Study of DC Transients in R-L-C Circuits | Basic Electrical Technology - Electrical Engineering (EE) PDF Download

Objectives

• Be able to write differential equation for a dc circuits containing two storage elements in presence of a resistance.
• To develop a thorough understanding how to find the complete solution of second order differential equation that arises from a simple RLC − − circuit.
• To understand the meaning of the terms (i) overdamped (ii) criticallydamped, and (iii) underdamped in context with a second order dynamic system.
• Be able to understand some terminologies that are highly linked with the performance of a second order system.

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 V_s . 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 2^nd-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 v_cn (t) and the steady state response v_cf (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 v_cn (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 v_cn(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⁺) = i_L (0^-) = 2 A; v_c (0⁺) = v_c (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, v_L (0⁺)=v_R (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 v_c(0^-) = 6V=v_c(0⁺) = v_bd(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 v_cd (0⁺)=6V and this implies that v_ca(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⁺)) = v_bd (0⁺) -v_cd (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^-) = i_L(0^-) = 0, v(0^-) = 0 and v_C(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 i_c(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 v_cn (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 v_c (0⁺)=10 volt., and the steady state capacitor voltage v_c(∞)=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 i_L(t) (ii) voltage across the capacitor v_c (t)