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In this chapter we shall study transient response of the RL, RC series and RLC circuits with external DC excitations. Transients are generated in Electrical ...
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B.Tech (EEE) R- 20
Malla Reddy College of Engineering and Technology (MRCET)
Malla Reddy College of Engineering and Technology ( MRCET )
UNIT- 1
TRANSIENT ANALYSIS (FIRST AND SECOND ORDER CIRCUITS)
Introduction
Transient Response of RL, RC series and RLC circuits for DC
excitations
Initial conditions
Solution using Differential equations approach
Solution using Laplace transformation
Summary of Important formulae and Equations
Illustrative examples
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It is evident that the current i ( t ) is zero before t = 0. and we have to find out current i ( t ) for time t >0. We
will find i ( t ) for time t >0 by writing the appropriate circuit equationand then solving it by separation of
the variables and integration.
Applying Kirchhoff’s voltage law to the above circuit we get :
V = v R
(t)+ v L
(t)
i (t) = 0 for t <0 and
Using the standard relationships of Voltage and Current for the Resistors and Inductors we can rewrite
the above equations as
V = Ri + Ldi/dt for t >
One direct method of solving such a differential equation consists of writing the equation in such a way
that the variables are separated, and then integrating each side of the equation. The variables in the
above equation are i and t. Thisequation is multiplied by dt andarranged with the variables separated as
shown below:
Ri. dt + Ldi = V. dt
i.e Ldi= (V – Ri)dt
i.e Ldi / (V – Ri) = dt
Next each side is integrated directly to get :
- (L/R ) ln(V− Ri) =t + k
Where k is the integration constant. In order to evaluate k , an initial condition must be invoked. Prior to
t = 0, i (t) is zero, and thus i (0−) = 0. Since the current in an inductorcannot change by a finite amount in
zero time without being associated withan infinite voltage, we have i (0+) = 0. Setting i = 0 at t = 0, in the
above equation we obtain
and, hence,
Rearranging we get
**- (L/R ) ln(V) = k
ln[ (V− Ri) /V] = − (R/L)t
Taking antilogarithm on both sides we get
From which we can see that
(V–Ri)/V= e
−Rt/L
i(t) = (V/R)–(V/R)e
−Rt/L
for t >
Thus, an expression for the response valid for all time t would be
i(t) = V/R [1− e
−Rt/L
This is normally written as:
i(t) = V/R [1− e
−t./τ
where ‘τ’ is called the time constant of the circuitand it’s unit is seconds.
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The voltage across the resistance and the Inductor for t >0 can be written as :
v R
(t) =i(t).R = V [1− e
−t./τ
v L
(t) = V −v R
(t) = V −V [1− e
−t./τ
] = V (e
−t./τ
A plot of the current i(t) and the voltages vR(t) & vL(t) is shown in the figure below.
Fig: Transient current and voltages in the Series RL circuit.
At t = ‘τ’ the voltage across the inductor will be
v L
(τ) = V (e
−τ /τ
) = V/e = 0.36788 V
and the voltage across the Resistor will be v R
(τ) = V [1− e
−τ./τ
The plots of current i(t) and the voltage across the Resistor v R
(t) are called exponential growth curves
and the voltage across the inductor v L
(t) is called exponential decay curve.
RC CIRCUIT with external DC excitation:
below. If the capacitor is not charged initially i.e. it’s voltage is zero ,then after the switch S is closed at
time t=0 , the capacitor voltage builds up gradually and reaches it’s steady state value of V volts after a
finite time. The charging current will be maximum initially (since initially capacitor voltage is zero and
voltage acrossa capacitor cannot change instantaneously) and then it will gradually comedown as the
capacitor voltagestarts building up. The current and the voltage during such charging periods are called
Transient Current and Transient Voltage.
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Now integrating both sides w.r.t their variables i.e. ‘ v C
(t )’ on the LHS and‘ t’ on the RHS we get
−RC ln [V − v C
(t)] = t+ k
where ‘ k ‘is the constant of integration. In order to evaluate k , an initial condition must be invoked. Prior
to t = 0, v C
(t) is zero, and thus v C
(t)(0−) = 0. Since the voltage across a capacitor cannot change by a finite
amount in zero time, we have v C
(t)(0+) = 0. Setting v C
(t)= 0 at t = 0, in the above equation we obtain :
−RC ln [V] = k
and substituting this value of k = −RC ln [V] in the above simplified equation −RC ln [V − v C
(t)] = t+ k
we get :
−RC ln [V − v C
(t)] = t−RC ln [V]
i.e. −RC ln [V − v C
(t)] + RC ln [V] = t i.e. −RC [ln {V − v C
(t)}− ln (V)]= t
i.e. [ln {V − v C
(t)}] − ln [V]} = −t/RC
i.e. ln [{V − v C
(t)}/(V)] = −t/RC
Taking anti logarithm we get [{V − v C
(t)}/(V)] = e
−t/RC
i.e v C
(t) = V(1− e
−t/RC
which is the voltage across the capacitor as a function of time.
The voltage across the Resistor is given by : v R
(t) = V−v C
(t) = V−V(1 − e
−t/RC
) = V.e
−t/RC
And the current through the circuit is given by: i(t) = C.[dv C
(t)/dt] = (CV/CR )e
−t/RC
=(V/R )e
−t/RC
Or the othe other way: i(t) = v R
(t) /R = ( V.e
−t/RC
) /R = (V/R )e
−t/RC
In terms of the time constant τ the expressions for v C
(t) , v R
(t) and i(t) are given by :
v C
(t) = V(1 − e
−t/RC
v R
(t) = V.e
−t/RC
i(t) = (V/R )e
−t/RC
The plots of current i(t) and the voltages across the resistor v R
(t) and capacitor v C
(t) are shown in the
figure below.
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Fig : Transient current and voltages in RC circuit with DC excitation.
At t = ‘τ’ the voltage across the capacitor will be :
v C
(τ) = V [1− e
−τ/τ
the voltage across the Resistor will be:
v R
(τ) = V (e
−τ /τ
) = V/e = 0.36788 V
and the current through the circuit will be:
i(τ) = (V/R) (e
−τ /τ
) = V/R. e = 0.36788 (V/R)
Thus it can be seen that after one time constant the charging current has decayed to approximately
36.8% of it’s value at t=0. At t= 5 τ charging current will be
i(5τ) = (V/R) (e
−5τ /τ
) = V/R. e
5
This value is very small compared to the maximum value of (V/R) at t=0 .Thus it can be assumed that the
capacitor is fully charged after 5 time constants.
The following similarities may be noted between the equations for the transients in the LC and RC
circuits:
The transient voltage across the Inductor in a LC circuit and the transient current in the RC
circuit have the same form k.(e
−t /τ
The transient current in a LC circuit and the transient voltage across the capacitor in the RC
circuit have the same form k.(1−e
−t /τ
But the main difference between the RC and RL circuits is the effect of resistance on the duration of the
transients.
In a RL circuit a large resistance shortens the transient since the time constant τ =L/R
becomessmall.
Where as in a RC circuit a large resistance prolongs the transient since the time constant τ = RC
becomes large.
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R
L
R
L
t/τ
- t/τ
R
- t/τ
L
- t/τ
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The Concept of Natural Response and forced response:
The RL and RC circuits we have studied are with external DC excitation. These circuits without the
external DC excitation are called source free circuits and their Response obtained by solving the
corresponding differential equations is known by many names. Since this response depends on the
general nature of the circuit (type of elements, their size, their interconnection method etc.,) it is often
called a Natural response. However any real circuit we construct cannot store energy forever. The
resistances intrinsically associated with Inductances and Capacitors will eventually dissipate the stored
energy into heat. The response eventually dies down,. Hence it is also called Transient response. As per
the mathematician’s nomenclature the solution of such a homogeneous linear differential equation is
called Complementary function.
When we consider independent sources acting on a circuit, part of the response will resemble the
nature of the particular source. (Or forcing function) This part of the response is called particular
solution. , the steady state response or forced response. This will be complemented by the
complementary function produced in the source free circuit. The complete response of the circuit is
given by the sum of the complementary function and the particular solution. In other words:
TheComplete response = Natural response + Forced response
There is also an excellent mathematical reason for considering the complete response to be composed
of two parts—the forced response and the natural response. The reason is based on the fact that the
solution of any linear differential equation may be expressed as the sum of two parts: the
complementarysolution (natural response) and the particular solution (forced response).
Determination of the Complete Response:
Let us use the same RL series circuit with external DC excitation to illustrate how to determine the
complete response by the addition of the natural and forced responses. The circuit shown in the figure
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Amoregeneral solutionapproach:
The method of solving the differential equation by separating the variables or by evaluating the
complete response as explained above may not be possible always. In such cases we will rely on a
verypowerful method, the success of which will depend upon our intuition or experience. We simply
guess or assume a form for the solution and then test our assumptions, first by substitution in the
differential equation, and then by applying the given initial conditions. Since we cannot be expected to
guess the exact numerical expression for the solution, we will assume a solution containing several
unknown constants and select the values for these constants in order to satisfy the differential equation
and the initial conditions.
Many of the differential equations encountered in circuit analysis have a solution which may be
represented by the exponential function or by the sum of several exponential functions.Hence Let us
assume a solution for the following equation corresponding to a source free RL circuit
in exponential form as
[ di/dt+ (R i /L)] = 0
i (t) = A.e
s1t
where A and s1 are constants to be determined. Now substituting this assumed solution in the original
governing equation we have:
A. s1. e
s1t
+ A .e
s1t
Or
(s1 + R/L). A.e
s1t
In order to satisfy this equation for all values of time, it is necessary that A = 0, or s1 = −∞, or s1 = −R/L.
But if A = 0 or s1 = −∞, then everyresponse is zero; neither can be a solution to our problem. Therefore,
wemust choose
And our assumed solution takes on the form:
s1 = −R/L
i (t) = A.e−
Rt/L
The remaining constant must be evaluated by applying the initial condition i (0) = I 0
. Thus, A = I 0
, and the
final form of the assumed solution is(again):
i (t) = I 0
.e
−Rt/L
A Direct Route: The Characteristic Equation:
In fact, there is a more direct route that we can take. To obtain the solution for the first order DE we
solved s1 + R/L= 0 which is known as the characteristic equation and then substituting this value of s1=-
R/L in the assumed solution i (t) = A.e
s1t
which is same in this direct method also. We can obtain the
characteristic equation directly from the differential equation, without the need for substitution of our
trial solution. Consider the general first-order differential equation:
a(d f/dt) + bf = 0
where a and b are constants. We substitute s for the differentiation operator d/dt in the original
differential equation resulting in
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a(d f/dt) + bf = (as + b) f = 0
From this we may directly obtain the characteristic equation: as + b = 0
which has the single root s = −b/a. Hence the solution to our differential equationis then given by :
f = A.e
−bt/a
This basic procedure can be easily extended to second-order differential equations which we will
encounter for RLC circuits and we will find it useful since adopting the variable separation method is
quite complex for solving second order differential equations.
Earlier, we studied circuits which contained only one energy storage element, combined with a passive
network which partly determined how long it took either the capacitor or the inductor to
charge/discharge. The differential equations which resulted from analysis were always first-order. In this
chapter, we consider more complex circuits which contain both an inductor and a capacitor. The result is
a second-order differential equation for any voltage or current of interest. What we learned earlier is
easily extended to the study of these so-called RLC circuits, although now we need two initial conditions
to solve each differential equation. There are two types of RLC circuits: Parallel RLC circuits and Series
circuits. Such circuits occur routinely in a wide variety of applications and are very important and hence
we will study both these circuits.
Parallel RLC circuit:
Let us first consider the simple parallel RLC circuit with DC excitation as shown in the figure below.
Fig:Parallel RLC circuit with DC excitation.
For the sake of simplifying the process of finding the response we shall also assume that the initial
current in the inductor and the voltage across the capacitor are zero. Then applying theKirchhoff’s
current law (KCL)( i = i C
+i L
) to the common node we get the following integrodifferential equation:
𝐭
(V−v)/R = 1/L ∫
𝐭𝐨
𝐯𝐝𝐭 ’ + C.dv/dt
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Now two new terms are defined as below :
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which is termed as resonant frequency and
ω0 = 1/√LC
α = 1/2RC
which is termed asthe exponential damping coefficient
α the exponential damping coefficient is a measure of howrapidly the natural response decays or damps
out to its steady, final value(usually zero). And s, s 1
, and s 2
, are called complex frequencies.
We should note that s 1
, s 2
, α, and ω 0
are merely symbols used to simplifythe discussion of RLC circuits.
They are not mysterious new parameters of any kind. It is easier, for example, to say “ alpha ” than it is
to say “ the reciprocalof 2RC .”
Now we can summarize these results.
The response of the parallel RLC circuit is given by :
where
and
v(t) = A 1
e
s1t
2
e
s2t..........
s 1
= −α + √ α
2
- ω 0
2 ..............
s 2
= −α − √ α
2
- ω 0
2 ............
α = 1/2RC ................................. [4]
ω0 = 1/ √LC .......... [5]
1
and A 2
must be found by applying the given initial conditions.
We note three basic scenarios possible with the equations for s1 and s2 depending on the relative
values of α and ω 0
(which are in turn dictated by the values of R, L, and C ).
CaseA:
α > ω0, i.e when (1/2RC)
2
>1/LCs 1
and s 2
will both be negative real numbers, leading to what is referred
to as an over damped response given by :
v(t) = A 1
e
s1t
2
e
s2t
Since s 1 and s 2 are both negative real numbersthis is the (algebraic) sumof two decreasing exponential
terms. Since s2 is a larger negative number it decays faster and then the response is dictated by the first
term A 1
e
s1t
CaseB :
α = ω 0
, , i.e when (1/2RC)
2
=1/LC , s 1
and s 2
are equal which leads to what is called a critically damped
response given by :
v(t) = e
−αt
1
t + A 2
Case C :
