Objectives
Introduction
In the preceding lesson, our discussion focused extensively on dc circuits having resistances with either inductor (L) or capacitor (C) (i.e., single storage element) but not both. Dynamic response of such first order system has been studied and discussed in detail. The presence of resistance, inductance, and capacitance in the dc circuit introduces at least a second order differential equation or by two simultaneous coupled linear first order differential equations. We shall see in next section that the complexity of analysis of second order circuits increases significantly when compared with that encountered with first order circuits. Initial conditions for the circuit variables and their derivatives play an important role and this is very crucial to analyze a second order dynamic system.
Response of a series R-L-C circuit due to a dc voltage source
Consider a series R - L - C circuit as shown in fig.11.1, and it is excited with a dc voltage source Vs . Applying KVL around the closed path for t > 0,
The current through the capacitor can be written as
Substituting the current ‘i (t)’expression in eq.(11.1) and rearranging the terms,
The above equation is a 2nd-order linear differential equation and the parameters associated with the differential equation are constant with time. The complete solution of the above differential equation has two components; the transient response vcn (t) and the steady state response vcf (t). Mathematically, one can write the complete solution as
Since the system is linear, the nature of steady state response is same as that of forcing function (input voltage) and it is given by a constant value A. Now, the first part vcn (t) of the total response is completely dies out with time while R > 0 and it is defined as a transient or natural response of the system. The natural or transient response (see Appendix in Lesson-10) of second order differential equation can be obtained from the homogeneous equation (i.e., from force free system) that is expressed by
The characteristic equation of the above homogeneous differential equation (using the operator
and solving the roots of this equation (11.5) one can find the constants α1 and α of the exponential terms that associated with transient part of the complete solution (eq.11.3) and they are given below.
The roots of the characteristic equation (11.5) are classified in three groups depending upon the values of the parameters R, L and C of the circuit.
Case-A (overdamped response): When this implies that the roots are distinct with negative real parts. Under this situation, the natural or transient part of the complete solution is written as
and each term of the above expression decays exponentially and ultimately reduces to zero as t → ∞ and it is termed as overdamped response of input free system. A system that is overdamped responds slowly to any change in excitation. It may be noted that the exponential term takes longer time to decay its value to zero than the term . One can introduce a factor that provides an information about the speed of system response and it is defined by damping ratio
Case-B (critically damped response): When this implies that the roots of eq.(11.5) are same with negative real parts. Under this situation, the form of the natural or transient part of the complete solution is written as
where the natural or transient response is a sum of two terms: a negative exponential and a negative exponential multiplied by a linear term. The expression (11.9) that arises from the natural solution of second order differential equation having the roots of characteristic equation are same value can be verified following the procedure given below.
The roots of this characteristic equation (11.5) are same when and the corresponding homogeneous equation (11.4) can be rewritten as
The solution of the above first order differential equation is well known and it is given by
Using the value of f in the expression we can get,
Integrating the above equation in both sides yields,
In fact, the term decays exponentially with the time and tends to zero as t →∞ . On the other hand, the value of the term equation (11.9) first increases from its zero value to a maximum value at a time and then decays with time, finally reaches to zero. One can easily verify above statements by adopting the concept of maximization problem of a single valued function. The second order system results the speediest response possible without any overshoot while the roots of characteristic equation (11.5) of system having the same negative real parts. The response of such a second order system is defined as a critically damped system’s response. In this case damping ratio
Case-C (underdamped response): this implies that the roots of eq.(11.5) are complex conjugates and they are expressed as The form of the natural or transient part of the complete solution is written as
For real system, the response vcn(t) must also be real. This is possible only if A1 and A2 conjugates. The equation (11.11) further can be simplified in the following form:
where β = real part of the root , γ = complex part of the root, Truly speaking the value of K and θ can be calculated using the initial conditions of the circuit. The system response exhibits oscillation around the steady state value when the roots of characteristic equation are complex and results an under-damped system’s response. This oscillation will die down with time if the roots are with negative real parts. In this case the damping ratio
Finally, the response of a second order system when excited with a dc voltage source is presented in fig.L.11.2 for different cases, i.e., (i) under-damped (ii) over-damped (iii) critically damped system response.
Example: The switch was closed for a long time as shown in fig.11.3. Simultaneously at , the switch is opened and is closed Find
Solution: When the switch S1 is kept in position ‘1’ for a sufficiently long time, the circuit reaches to its steady state condition. At time t = 0- , the capacitor is completely charged and it acts as a open circuit. On other hand,
the inductor acts as a short circuit under steady state condition, the current in inductor can be found as
Using the KCL, one can find the current through the resistor subsequently the voltage across the capacitor
Note at t = 0+ not only the current source is removed, but 100Ω resistor is shorted or removed as well. The continuity properties of inductor and capacitor do not permit the current through an inductor or the voltage across the capacitor to change instantaneously. Therefore, at t=0+ the current in inductor, voltage across the capacitor, and the values of other variables at t = 0+ can be computed as
iL (0+) = iL (0-) = 2 A; vc (0+) = vc (0-) = 200 volt.
Since the voltage across the capacitor at t = 0+ is 200 volt, the same voltage will appear across the inductor and the 50Ω resistor. That is, vL (0+)=vR (0+)=200volt. and hence, the current . Applying KCL at the bottom terminal of the capacitor we obtain and subsequently,
Example: The switch ‘S’ is closed sufficiently long time and then it is opened at time ‘t = 0’ as shown in fig. 11.4. Determine when
R1 =R2= 3Ω .
Solution: At t=0- (just before opening the switch), the capacitor is fully charged and current flowing through it totally blocked i.e., capacitor acts as an open circuit). The voltage across the capacitor is vc(0-) = 6V=vc(0+) = vbd(0+) and terminal ‘b’ is higher potential than terminal ‘d’. On the other branch, the inductor acts as a short circuit (i.e., voltage across the inductor is zero) and the source voltage 6V will appear across the resistance R2 . Therefore, the current through inductor Note at (since the voltage drop across the resistance and vcd (0+)=6V and this implies that vca(0+) = 6V = voltage across the inductor (note, terminal ‘ c ’ is + ve terminal and inductor acts as a source of energy).
Now, the voltage across the terminals ‘b’ and ‘c’ (v0(0+)) = vbd (0+) -vcd (0+)= 0 V .
The following expressions are valid at t = 0+
(note, voltage across the capacitor will decrease with time i.e., (We have just calculated the voltage across the inductor at t=0+ as
Example: Refer to the circuit in fig.11.5(a). Determine,
Solution: When the switch was in ‘off’ position i.e., t < 0
i(0-) = iL(0-) = 0, v(0-) = 0 and vC(0-) = 0
The switch ‘S1’ was closed in position ‘1’ at time t = 0 and the corresponding circuit is shown in fig 11.5 (b).
(i) From continuity property of inductor and capacitor, we can write the following expression for t = 0+
(ii) KCL at point ‘a’
At t = 0+ , the above expression is written as
We know the current through the capacitor ic(t) can be expressed as
Note the relations
change in voltage drop in 6Ω resistor = change in current through 6 Ω
Applying KVL around the closed path ‘b-c-d-b’, we get the following expression.
At, t = 0+ the following expression
(iii) At t = α , the circuit reached its steady state value, the capacitor will block the flow of dc current and the inductor will act as a short circuit. The current through 6 Ω and 12 Ω resistors can be formed as
Example: he switch has been closed for a sufficiently long time and then it is opened at (see fig.11.6(a)). Find the expression for for inductor values of and and for each case.
Solution: At t = 0- (before the switch is opened) the capacitor acts as an open circuit or block the current through it but the inductor acts as short circuit. Using the properties of inductor and capacitor, one can find the current in inductor at time t = 0+ as
(note inductor acts as a short circuit) and voltage across the The capacitor is fully charged with the voltage across the 5Ω resistor and the capacitor voltage at t = 0+ is given by
The circuit is opened at time t =0 and the corresponding circuit diagram is shown in fig. 11.6(b).
Let us assume the current flowing through the circuit i(t) is and apply KVL equation around the closed path is
The solution of the above differential equation is given by
The solution of natural or transient response vcn (t) is obtained from the force free equation or homogeneous equation which is
The characteristic equation of the above homogeneous equation is written as
The roots of the characteristic equation are given as
and the roots are equal with negative real sign. The expression for natural response is given by
The forced or the steady state response is the form of applied input voltage and it is constant ‘A’. Now the final expression for
The initial and final conditions needed to evaluate the constants are based on (Continuity property).
It may be seen that the capacitor is fully charged with the applied voltage when t = ∞ and the capacitor blocks the current flowing through it. Using t =∞ in equation (11.19) we get,
Using the value of A in equation (11.20) and then solving (11.20) and (11.21) we get,
The total solution is
The circuit responses (critically damped) for L =0.5H are shown fig.11.6 (c) and fig.11.6(d).
Case-2: L = 0.2H ,R = 1Ω and C = 2F
It can be noted that the initial and final conditions of the circuit are all same as in case-1 but the transient or natural response will differ. In this case the roots of characteristic equation are computed using equation (11.17), the values of roots are
The total response becomes
Using the initial conditions that obtained in case-1 are used in equations (11.23)-(11.24) with A=12 ( final steady state condition) and simultaneous solution gives
The total response is
The system responses (overdamped) for L =0.2 H are presented in fig.11.6(c) and fig.11.6 (d).
Case-3: L = 8.0H,R = 1Ω and C = 2F
Again the initial and final conditions will remain same and the natural response of the circuit will be decided by the roots of the characteristic equation and they are obtained from (11.17) as
The expression for the total response is
(note, the natural response is written from eq.(11.12) when roots are complex conjugates and detail derivation is given there.)
Again the initial conditions used in equations (11.26)-(11.27) with A=12 (final steady state condition) and simultaneous solution gives
The total response is
The system responses (under-damped) for L =8.0 H are presented in fig.11.6(c) and fig. 11.6(d).
Remark: One can use t = 0 and t = ∞ in eq. 11.22 or eq. 11.25 or eq. 11.28 to verify whether it satisfies the initial and final conditions ( i.e., initial capacitor voltage vc (0+)=10 volt., and the steady state capacitor voltage vc(∞)=12 volt.) of the circuit.
Example: The switch ‘S1’ in the circuit of Fig. 11.7(a) was closed in position ‘1’ sufficiently long time and then kept in position
Solution: When the switch was in position ‘1’, the steady state current in inductor is given by
Using the continuity property of inductor and capacitor we get
The switch ‘S1’ is kept in position ‘2’ and corresponding circuit diagram is shown in Fig.11.7 (b)
Applying KCL at the top junction point we get,
The roots of the characteristics equation of the above homogeneous equation can obtained for
the values of roots of characteristic equation are given as
The transient or neutral solution of the homogeneous equation is given by
To determine A1 and A2 , the following initial conditions are used.
Solving equations (11.31) and (11,32) we get ,
The natural response of the circuit is
the roots of the characteristic equation are
The natural response becomes 1
Where k and θ are the constants to be evaluated from initial condition.
Using equation (11.34) and the values of β and γ in equation (11.35) we get,
From equation (11.34) and (11.36) we obtain the values of θ and k as
∴ The natural or transient solution is
the roots of characteristic equation are respectively. The natural solution is given by
where constants are computed using initial conditions.
The natural response is then
Following the procedure as given in case-1 one can obtain the expressions for (i) current in inductor iL(t) (ii) voltage across the capacitor vc (t)
57 docs|62 tests
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1. What is a DC transient in an R-L-C circuit? |
2. How does an inductor affect the DC transient response in an R-L-C circuit? |
3. What is the significance of the time constant in determining the DC transient response of an R-L-C circuit? |
4. How does a capacitor affect the DC transient response in an R-L-C circuit? |
5. What are some practical applications of studying DC transients in R-L-C circuits? |
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