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NETWORK THEORY SUBJECT ELECTRICAL ENGINERING STUDENTS
Typology: Lecture notes
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Recognized under 2(f) and 12 (B) of UGC ACT 1956 (Affiliated to JNTUH, Hyderabad, Approved by AICTE - Accredited by NBA & NAAC – ‘A’ Grade - ISO 9001:2015 Certified) Maisammaguda, Dhulapally (Post Via. Kompally), Secunderabad – 500100, Telangana State, India
B.Tech (EEE) R- 17
II Year B.Tech EEE-II Sem L^ T/P/D^ C 4 - /-/- 4
(R17A0209) NETWORK THEORY
Objectives:
This course introduces the analysis of transients in electrical systems, to understand three phase circuits, to evaluate network parameters of given electrical network, to draw the locus diagrams and to know about the network functions
To prepare the students to have a basic knowledge in the analysis of Electric Networks
UNIT-I D.C Transient Analysis: Transient response of R-L, R-C, R-L-C circuits (Series and parallel combinations) for D.C. excitations, Initial conditions, Solution using differential equation and Laplace transform method. UNIT - II A.C Transient Analysis: Transient response of R-L, R-C, R-L-C circuits (Series and parallel combinations) for sinusoidal excitations, Initial conditions, Solution using differential equation and Laplace transform method. UNIT - III Three Phase Circuits: Phase sequence, Star and delta connection, Relation between line and phase voltages and currents in balanced systems, Analysis of balanced and Unbalanced three phase circuits UNIT – IV Locus Diagrams: Series and Parallel combination of R-L, R-C and R-L-C circuits with variation of various parameters. Resonance: Resonance for series and parallel circuits, concept of band width and Q factor. UNIT - V Network Parameters: Network functions driving point and transfer impedance function networks- poles and zeros – necessary conditions for driving point function and for transfer function Two port network parameters – Z, Y, ABCD and hybrid parameters and their relations– 2 - port network parameters using transformed variables.
TEXT BOOKS:
Introduction:
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 circuits due to abrupt changes in the operating conditions when energy storage elements like Inductors or capacitors are present. Transient response is the dynamic response during the initial phase before the steady state response is achieved when such abrupt changes are applied. To obtain the transient response of such circuits we have to solve the differential equations which are the governing equations representing the electrical behavior of the circuit. A circuit having a single energy storage element i.e. either a capacitor or an Inductor is called a Single order circuit and it’s governing equation is called a First order Differential Equation. A circuit having both Inductor and a Capacitor is called a Second order Circuit and it’s governing equation is called a Second order Differential Equation. The variables in these Differential Equations are currents and voltages in the circuit as a function of time.
A solution is said to be obtained to these equations when we have found an expression for the dependent variable that satisfies both the differential equation and the prescribed initial conditions. The solution of the differential equation represents the Response of the circuit. Now we will find out the response of the basic RL and RC circuits with DC Excitation.
RL CIRCUIT with external DC excitation: Let us take a simple RL network subjected to external DC excitation as shown in the figure. The circuit consists of a battery whose voltage is V in series with a switch, a resistor R , and an inductor L. The switch is closed at t = 0.
Fig 1.1: RL Circuit with external DC excitation
When the switch is closed current tries to change in the inductor and hence a voltage VL(t) is induced across the terminals of the Inductor in opposition to the applied voltage. The rate of change of current decreases with time which allows current to build up to it’s maximum value.
i(t) = V/R [1 − e − t. /τ^ ]
where ‘τ’ is called the time constant of the circuit and it’s unit is seconds.
The voltage across the resistance and the Inductor for t >0can be written as :
vR(t) =i(t).R = V [1 − e − t ./τ^ ]
vL(t) = V − vR(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 1.2: Transient current and voltages in the Series RL circuit.
At t = ‘τ’ the voltage across the inductor will be
vL( τ ) = V (e −τ /τ ) = V/e = 0.36788 V
and the voltage across the Resistor will be vR( τ ) = V [1 − e −τ./τ^ ] = 0.63212 V
The plots of current i(t) and the voltage across the Resistor vR(t) are called exponential growth curves and the voltage across the inductor vL(t)is called exponential decay curve.
RC CIRCUIT with external DC excitation:
A series RC circuit with external DC excitation V volts connected through a switch is shown in the figure 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 across a capacitor cannot change instantaneously) and then it will gradually comedown as the capacitor voltage starts building up. The current and the voltage during such charging periods are called Transient Current and Transient Voltage.
Fig 1.3: RC Circuit with external DC excitation
Applying KVL around the loop in the above circuit we can write
V = vR(t) + vC(t)
Using the standard relationships of voltage and current for an Ideal Capacitor we get
vC(t) = (1/C ) 𝒊 𝒕 𝒅𝒕 or i(t) = C.[dvC(t)/dt]
and using this relation, vR(t) can be written asvR(t) = Ri(t) = R. C.[dvC(t)/dt]
Using the above two expressions for vR(t) and vC(t)the above expression for V can be rewritten as :
V = R. C.[dvC(t)/dt] + vC(t)
Or finally dvC(t)/dt + (1/RC). vC(t) = V/RC
The inverse coefficient of vC(t) is known as the time constant of the circuit τ and is given by τ = RC and it’s units are seconds.
The above equation is a first order differential equation and can be solved by using the same method of separation of variables as we adopted for the LC circuit.
Multiplying the above equation dvC(t)/dt + (1/RC). vC(t) = V/RC
both sides by ‘dt ’ and rearranging the terms so as to separate the variables vC(t) and t we get:
dvC(t)+ (1/RC). vC(t). dt = (V/RC).dt
vR(t) = V.e − t/RC
i(t) = (V/R )e − t/RC
The plots of current i(t) and the voltages across the resistor vR(t)and capacitor vC(t)are shown in the figure below.
Fig 1.4 : Transient current and voltages in RC circuit with DC excitation.
At t = ‘τ’ the voltage across the capacitor will be:
vC( τ ) = V [1 − e −τ/τ^ ] = 0.63212 V
the voltage across the Resistor will be:
vR( 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 −^5 τ /τ ) = V/R. e^5 = 0.0067(V/R)
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 becomes small. Where as in a RC circuit a large resistance prolongs the transient since the time constant τ = RC becomes large. Discharge transients: Consider the circuit shown in the figure below where the switch allows
both charging and discharging the capacitor. When the switch is position 1 the capacitor gets
charged to the applied voltage V. When the switch is brought to position 2, the current
discharges from the positive terminal of the capacitor to the negative terminal through the
resistor R as shown in the figure (b). The circuit in position 2 is also called source free circuit since there is no any applied voltage.
Fig 1.5: RC circuit (a) During Charging (b) During Discharging
The current i 1 flow is in opposite direction as compared to the flow of the original charging
current i. This process is called the discharging of the capacitor. The decaying voltage and the
current are called the discharge transients. The resistor, during the discharge will oppose the
flow of current with the polarity of voltage as shown. Since there is no any external voltage
source, the algebraic sum of the voltages across the Resistance and the capacitor will be zero
(applying KVL) .The resulting loop equation during the discharge can be written as
vR(t)+vC(t) = 0 or vR(t) = - vC(t)
We know that vR(t) = R.i(t) = R. C.dvC(t) /dt. Substituting this in the first loop equation we
get R. C.dvC(t)/dt + vC(t) = 0
The solution for this equation is given by vC(t) = Ke-t/τ^ where K is a constant decided by the
initial conditions and τ =RC is the time constant of the RC circuit
The solution for this equation is given by i(t) = Ke-t/τ^ where K is a constant decided by the initial
conditions and τ =L/R is the time constant of the RL circuit.
The value of the constant K is found out by invoking the initial condition i(t) = V/R @t = 0 Then
we get K = V/R and hence i(t) = (V/R). e- t/τ^ ; vR(t) = R.i(t)= Ve- t/τ^ and vL(t) = - Ve- t/τ^ The
plots of the voltages across the Resistor and the Inductor and the decaying current through the
circuit are shown in the figure below.
Fig 1.8: Plot of Decay transients in RL circuit
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:
The Complete 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 complementary solution (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
Fig 1.9: RL circuit with external DC excitation
was analyzed earlier, but by a different method. The desired response is the current i (t) , and now we first express this current as the sum of the natural and the forced current,
i = in+ i (^) f
The functional form of the natural response must be the same as that obtained without any sources. We therefore replace the step-voltage source by a short circuit and call it the RL source free series loop. And in can be shown to be :
in= Ae−Rt /L
where the amplitude A is yet to be determined; since the initial condition applies to the complete response, we cannot simply assume A = i ( 0 ) .We next consider the forced response. In this particular problem the forced response is constant, because the source is a constant V for all positive values of time. After the natural response has died out, there can be no voltage across the inductor; hence the all y the applied voltage V appears across R , and the forced response is simply
i (^) f = V / R
where A ands1 are constants to be determined. Now substituting this assumed solution in the original governing equation we have:
A. s1. es1t+ A .es1t^. R/L = 0
or
(s1 + R/L). A.es1t= 0
In order to satisfy this equation for all values of time, it is necessary that A = 0, or s1 = −∞ , or s = − R/L. But if A = 0 or s1 = −∞ , then every response is zero; neither can be a solution to our problem. Therefore, we must choose
s1 = − R/L
And our assumed solution takes on the form:
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
solveds1 + R/L= 0 which is known as the characteristic equation and then substituting this value of s1=- R/Lin the assumed solution i (t) = A.es1t^ 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
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 equation is 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.
RLC CIRCUITS:
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 1.10: 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 the Kirchhoff’s current law (KCL)( i = iC +iL )to the common node we get the following integrodifferential equation:
Definition of Frequency Terms:
The form of the natural response as given above gives very little insight into the nature of the curve we might obtain if v(t) were plotted as a function of time. The relative amplitudes of A 1 and A 2 , for example, will certainly be important in determining the shape of the response curve. Further the constants s 1 and s 2 can be real numbers or conjugate complex numbers, depending upon the values of R , L , and C in the given network. These two cases will produce fundamentally different response forms. Therefore, it will be helpful to make some simplifying substitutions in the equations for s 1 and s 2.Since the exponents s 1 t and s 2 t must be dimensionless, s 1 and s 2 must have the unit of some dimensionless quantity “per second.” Hence in the equations for s 1 and s 2 we see that the units of 1 / 2 RC and 1 / √ LC must also be s −^1 (i.e., seconds −^1 ). Units of this type are called frequencies.
Now two new terms are defined as below:
ω 0 = 1/ √ LC
which is termed as resonant frequency and
α = 1/2RC
which is termed as the exponential damping coefficient
α the exponential damping coefficient is a measure of how rapidly 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 simplify the 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 reciprocal of 2RC .”
Now we can summarize these results.
The response of the parallel RLC circuit is given by :
v(t) = A 1 es1t+ A 2 es2t *…….. 1+
where
s 1 = −α +√ α^2 –ω 02 *…….. 2+
s 2 = −α − √ α^2 – ω 02 *…….. 3+
α = 1/2RC *…….. 4+
and
ω 0 = 1/ √ LC *…….. 5+
A 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).
Case A:
α > ω 0,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 es1t+ A 2 es2t
Sinces 1 and s 2 are both negative real numbersthis is the (algebraic) sumof two decreasing exponential terms. Sinces2 is a larger negative number it decays faster and then the response is dictated by the first term A 1 es1t.
Case B:
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 −α t(A 1 t + A 2 )
Case C :
α < ω 0,i.e when (1/2RC)^2 <1/LC both s 1 and s 2 will have nonzero imaginary components, leading to what is known as an under damped response given by :
v(t) = e −α t(A 1 cos ω d t + A 2 sin ω d t)
where ω d is called natural resonant frequency and is given by:
ω d = √ ω 02 – α^2
We should also note that the general response given by the above equations [1] through [5] describe not only the voltage but all three branch currents in the parallel RLC circuit; the constants A 1 and A 2 will be different for each, of course.
