Study of DC transients in R-L-C Circuits

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.

 

L.11.1 Introduction

In the preceding lesson, our discussion focused extensively on dc circuits having resistances with either inductor () or capacitor () (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. LC

L.11.2 Response of a series R-L-C circuit due to a dc voltage source

Consider a series RLcircuit as shown in fig.11.1, and it is excited with a dc voltage source C−−sV. Applying around the closed path for , KVL0t>

()()()cditLRitvtdt++= (11.1)

The current through the capacitor can be written as

 ()()cdvtitCdt=

Substituting the current ''expression in eq.(11.1) and rearranging the terms, ()it

22()()()cccdvtdvtLCRCvtdtdt++ (11.2)

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 and the steady state response. Mathematically, one can write the complete solution as ()cnvt()cfvt

(1212()()()ttccncfvtvtvtAeAeAαα=+=++ (11.3)

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. Now, the first part of the total response is completely dies out with time while 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 A()cnvt0R>

22()()()0cccdvtdvtLCRCvtdtdt++=22()()1()0cccdvtdvtRvtdtLdtLC⇒++

22()()()0cccdvtdvtabcvtdtdt++ (where 11,RabandcLL===) (11.4)

The characteristic equation of the above homogeneous differential equation (using the operator 222,dddtdtαα== and ()0cvt≠) is given by

22100RabcLLCαααα++=⇒++= (where 11,RabandcLL===) (11.5)

and solving the roots of this equation (11.5) one can find the constants 1andαα of the exponential terms that associated with transient part of the complete solution (eq.11.3) and they are given below.

22122211;2222112222RRbbacLLLCaaRRbbacLLLCaaαα⎛⎞⎛⎛⎞⎛⎞⎜⎟⎜=−+−=−+−⎜⎟⎜⎟⎜⎟⎜⎝⎠⎝⎠⎝⎠⎝⎛⎞⎛⎛⎞⎛⎞⎜⎟⎜=−−−=−−−⎜⎟⎜⎟⎜⎟⎜⎝⎠⎝⎠⎝⎠⎝ (11.6)

where, 1RbandcLL==.

The roots of the characteristic equation (11.5) are classified in three groups depending upon the values of the parameters ,,RLand of the circuit. C

Case-A (overdamped response): When 2102RLLC⎛⎞−>⎜⎟⎝⎠, 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

112()tcnvtAeAeα=+ (11.7)

and each term of the above expression decays exponentially and ultimately reduces to zero as 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 t→∞11tAeαtakes longer time to decay its value to zero than the term21tAeα. One can introduce a factorξ that provides an information about the speed of system response and it is defined by damping ratio

()122RActualdampingbLcriticaldampingacLCξ=== (11.8)

Case-B ( critically damped response): When 2102RLLC⎛⎞−=⎜⎟⎝⎠, 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

()12()tcnvtAtAeα=+ (where 2RLα=−) (11.9)

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 122RLααα=== when 221022RRLLCLLC⎛⎞⎛⎞−=⇒=⎜⎟⎜⎟⎝⎠⎝⎠ and the corresponding homogeneous equation (11.4) can be rewritten as

22222()()12(2()()2(ccccccdvtdvtRvtdtLdtLCdvtdvtorvtdtdtαα++++

()()()()0ccccdvtdvtdorvtvtdtdtdtααα⎛⎞⎛⎞+++=⎜⎟⎜⎟⎝⎠⎝⎠

0dforfdtα+= where ()()ccdvtfvtdtα=+

The solution of the above first order differential equation is well known and it is given by

1tfAeα=

Using the value of f in the expression ()()ccdvtfvtdtα=+ we can get,

11()()()()tttccccdvtdvtvtAeeevtAdtdtααααα−+=⇒+= ()1()tcdevtAdtα⇒=

Integrating the above equation in both sides yields,

()12()tcnvtAtAeα=+

In fact, the term 2tAeα(with 2RLα=−) decays exponentially with the time and tends to zero as . On the other hand, the value of the term t→∞1tAteα(with 2RLα=−) in equation (11.9) first increases from its zero value to a maximum value112LAeR− at a time 12LtRRα⎛⎞=−=−−=⎜⎟⎝⎠ 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 ()122RActualdampingbLcriticaldampingacLCξ=== (11.10)

Case-C (underdamped response): When2102RLLC⎛⎞−<⎜⎟⎝⎠ , this implies that the roots of eq.(11.5) are complex conjugates and they are expressed as 221211;2222RRRRjjjLLCLLLCLαβγαβ⎛⎞⎛⎛⎞⎛⎞⎜⎟⎜=−+−=+=−−−=−⎜⎟⎜⎟⎜⎟⎜⎝⎠⎝⎠⎝⎠⎝. The form of the natural or transient part of the complete solution is written as

()()121212()jttcnvtAeAeAeAeβγβγαα+=+=+ =()()()()1212cossinteAAtjAAtβγ⎡⎤++−⎣⎦ (11.11)

=()()12cossinteBtBtβγ⎡⎤+⎣⎦ where ()112212;BAABjAA=+=−

For real system, the response must also be real. This is possible only if ()cnvt12AandA conjugates. The equation (11.11) further can be simplified in the following form:

()sinteKtβγθ+ (11.12)

where β=real part of the root , γ=complex part of the root, 2211122tanBKBBandBθ−⎛=+=⎜⎝⎠. Truly speaking the value of Kandθ 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

()122RActualdampingbLcriticaldampingacLCξ=== (11.13)

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: L.11.1 The switch was closed for a long time as shown in fig.11.3. Simultaneously at , the switch is opened and is closed Find 1S0t=1S2S()(0);()(0);Lcaibv++()(0);Rci+(0)()(0);()(0);()cLcdvdveifdt+++.

Solution: When the switch is kept in position '1' for a sufficiently long time, the circuit reaches to its steady state condition. At time 1S0t, 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

50(0)6210050LiA−=×=+

Using the KCL, one can find the current through the resistor and subsequently the voltage across the capacitor (0)624RiA−=−=(0)450200.cvv−=×=

Note at not only the current source is removed, but 1000t+=Ω 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 the current in inductor, voltage across the capacitor, and the values of other variables at 0t+=0t+= can be computed as

(0)(0)2LLiiA; vv

Since the voltage across the capacitor at 0t+= is 20, the same voltage will appear across the inductor and the 50 resistor. That is, and hence, the current 0voltΩ(0)(0)200.LRvvvo++==()(0)Ri+in resistor = 50Ω200450A=. Applying KCL at the bottom terminal

of the capacitor we obtain and subsequently, (0)(42)6ci+=−+=−(0)(0)6600./sec.0.01ccdvivoltdtC++−===−

Example: L.11.2 The switch '' is closed sufficiently long time and then it is opened at time '' as shown in fig.11.4. Determine S0t=00000()()()()(0)()()(0),()()tcLLttdvtdvtditiviiiiiiandivvdtdtdt+++=++==when . 123RR==Ω

Solution: At (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 0t−=(0)6(0)ccvVv−+== =(0)bdv+ and terminal 'b' is higher potential than terminal ''. On the other branch, the inductor acts as a short circuit (i.e., voltage across the inductor is zero) and the source voltage will appear across the resistanced6V2R. Therefore, the current through inductor 6(0)2(0i. Note at , = 0 (since the voltage drop across the resistance 0t+=(0)adv+13R=Ω = ) and and this implies that = voltage across the inductor ( note, terminal 'c' is + ve terminal and inductor acts as a source of energy ). 6abvV=−(0)6cdv+=(0)6cav+=

Now, the voltage across the terminals '' and 'c' (b0(0)v+) = (0)(0)bdcdvv++−= . The following expressions are valid at 0V0t+= 00(0)21/sec. (note, voltage across the capacitor will

decrease with time i.e., 01/sectdvvoltdt+==− ). We have just calculated the voltage across the inductor at as 0t+=

00()()6(0)612/sec.0.5LLcattditditvLVAdtdt+++====⇒==

Now, ()02(0)(0)(0)112335/sec.cLdvdvdiRvdtdtdt+++=−=−×=−

Example: L.11.3 Refer to the circuit in fig.11.5(a). Determine,

(i) (ii) (0),(0)(0)Liiandv++()0(0)didvanddtdt++ (iii) ()(),Lii∞∞ ()andv∞ (assumed ) (0)0;(0)0cLvi==

Solution: When the switch was in 'off' position i.e., t < 0

----LCi(0) = i(0)=0, v(0)=0 and v(0)=0

The switch '' was closed in position '1' at time t = 0 and the corresponding circuit is shown in fig 11.5 (b). 1S

(i) From continuity property of inductor and capacitor, we can write the following expression for t = 0⁺

+-+-LLcci(0) = i(0)=0, v(0)= v(0)=0 1(0)(0)06civ++⇒==

++Lv(0)= i(0) 6=0 volt×.

 (ii) KCL at point 'a'

8()()()cLititit=++

At 0t+= , the above expression is written as

8 (0)(0)(0)cLiii++=++(0)8ciA+⇒=

We know the current through the capacitor can be expressed as ()cit

ccdv(t)i(t)=Cdt

++ccdv(0)i(0)=Cdt

+cdv(0)1 = 8 × =2 volt./sec.dt4∴.

Note the relations

()0cdvdt+=change in voltage drop in 6Ω resistor = change in current through resistor =6Ω6×()06didt+×()026didt+⇒= 1./sec.3amp=

Applying KVL around the closed path 'b-c-d-b', we get the following expression.

()()()cLvtvtvt=+

At, the following expression 0t+=

(0)(0)(0)12(0)(0)0(0)012(0)000cLLLLLLvvididivvLdtdt+++++++=+×=+×⇒=⇒=⇒

+Ldi(0)=0dt and this implies +Ldi(0)12=120=0 v/secdt×=+dv(0)= 0dt

Now, at ()()LvtRitalso=0t+=

(0)(0)(0)120/sec.LLdvdidiRvdtdtdt+++===

 

(iii) Attα=, 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 and 12 Ω resistors can be formed as 6Ω

L12×816i()===5.333A, i()=8-5.333=2.667A183∞∞

()32.cvv∞=

Example: L.11.4 The switch has been closed for a sufficiently long time and then it is opened at (see fig.11.6(a)). Find the expression for (a) , (b) for inductor values of ()1S0t=()cvt(),cit0t>0.5()0.2iLHiiLH==()1.0iiiLH= and plot and for each case. ()cvtvst−−()itvst−−

Solution: At (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 0t−=0t+= as

12(0)(0)215LLii+−===+ (note inductor acts as a short circuit) and voltage across the resistor = The capacitor is fully charged with the voltage across the resistor and the capacitor voltage at 5Ω2510.volt×=5Ω0t+= is given by

(0)(0)10.ccvvvo+−==The circuit is opened at time 0t= and the corresponding circuit diagram is shown in fig. 11.6(b).

Case-1: 0.5,12LHRandCF==Ω=

Let us assume the current flowing through the circuit is and apply KVL equation around the closed path is ()it

 

 

 

22()()()()()()ccscsdvtdvtditVRitLvtVRCLCvtdtdtdt=++⇒=++ (note, ()()cdvtitCdt=)

22()()1()ccsdvtdvtRVdtLdtLC=+++ (11.14)

The solution of the above differential equation is given by

()()()ccncfvtvtvt=+ (11.15)

The solution of natural or transient response is obtained from the force free equation or homogeneous equation which is ()cnvt

22()()1()0cccdvtdvtRvtdtLdtLC++ (11.16)

The characteristic equation of the above homogeneous equation is written as 210RLLCαα++= (11.17)

The roots of the characteristic equation are given as

2111.022RRLLLCα⎛⎞⎛⎞⎜⎟=−+−=−⎜⎟⎜⎟⎝⎠⎝⎠; 2211.022RRLLLCα⎛⎞⎛⎞⎜⎟=−−−=−⎜⎟⎜⎟⎝⎠⎝⎠

and the roots are equal with negative real sign. The expression for natural response is given by

()12()tcnvtAtAeα=+ (where 121ααα===−) (11.18)

The forced or the steady state response is the form of applied input voltage and it is constant ''. Now the final expression for is ()cfvtA()cvt

()()1212()tcvtAtAeAAtAeAα−=++=++ (11.19)

The initial and final conditions needed to evaluate the constants are based on

(0)(0)10;(0)(0)2ccLLvvvoltii+−+−==== (Continuity property).

 

 

 

At ; 0t+=

10220()ctvtAeAAA+−×==+=+ (11.20) 210AA⇒+=

Forming ()cdvtdt(from eq.(11.19)as

()()121121()tttcdvtAtAeAeAtAeAedtααα−−=++=−++

12120()1ctdvtAAAAdt+==−⇒−= (11.21)

(note, (0)(0)(0)(0)21/sec.cccLdvdvCiivoltdtdt++++===⇒=)

It may be seen that the capacitor is fully charged with the applied voltage when and the capacitor blocks the current flowing through it. Using t=∞t=∞ in equation (11.19) we get,

()12cvAA∞=⇒=

Using the value of in equation (11.20) and then solving (11.20) and (11.21) we get,. A121;2AA=−=−

The total solution is

()()()()()212122;()()2221ttcttcvttetedvtitCteetedt−−−−−=−++=−+⎡⎤==×+−=×+⎣⎦ (11.22)

The circuit responses (critically damped) for 0.5LH=are shown fig.11.6 (c) and fig.11.6(d).

Case-2: 0.2,12LHRandCF==Ω=

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

120.563;4.436αα=−=−

The total response becomes

124.4360.5631212()tttcvtAeAeAAeAeAαα−−=++=++ (11.23)

124.4360.536112212()4.4350.563ttttcdvtAeAeAeAedtαααα−=+=−− (11.24) Using the initial conditions( (0)10cv+=, (0)1/seccdvvoltdt+=) that obtained in case-1 are used in equations (11.23)-(11.24) with 12A=( final steady state condition) and simultaneous solution gives

120.032;2.032AA==−

 

 

 

The total response is

4.4360.5630.5634.436()0.0322.03212()()21.140.14ttctcvteedvtitCeedt−−−−=−+⎡⎤==−⎣⎦ (11.25)

The system responses (overdamped) for 0.2LH=are presented in fig.11.6(c) and fig.11.6 (d).

Case-3: 8.0,12LHRandCF==Ω=

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

120.0630.243;0.0630.242jjjjαβγαβγ=+=−+=−=−−

The expression for the total response is

()()()()sintccncfvtvtvteKtAβγθ=+=++ (11.26)

(note, the natural response ()()sintcnvteKtβγθ= is written from eq.(11.12) when roots are complex conjugates and detail derivation is given there.)

()(()sincostcdvtKettdtββγθγγθ⎡=++⎣ (11.27)

Again the initial conditions ((0)10cv+=, (0)1/seccdvvoltdt+=) that obtained in case-1 are used in equations (11.26)-(11.27) with 12A=(final steady state condition) and simultaneous solution gives

()04.13;28.98degKrθ==−

The total response is

()()()()(0.06300.06300.06300()sin124.13sin0.24228.9912()124.13sin0.24228.99()()20.999*cos0.24228.990.26sin0.24228.99ttctctcvteKtetvtetdvtitCettdtβγθ−−−=++=−+=+−⎡⎤==−−−⎣⎦ (11.28)

The system responses (under-damped) for 8.0LH= are presented in fig.11.6(c) and fig. 11.6(d).

 

Version 2 EE IIT, Kharagpur Version 2 EE IIT, Kharagpur

 

 

Remark: One can use 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 , and the steady state capacitor voltage 0tandt=(0)10.cvv+=()12.cvv∞=) of the circuit.

Example: L.11.5 The switch '' in the circuit of Fig. 11.7(a) was closed in position '1' sufficiently long time and then kept in position '2'. Find (i) (ii) for t ≥ 0 if C is (a) 1S()cvt()cit19F (b) 14F (c) 18F .

Solution: When the switch was in position '1', the steady state current in inductor is given by

---LcL30i(0)==10A, v(0)=i(0)R=10×2=20 volt.1+2

Using the continuity property of inductor and capacitor we get

+-+-LLcci(0)=i(0)=10, v(0)=v(0)=20 volt.

The switch '' is kept in position '2' and corresponding circuit diagram is shown in Fig.11.7 (b) 1S

Applying KCL at the top junction point we get,

Lv(t)c+i(t)+i(t)=0cR

 

 

 

Lv(t)dv(t)cc+C+i(t)=0Rdt

LLL2di(t)di(t)L+C.L+i(t)=02Rdtdt [note: ()()LcditvtLdt=]

or LLL2di(t)di(t)11++i(t2RCdtLCdt (11.29)

The roots of the characteristics equation of the above homogeneous equation can obtained for 19CF=

221194×919+4LC+RC22RC2α==22⎛⎞⎛⎞−−−−⎜⎟⎜⎟⎝⎠⎝⎠−

222194×9194LCRC22RC2α==22⎛⎞⎛⎞−−−−−−⎜⎟⎜⎟⎝⎠⎝⎠−

Case-1()1.06,overdampedsystemξ=:1C=F9, the values of roots of characteristic equation are given as

121.5,3.0αα=−=−

The transient or neutral solution of the homogeneous equation is given by

- 1.5t -3.0tL12i(t)=Ae+Ae (11.30)

To determine 1A and 2A, the following initial conditions are used.

At ; 0t+=

+-LL112i(0)=i(0)=10AAAA+=+ (11.31)

++-+LccLt = 0di(t)v(0)=v(0)=v(0)=Ldt

- 1.5t- 3.0t120=2×-1.5 e-3.0eA⎡⎤××⎣⎦ (11.32)

[]121=2-1.5A-3A=- 3A- 6A

Solving equations (11.31) and (11,32) we get , 2116.66,26.666AA=−=.

The natural response of the circuit is

1.53.01.53.0L8050i26.6616.6633ttteeee−−−=−=−

 

 

 

1.53.0LdiL226.661.516.663.0dtttee−−⎡⎤=×−−×−⎣⎦

()(- 3.0t- 1.5tc- 3.0t- 1.5t- 1.5t- 3.0t()(t)=100e- 80e()1()300.0e120e13.33e33.33e9Lccvtvdvtitcdt⎡⎤=⎣⎦==−+=−

Case-2 ()0.707,underdampedsystemξ=: For 1C=F4, the roots of the characteristic equation are

121.01.01.01.0jjjjαβγαβγ=−+=+=−−=−

The natural response becomes 1

β tLi(t)=k e sin(t+)γθ (11.33)

Where and θ are the constants to be evaluated from initial condition. k

At , from the expression (11.33) we get, 0t+=

()+Li0=k sinθ

10=k sinθ (11.34)

+ +βtβtt = 0 t = 0di(t)L=2kβ esin(t+)+e cos(t+)dtγθγγθ⎡⎤×⎣⎦ (11.35)

Using equation (11.34) and the values of andβγ in equation (11.35) we get,

202(cos)cosksnkβθγθθ=+= (note:1,1sin10andkβγθ=−==) (11.36)

From equation ( 11.34 ) and ( 11.36 ) we obtain the values of θ and as k

-1o11tan==tan=26.5622θθ⎛⎞⇒⎜⎟⎝⎠ and 1022.36sinkθ==

∴ The natural or transient solution is

()-toLi(t)=22.36 e sint+26.56

[] βtcdi(t)L=v(t)=2kβ sin (t+θ)+ cos (t+θ)edtγγγ××

oo=44.72cos (t+26.56) - sin (t+26.56)te−⎡⎤×⎣⎦

{oo()1()44.72cos (t+26.56) - sin (t+26.56)e422.36cos(26.56)cctdvtditcdtdtte−⎡⎤==×⎣⎦=−+

 

 

 

Case-3()1,criticallydampedsystemξ=: For 1C=F8; the roots of characteristic equation are 122;2αα=−=− respectively. The natural solution is given by

()12()tLitAtAeα=+ (11.37)

where constants are computed using initial conditions.

At ; from equation ( 11.37) one can write 0t+=

L22i(0)10AA+=⇒=

()()++2110t=01210121t=0di(t)L =2dt2di(t)L (0)202230dttttttttcAeAteAeAAeAtevAAAααααααααα++==+⎡⎤×++⎣⎦⎡⎤=×++⎣⎦===−⇒=

The natural response is then

()2()1030tLitte−=+

()2Ldi(t)L 21030dttdtedt−⎡⎤=×+⎣⎦

Ldi(t)L dt= ()cvt[]2=21060tte−−

()22()1()2106020308ttccdvtditcteetedtdt−−⎡⎤⎡==××−=−+⎣⎦⎣

Case-4 : For ()2,overdampedsystemξ=1C=32F

Following the procedure as given in case-1 one can obtain the expressions for (i) current in inductor (ii) voltage across the capacitor ()Lit()cvt

1.0814.93()11.51.5ttLitee−−=−

14.931.08()()44.824.8ttcditLvteedt−−⎡⎤==−⎣⎦

14.931.081.0814.93()1()44.824.8320.83720.902ttccttdvtditceedtdtee−−−−⎡⎤==×−⎣⎦=−

L.11.3 Test your understanding (Marks: 80)

T.11.1 Transient response of a second-order ------------------ dc network is the sum of two real exponentials. [1]

 

 

 

T.11.2 The complete response of a second order network excited from dc sources is the sum of -------- response and ---------------- response. [2]

T.11.3 Circuits containing two different classes of energy storage elements can be described by a ------------------- order differential equations. [1]

T.11.4 For the circuit in fig.11.8, find the following [6]

(0)(0)(0)(0)()(0)()(0)()()()()ccLLccdvdvdidiavbvcdefdtdtdtdt−+−+−+

(Ans.() )6.()6.()0/sec.()0/sec.()0/sec.()3./sec.avoltbvoltcVdVeampfamp

T.11.5 In the circuit of Fig. 11.9,

Find,

 

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(0)(0)()(0)(0)()()()()RLRLRdvdvavandvbandcvandvdtdt++++∞∞ [8]

(Assume the capacitor is initially uncharged and current through inductor is zero).

(Ans. () )0,0()0,2./.()32,0aVVbVVoltSeccVV

T.11.6 For the circuit shown in fig.11.10, the expression for current through inductor

is given by ()2()10300tLittefort−=+≥

Find, (the values of )a,LC (initial condition )b(0)cv− the expression for . ()c()0cvt>

(Ans. ()21()2,()(0)20()()20120.8tccaLHCFbvVcvtteV−−====−) [8]

T.11.7 The response of a series RLC circuit are given by

4.4360.5630.5634.436()120.0322.032()2.280.28−−−−=+−=−ttcttLvteeitee

where are capacitor voltage and inductor current respectively. Determine (a) the supply voltage (b) the values ()()cvtandit,,RLC of the series circuit. [4+4]

(Ans. () )12()1,0.22aVbRLHandCF=Ω==

T.11.8 For the circuit shown in Fig. 11.11, the switch ''was in position '1' for a long time and then at it is kept in position '2'. S0t=

 

 

 

Find,

()(0);()(0);()(0);()();LcRLaibvcvdi−++∞ [8]

Ans.

()(0)10;()(0)400;()(0)400()()20LcRLaiAbvVcvVdiA−++===∞=−

T.11.9 For the circuit shown in Fig.11.12, the switch '' has been in position '1' for a long time and at it is instantaneously moved to position '2'. S0t=

Determine and sketch its waveform. Remarks on the system's ()0itfort≥

response. [8]

(Ans. 771().33ttiteeamps−−⎛⎞=−⎜⎟⎝⎠)

T.11.10 The switch '' in the circuit of Fig.11.13 is opened at S0t= having been closed for a long time.

 

 

 

Determine (i) (ii) how long must the switch remain open for the voltage to be less than 10% ot its value at ()0cvtfort≥()cvt0t=? [10]

(Ans. (i) ) ()10()()16240()0.705sec.tcivtteii−=+

T.11.11 For the circuit shown in Fig.11.14, find the capacitor voltage and inductor current for all [10] ()cvt()Lit(00)ttandt<≥

Plot the wave forms and for . ()cvt()Lit0t≥

(Ans. ()0.50.5()10sin(0.5);()5cos(0.5)sin(0.5)−−==−ttctLvetittte)

T.11.12 For the parallel circuit shown in Fig.11.15, Find the response RLC

 

 

 

of respectively. [10] ()()Litandvt

(Ans. ()22()4412.;()48.ttLcitetampsvttevolt−−⎡⎤=−+=⎣⎦)