, for this RC circuit, the value of Tis
00:22:52.759 --> 00:22:53.940
nothing but RC.
00:22:53.940 --> 00:22:59.580
So, overall equation will be
V-V*e^(-t/RC)
00:22:59.580 --> 00:23:05.549
This is the response of the RC Circuit for
the forced excitation.
00:23:05.549 --> 00:23:10.580
Similarly, let's try to find out the source
free response for the RC Circuit.
00:23:10.580 --> 00:23:15.970
So, we have seen that for the source free
response, final value or value across the
00:23:15.970 --> 00:23:19.650
capacitor at time t is equal to infinity is
nothing but zero.
00:23:19.650 --> 00:23:24.480
While the initial value of the voltage across
the capacitor is nothing but V volt.
00:23:24.480 --> 00:23:27.789
Again the final value will be 0.
00:23:27.789 --> 00:23:39.940
And that will get multiplied by e^(-t/T)
So, we will get source free response as V*e^(-t/RC)
00:23:39.940 --> 00:23:47.019
So, in this way using this equation, you can
easily find the value of either capacitor
00:23:47.019 --> 00:23:50.759
voltage or inductor current in very short
00:23:50.759 --> 00:23:55.600
So, now so far whatever circuits that we have
discussed, we have assumed that the circuit
00:23:55.600 --> 00:23:59.200
contains only single resistor and one reactive
element.
00:23:59.200 --> 00:24:01.700
either inductor or capacitor.
00:24:01.700 --> 00:24:06.669
But it might happen that your circuit may
contain more than one resistors in the circuit.
00:24:06.669 --> 00:24:08.139
Let's take one example.
00:24:08.139 --> 00:24:12.590
So, if you see this circuit, in this circuit
we have total three resistors.
00:24:12.590 --> 00:24:15.320
This R1, R2 and R3.
00:24:15.320 --> 00:24:20.879
So, in such scenarios what we do, we used
to find the equivalent resistance for the
00:24:20.879 --> 00:24:22.039
given circuit.
00:24:22.039 --> 00:24:30.130
So, first of all, to find this equivalent
resistance, draw the circuit at time t=0+
00:24:30.130 --> 00:24:32.669
So, at time t=0+, this switch will get closed.
00:24:32.669 --> 00:24:35.619
So, the equivalent circuit will look loke
this.
00:24:35.619 --> 00:24:42.109
Now, in this equivalent circuit, at time t=0+,
find the equivalent Thevenin's resistance
00:24:42.109 --> 00:24:43.879
00:24:43.879 --> 00:24:49.120
So, to find that just remove this capacitor
from the circuit and the find the equivalent
00:24:49.120 --> 00:24:51.269
resistance across this two terminals.
00:24:51.269 --> 00:24:55.679
And we already know, to find the Thevenin's
equivalent resistance, we used to remove all
00:24:55.679 --> 00:24:57.490
the sources inside the circuit.
00:24:57.490 --> 00:25:02.649
So, this voltage source will also get short-circuited.
00:25:02.649 --> 00:25:06.200
And if you see now this R1 will also get short-circuited.
00:25:06.200 --> 00:25:12.799
So, the equivalent resistance across this
two terminal is nothing but R2 in parallel
00:25:12.799 --> 00:25:13.799
with R3.
00:25:13.799 --> 00:25:19.220
So, this will be equivalent resistance across
the capacitor.
00:25:19.220 --> 00:25:26.690
And time constant RC can be given as Req*C
That is (R2||
00:25:26.690 --> 00:25:30.820
R3)*C.
So, in this way when your circuit contains
00:25:30.820 --> 00:25:34.929
more than one resistive element then you can
00:25:34.929 --> 00:25:36.529
across this reactive element.
00:25:36.529 --> 00:25:41.249
So, now let's see the summary or procedure
to find the transient equations for this RC
00:25:41.249 --> 00:25:42.470
and RL circuits.
00:25:42.470 --> 00:25:47.399
So, first of all, draw the circuit at time
t=0+, that means how the circuit looks like
00:25:47.399 --> 00:25:50.159
at time t=0+.
00:25:50.159 --> 00:25:54.129
The second step is to find the initial condition
for the given circuit, that means what is
00:25:54.129 --> 00:26:01.359
capacitor voltage at time t=0+ if it is RC
circuit and find the equivalent initial value
00:26:01.359 --> 00:26:04.580
of the
inductor current if it is RL circuit.
00:26:04.580 --> 00:26:08.880
Then after if your circuit contains more than
one resistive element, then find the equivalent
00:26:08.880 --> 00:26:11.179
resistance of the circuit.
00:26:11.179 --> 00:26:15.710
And your time constant t will be nothing but
Req*Ceq
00:26:15.710 --> 00:26:22.630
And if your circuit is RL circuit, then your
Time constant T can be given as Leq/Req
00:26:22.630 --> 00:26:27.210
Then after finding the final values of this
capacitor voltage or inductor current at time
00:26:27.210 --> 00:26:29.369
t is equal to infinity.
00:26:29.369 --> 00:26:33.739
And then after finding the particular integral
and the complementary function for the given
00:26:33.739 --> 00:26:34.739
00:26:34.739 --> 00:26:36.679
Or you can use shortcut method.
00:26:36.679 --> 00:26:41.080
That is [final value] + [initial value - final
value]*e^(-t/T)
00:26:41.080 --> 00:26:47.960
So, using this you can find the total solution
for the given RC or RL circuit.
00:26:47.960 --> 00:26:53.100
So, I hope in this video you understood how
to find the transient equation for the RC
00:26:53.100 --> 00:26:58.259
and RL circuit for the source free response
as well as for forced DC excitation.
00:26:58.259 --> 00:27:12.519
So, in the next video, we will solve some
problems for this first order RC and RL circuits.

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