Initial-Conditions€¦ · and evaluation of condition at t= 1are known as nal conditions. The following are the objectives of studying the behavior of the circuit for Initial-Conditions:

Chapter 1

Initial-Conditions

1.0.1 Introduction:

Most of the transmission lines, electrical circuits and communication networks are made up of network elementslike resistor R,inductor, L and Capacitor C. These networks are connected by voltage and current sources. It ismost useful to understand the behavior of the network when we switched on the network by supplying voltagesource. It is most important to determine the transient response of R-L, R-C, R-L-C series circuits for d.c anda.c excitations.

Assuming that at reference time t = 0, the switch in the circuit is closed and also assuming that switchact in zero time. To differentiate between the time immediately before and immediately after the operationof a switch, t = 0− and t = 0− signs are used. The condition existing just before the switch is operated willbe designated as i(0−),v(0−), q(0−) and the conditions existing after closing of a switch is designated as asi(0+),v(0+), q(0+). Also initial conditions of a network depend on the past history of the network prior theclosing of the network at to t = 0− and the network structure at t = 0+, after switching.

The evaluation of voltages and currents and their derivatives at t = 0+, are known as initial conditionsand evaluation of condition at t =∞ are known as final conditions.

The following are the objectives of studying the behavior of the circuit for Initial-Conditions:

• The most important reason is that the initial and final conditions must be known evaluate the arbitraryconstants that appear in the general solution of a differential equation.

• The initial conditions give knowledge of the behavior of the circuit elements at the instant of switching

• The final conditions give knowledge of the behavior of the circuit elements after the settling of circuit att =∞

———————————————————————————————————————-

1.1 initial Conditions

Resistor

Consider a circuit which consists of resistor Rconnected as shown in Figure 1.1. The circuit resistorR is connected by a voltage source V in series withswitch K as shown in Figure.

RV Ri

K

Figure 1.1: Series resonance circuit

When the switch K is closed at t=0 the current Iis flowing in a circuit and is given by

I =V

R

1

1.1. INITIAL CONDITIONS Chapter 1. Initial-Conditions

Inductor

Consider a circuit which consists of inductor Lconnected as shown in Figure 1.2. The inductor L isconnected by a voltage source V in series with switchK as shown in Figure. When the switch K is closed att=0 the current flowing in a inductor at t = 0+ is zerothe inductor acts as a open circuit at t = 0+ which isas shown in Figure 1.3.

LV Li

K

Figure 1.2: Inductor circuit

LV Li

K

O CV Li

K 0t

Figure 1.3: Inductor circuit at t = 0+

The final-condition equivalent circuit of aninductor is derived from the basic relationship.

v = Ldi

dt

Under steady state condition, rate of change of currentflowing in inductor is di

dt = 0. This means, v = 0 andhence L acts as short at t =∞. The equivalent circuitsof an inductor at t =∞ is as shown in Figure 1.4

LV Li

K

S CV Li

K t 0t

Figure 1.4: Inductor circuit at t = 0+

Capacitor

Consider a circuit which consists of capacitor Cconnected as shown in Figure 1.5. The capacitor is

connected by a voltage source V in series with switchK as shown in Figure 1.5. When the switch K is

closed at t=0 capacitor C acts as short circuit andcurrent flows in a capacitor instantaneously.

CV Ci

K

S CV Ci

K 0t

Figure 1.5: Capacitor circuit at t = 0+

If the capacitor is initially charged with chargeq0 coulombs at t=0-, then at t=0+ the capacitor isequivalent to voltage source v0 = q0

c which is as shownin Figure 1.6

CV Ci

K

V Ci

K 0t

0 oqv C +-+-

0 oqv C

Figure 1.6: Capacitor circuit at t = 0+

The final condition of capacitor circuit is derivedfrom the following relationship. The voltage acrosscapacitor is

v = Cdv

dt

Under steady state condition, rate of change ofvoltage capacitor is dv

dt = 0. This means, v = 0 andhence C acts as open circuit at t =∞. The equivalentcircuits of a capacitor at t = ∞ is as shown in Figure1.7

CV Ci

K

V Ci

K t

O C

Figure 1.7: Capacitor circuit at t =∞If the capacitor is initially charged with voltage v0

then the final condition at t =∞ of a capacitor circuitis replaced with voltage source v0 with open circuitwhich is as shown in Figure 1.8

v = Cdv

dt

CV Ci

K

V Ci

K t

0 oqv C+-+-

0 oqv CO C

Figure 1.8: Capacitor circuit at t =∞

Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 2


Table 1.1: Initial and Final Conditions

at t=0- at t=0+ at t =∞R R R

L O. C S.C

LOI OI OI

C S.C O. CC0q+- 0qV

C=+-

0qV C+- O. C

Procedure for Evaluating Initial Conditions:

1. Before closing or opening the switch at t=0- find the history of the network, at t=0- find i(0-), v(0-), i.e.,current through inductor and voltage across the capacitor before switching

2. Draw the circuit after switching operation at t=0+.

3. Replace inductor with open circuit or by current source having source

4. Replace capacitor with short circuit or with a voltage source vc = q0c if it has an initial charge q0.

5. Find i(0+), and v(0+) at t=0+

6. Obtain an expression for didt and find di

dt at t=0+

7. Obtain an expression for d2idt2

and find d2idt2

at t=0+

8. Similarly determine voltages across circuit elements and its derivatives.



Q 1) In the circuit shown in Figure 1.9 the switch Ks closed at t=0, with capacitor uncharged. Find thevalues i, di

dt and d2idt2

at t=0+, for element values asfollows V=100 V R = 1000 Ω and C = 1 µF .

R

V i

K

C

Figure 1.9: Example

Solution:

KVL for the given circuit is

V = Ri+1

C

∫idt (1.1)

At t=0+ the capacitor acts as short circuit which is asshown in Figure 1.10

V = Ri(0+) (1.2)

i(0+) =V

R=

100

1000= 0.1A (1.3)

R

V i S C0t

Figure 1.10: Example

Differentiating equation (1)

0 = Rdi

dt+

i

C(1.4)

Substituting initial conditions

0 = Rdi

dt(0+) +

i(0+)

Cdi

dt(0+) = − i(0+)

RCdi

dt(0+) = − 0.1

1000× 1× 10−6− 100A/sec

d2i

dt2(0+) = − 1

RC

di

dt(0+)

= − .1

1000× 1× 10−6(−100)

= − .1

1000× 1× 10−6(−100)

= 1× 105A/sec2

Q 2) In the circuit shown in Figure 1.11 the switch Ks closed at t=0, with zero current in the conductor.Find the values i, di

dt and d2idt2

at t=0+, for elementvalues as follows V=100 V R = 10 Ω and L = 1 H.

R

V i

K

L


Solution:

V = Ri+ Ldi

dt(1.5)

at t=0+ the inductor acts as open circuit which is asshown in Figure 1.12

i(0+) = 0 (1.6)

R

V (0 )i +

K


From equation (1) and substituting initialconditions

Ldi

dt= V −Ri

Ldi

dt(0+) = V −Ri(0+) = 100− 0

di

dt(0+) =

100

1= 100A/sec

d2i

dt2(0+) = −R

L

di

dt(0+) = −10

1× 100

= −1000A/sec2

Q 3) In the circuit shown in Figure 1.13 V=10 vR = 10 Ω L = 1 H and C = 10 µF and vc(0) = 0, find

i(0+) = 0, didt(0+) and d2i

dt2(0+).

R

V i

K

C

L




Solution:

V = Ri+ Ldi

dt+

1

C

∫idt = 0 (1.7)

at t=0+ the inductor acts as open circuit and capacitoracts as short circuit which is as shown in Figure 1.31

R

V (0 )i +


i(0+) = 0 (1.8)

From equation (1) and (2)

V = Ri(0+) + Ldi

dt(0+) +

1

C

∫i(0+)dt

V = R× 0 + Ldi

dt(0+) + 0

Ldi

dt(0+) = V

di

dt(0+) =

V

L=

10

1= 10A/sec


Rdi

dt+ L

d2i

dt2+

i

C= 0 (1.9)


Rdi

dt(0+) + L

d2i

dt2(0+) +

i(0+)

C= 0 (1.10)

10× 10 + Ld2i

dt2(0+) +

i(0+)

C= 0

d2i

dt2(0+) =

−100

L=−100

1= −100A/sec2

Q 4) In the circuit shown in Figure 1.15 K is changedfrom position a to b at t=0. Solve for find i, di

dt andd2idt2

.

100 Vi

S

1 H

1000 Ω

0.1 Fμ

b

a


Solution: Before connecting to position b

V = Ri(0−) (1.11)

i(0−) =V

R=

100

1000= 0.1 A (1.12)

When switch is at position b,and at t=0+ circuit whichis as shown in Figure 1.16

i 1 H

1000 Ω

0.1 Fμ


Ri+ Ldi

dt+

1

C

∫idt = 0

Ri(0+) + Ldi

dt(0+) +

1

C

∫i(0+)dt = 0

It is given that capacitor is initially uncharged

1

C

∫i(0+)dt = vc(0+) = 0 (1.13)

Ri(0+) + Ldi

dt(0+) = 0

1000× (0.1) + Ldi

dt(0+) = 0

Ldi

dt(0+) = −100

di

dt(0+) =

−100

1= −100 A/sec


Rdi

dt+ L

d2i

dt2+

i

C= 0 (1.14)


Rdi

dt(0+) + L

d2i

dt2(0+) +

i(0+)

C= 0 (1.15)

1000× (−100) + 1d2i

dt2(0+) +

0.1

0.1× 10−6= 0

d2i

dt2(0+) = −9× 105A/sec2

Q 4-2) In the circuit shown in Figure 1.17 K is changedfrom position a to b at t=0, steady state conditionhaving been reached before switching. Find the valuesof find i, di

dt and d2idt2

b at t=0+.



ik

1 H10

1 Fba

20 V


Solution: Before connecting to position b 1.18(a)

V = Ri(0−) (1.16)

i(0−) =V

R=

20

10= 2 A (1.17)

When switch is at position b,and at t=0+ circuit whichis as shown in Figure 1.18 (b)

i

k 10 a

20 V i 1 H10

1 F(a)

20 V

(b)Figure 1.18: Example

Ri+ Ldi

dt+

1

C

∫idt = 0

Ri(0+) + Ldi

dt(0+) +

1

C

∫i(0+)dt = 0


1

C

∫i(0+)dt = vc(0+) = 0 (1.18)

Ri(0+) + Ldi

dt(0+) = 0

10× 2 + 1di

dt(0+) = 0

di

dt(0+) = −20 A/sec


Rdi

dt+ L

d2i

dt2+

i

C= 0 (1.19)


Rdi

dt(0+) + L

d2i

dt2(0+) +

i(0+)

C= 0 (1.20)

10× (−20) + 1d2i

dt2(0+) +

2

1× 10−6= 0

d2i

dt2(0+) = 200− 2× 106 = −1.9998× 106A/sec2

Q 4-3) In the circuit shown in Figure ?? K is changed

from position a to b at t=0, steady state conditionhaving been reached before switching. Find the valuesof find i, di

dt and d2idt2

b at t=0+.

ik

2 H20

2 Fba

40 V


Solution: Before connecting to position b 1.20(a)

V = Ri(0−) (1.21)

i(0−) =V

R=

40

20= 2 A (1.22)

When switch is at position b,and at t=0+ circuit whichis as shown in Figure 1.20 (b)

i

k 20 a

40 V i 2 H20

2 F(a) (b)


Ri+ Ldi

dt+

1

C

∫idt = 0

Ri(0+) + Ldi

dt(0+) +

1

C

∫i(0+)dt = 0


1

C

∫i(0+)dt = vc(0+) = 0 (1.23)

Ri(0+) + Ldi

dt(0+) = 0

20× 2 + 2di

dt(0+) = 0

di

dt(0+) =

−40

2= −20 A/sec


Rdi

dt+ L

d2i

dt2+

i

C= 0 (1.24)


Rdi

dt(0+) + L

d2i

dt2(0+) +

i(0+)

C= 0



20× (−20) + 2d2i

dt2(0+) +

2

2× 10−6= 0

2d2i

dt2(0+) = 400− 1× 106 = −0.9996× 106A/sec2

d2i

dt2(0+) =

−0.9996× 106

2= −0.4998× 106A/sec2

Q 5) In the circuit shown in Figure 1.24 the switch wasin position a for sufficiently long time to have achievedsteady state. At t=0, the switch was changed fromat to b. Determine IL and vc their first and secondderivatives at t=0+.

10 V

S 1 H

1 Ω1 Fμ

ba

5 V 2 ( )i t1( )i t


Solution: When switch is at position a at t=0-

circuit which is as shown in Figure 1.22

5 V 1 Ω1( )i t


V = Ri1(0−) (1.25)

i1(0−) =V

R=

5

1= 5 A (1.26)

Also

i2(0−) = 5 A (1.27)

vc(0−) = 5 V vc(0+) = 5 V (1.28)

When switch is at position b,and at t=0+ circuit whichis as shown in Figure 1.23

10 V

1 H

1 Ω1 Fμ 2 ( )i t1( )i t


10 = 1di1dt

+ vc(t)

di1dt

(0+) = 10− vc(0+) = 10− 5 = 5

The voltage across capacitor is

vc(t) =1

C

∫i1 − i2 (1.29)

dvc(t)

dt=

1

C(i1 − i2)

dvc(t)

dt(0+) =

1

C(5− 5) = 0


d2vc(t)

dt=

1

C

[di1dt− di2

dt

](1.30)


d2vc(t)

dt(0+) =

1

C

[di1dt

(0+)− di2(0+)

dt

]=

1

1[5− 0] = 5V/sec2

Q 6) In the circuit shown in Figure 1.24 the switchis opened at t=0, after the network has attained thesteady state with the switch closed. (a) Find anexpression for the voltage across the switch at t=0+,(b) If the parameters are adjusted such that i(0+)=1Aand di

dt(0+) = −1A/sec, what is the value of thederivative of the voltage across the switch.

V( )i t

K

LC1R

2R


Solution:

When switch is closed attains steady state is in which Lacts as short circuit and capacitor acts as open circuitwhich is as shown in Figure 1.25

V( )i t

LC1R

2R


V = R2i(0−) (1.31)

i(0−) =V

R2(1.32)

Also



vc(0−) = 0 V vc(0+) = 0 V (1.33)

When switch is opened at t=0 circuit the voltageacross the switch K is

vk = R1i(t) +1

C

∫idt

vk(0+) = R1i(0+) + 0 = R1V

R2= V

R1

R2

dvk(t)

dt= R1

di

dt+

i

Cdvk(t)

dt(0+) = R1(−1) +

1

C=

1

C−R1 V/s

Q 7) In the circuit shown in Figure 1.26 the steadystate is reached with switch is opened at t=0,after the switch k is closed. Find the values ofVa(0−) and Va(0+)

5 V10 Ω 20 Ω

10 Ω

10 Ω2 Fμ

avbv


Solution:

When switch is opened and when steady state isreached capacitor acts as open circuit there is nocurrent flows in a circuit which is as shown in Figure1.27

5 V10 Ω 20 Ω

10 Ω

10 Ω

avbv


va(0−) = 5 V vb(0−) = 5 V (1.34)

vb is a Voltage across capacitor which is

vb(0−) = 5 V vb(0+) = 5 V (1.35)

When switch is closed at t=0 apply KCL to nodea and by initial conditions

va − 5

10+va10

+va − vb

20= 0

va(0+)− 5

10+va(0+)

10+va(0+)− vb(0+)

20= 0

va(0+)− 5

10+va(0+)

10+va(0+)− 5

20= 0

va(0+)

10+va(0+)

10+va(0+)

20+−5

10+−5

20= 0

5va(0+)

20− −15

20= 0

5va(0+)

20=

15

20

va(0+) =15

5= 3V olts

Q 8) In the circuit shown in Figure 1.44 the steadystate is reached with switch is opened, at t=0 theswitch k is closed. Find voltage across capacitor,initial values of i1, i2,

di1dt ,

di2dt at t=0+ and find di1

dt (∞)where V=100V, R1 = 10Ω, R2 = 20Ω, R3 = 20Ω, L =1H,C = 1µF

V 1i

K

L C

1R2R 3R

2i


Solution:

When switch is opened and when steady state isreached capacitor acts as open circuit and inductoracts as short circuit which is as shown in Figure1.45(a).

V 1i

L

1R 2R 3R2i V 1i

L C2R 3R2i

(a) (b)

+-66.6 V




i1(0−) =V

R1 +R2=

100

30= 3.33A (1.36)

i2(0−) = 0A (1.37)

Voltage across capacitor is voltage across R2

vc(0−) = i1(0−)×R2 = 3.33× 20 = 66.67V

When switch is closed at t=0 R1 is short circuited.Inductor acts as current source with a value of 3.33 Aand capacitor acts as voltage source with a value of66.67 V which is as shown in Figure 1.45(b).

i2(0+) =100− 66.67

20= 1.67

V = R2i1 + Ldi1dt

V = R2i1(0+) + Ldi1(0+)

dt

1di1(0+)

dt= 100− 20× 3.33

di1(0+)

dt= 33.3A/sec

For the capacitor branch

V = R3i2 +1

C

∫i2dt

V = R3i2(0+) +1

C

∫i2(0+)dt

Differentiating we get

= R3di2(0+)

dt+

1

Ci2(0+)

di2(0+)

dt= − 1

R3Ci2(0+)

= − 1

20× 1× 10−61.678 = 83500 A/sec

di1(∞)

dt=

100

20= 5A/sec



JAN-2017-CBCS In the circuit shown in Figure 1.30V=10 v R = 10 Ω L = 1 H and C = 10 µF andvc(0) = 0, find i(0+), di

dt(0+) and d2idt2

(0+).

R

V ( )i t

K

C

L

Figure 1.30: 2017-CBCS-Question Paper

Solution:

V = Ri+ Ldi

dt+

1

C

∫idt = 0 (1.38)

at t=0+ the inductor acts as open circuit and capacitoracts as short circuit which is as shown in Figure 1.31

R

V (0 )i +


i(0+) = 0 (1.39)

From equation (1) and (2)

V = Ri(0+) + Ldi

dt(0+) +

1

C

∫i(0+)dt = 0

di

dt(0+) =

V

L

10

1= 10A/sec


Rdi

dt+ L

d2i

dt2+

i

C= 0 (1.40)


Rdi

dt(0+) + L

d2i

dt2(0+) +

i(0+)

C= 0 (1.41)

10× 10 + Ld2i

dt2(0+) +

i(0+)

C= 0

d2i

dt2(0+) =

−100

L=−100

1= −100A/sec2

JAN-2017-6b-CBCS In the network shown in Figure1.32 a steady state is reached with the switch K open.At t=0, the switch is closed. For the element givendetermine the values Va(0−) and Va(0+)

5 V10 Ω 20 Ω

10 Ω

10 Ω2 H

avbv


Solution:When switch is opened and steady state reached

inductor acts as short circuit which is as shown inFigure 1.33 (a)

5 V10 20

10

10

av bv

S C

(a)

5 V10 20

10

10

av bvS C

(b)


Va(0−)

[1

10+

1

20

]=

5

10(1.42)

Va(0−)[0.15] =5

10(1.43)

Va(0−) =0.5

0.15= 3.333V (1.44)

When switch is closed resistor 10 Ω is connected toa point Va steady state reached inductor acts as shortcircuit which is as shown in Figure 1.33 (b)

Va(0+)

[1

10+

1

10+

1

20

]=

5

10(1.45)

Va(0+)[0.25] =5

10(1.46)

Va(0+) =0.5

0.25= 2V (1.47)

JAN-2017-NON-CBCS For the circuit shown in Figure1.34 the switch S is changed from position 1 to 2 att=0. The circuit was under steady state before thisaction. Determine the value v and i at t=0+ and theirfirst and second derivatives.

50 V i

S

2 F2H

20 v2

1


Solution:



Before connecting to position 2 switch was atposition 1 at t=0- under steady state conditioncapacitor charges with voltage of v(0−) = 50 = v(0+)and after that it acts as an open circuit which is asshown in Figure 1.35

50 V i

S 20

(0 ) 50VCv 1

(a)

( )i t 2 F2H

20 2

(b)


At t = 0- , inductor is in open circuit and capacitoris after fully charging it is also in open circuit state.That is

i(0−) = 0 and also i(0+) = 0 (1.48)

When switch is at position 2, and at t=0+ the circuitdiagram is as shown in Figure 1.35

Applying KVL for the circuit we have

Ri+ Ldi

dt+

1

C

∫idt = 0

Ri+ Ldi

dt+ vc(t)dt = 0

At t = 0+ and vc(0+) = 50

Ri(0+) + Ldi

dt(0+) + vc(0+) = 0

20× 0 + 2di

dt(0+) + 50 = 0

di

dt(0+) =

−50

2= −25


Rdi

dt+ L

d2i

dt2+

i

C= 0 (1.49)


Rdi

dt(0+) + L

d2i

dt2(0+) +

i(0+)

C= 0

20× (−25) + 2d2i

dt2(0+) +

0

C= 0

d2i

dt2(0+) =

500

2= 250A/sec2

July-2016 For the circuit shown in Figure 1.36 theswitch K is changed from position A to B at t=0.

After having reached steady state in position A. Findi didt ,

d2idt2, andd3i

dt3at t=0+.

100 Vi

K

1 H

10c

Fμ=

A

B


Solution:

Before connecting to position B switch was atposition A at t=0- under steady state conditioncapacitor charges with voltage of v(0−) = 100 = v(0+)and after that it acts as an open circuit which is asshown in Figure 1.37(a)

100 V i

S

(0 ) 100VCv A

(a)

( )i t

1 H

10 FB

(b)

S C


At t = 0- , inductor is in short circuit and capacitoris after fully charged and it is also in open circuit state.

i(0−) = 0 and also i(0+) = 0 (1.50)

When switch is at position B, and at t=0+ the circuitdiagram is as shown in Figure 1.35 (b)


Ldi

dt+

1

C

∫idt = 0

Ldi

dt+ vc(t)dt = 0

At t = 0+ and vc(0+) = 100

Ldi

dt(0+) + vc(0+) = 0

1di

dt(0+) + 100 = 0

di

dt(0+) = −100


Ld2i

dt2+

i

C= 0 (1.51)




d2i

dt2(0+) +

i(0+)

C= 0

1d2i

dt2(0+) +

0

C= 0

d2i

dt2(0+) = 0A/sec2

d3i

dt3(0+) = 0A/sec2

July-2016-2 For the circuit shown in Figure 1.38 theswitch K is opened at t=0. Find i di

dt , vc,dvcdt at t=0+.

V=4V i K

R=2Ω L=2H C=1F


Solution: When switch is at position a at t=0-

circuit which is as shown in Figure 1.39

V=4V iO CR=2Ω S C

(a) (b)V=4V i

R=2Ω L=2H C=1F


When switch is at opened, at t=0+ circuit whichis as shown in Figure 1.39 (a)

V = Ri(0−) (1.52)

i(0−) =V

R=

4

2= 2 A (1.53)

When switch is at opened, at t=0+ circuit whichis as shown in Figure 1.39 (b)

Applying KVL

V =1

C

∫idt+ L

di

dt+Ri (1.54)

At t=0+

2 = 0 + 1× di(0+)

dt+ 1× i(0+) (1.55)

2 = 0 + 1× di(0+)

dt+ 1× i(0+)

di

dt(0+) = 2− 1× 2 = 0


vc(t) =1

C

∫i (1.56)

dvc(t)

dt=

1

Ci

dvc(t)

dt(0+) =

1

Ci(0+)

dvc(t)

dt(0+) =

1

12 = 2V

July-2015-6-b For the circuit shown in Figure 1.40 theswitch K is opened at t=0 after reaching the steadystate condition. Determine voltage across switch andits first and second derivatives at t=0+.

i

K

R=1Ω

L=1H1C= F2V=2V

kV


Solution: Before opening switch and at t=0-

circuit diagram is as shown in Figure 1.41 (a)

i R=1Ω

L=1H1C= F2V=2V

kV

(b)i R=1ΩV=2V

(a)

SC


V = Ri(0−)

i(0−) =V

R=

2

1= 2 A = i(0+)

When switch is opened at t=0+ circuit diagram isas shown in Figure 1.41 (b)

Applying KVL

V =1

C

∫idt+ L

di

dt+Ri (1.57)

At t=0+

2 = 0 + 1× di(0+)

dt+ 1× i(0+) (1.58)

2 = 0 + 1× di(0+)

dt+ 1× i(0+)

di

dt(0+) = 2− 1× 2 = 0



Differentiating Equation 1.57

0 =i

C+ L

d2i

dt2+R

di

dt

1× d2i1dt2

(0+) = −Rdidt

(0+)− i(0+)

Cd2i(0+)

dt2= −1× 0− 2

1/2(0+) = −4 A/sec2

Differentiating Equation 1.57

0 =1

C

di

dt+ L

d3i

dt3+R

d2i

dt2

1× d3i1dt3

(0+) = −Rd2i(0+)

dt2− 1

C

d(0+)i

dtd3i(0+)

d32= −1× (−4)− 0 = 4 A/sec2


Vk + Ldi

dt+R× i = 2 (1.59)

dVkdt

+ Ld2i

dt2+R× di

dt= 0

dvc(t)

dt(0+) =

1

C(5− 5) = 0


d2Vkdt2

+ Ld3i

dt3+R× d2i

dt2= 0 (1.60)

d2Vk(0+)

dt2= −Ld

3i(0+)

dt3+R× d2i(0+)

dt2

d2Vk(0+)

dt2= −1× 4− 1× (−4) = 0 V/sec2

DEC-2015 6-a For the circuit shown in Figure 1.42the switch K is changed from position A to B at t=0,the steady state having been reached before switching.Calculate i, di

dt , andd2idt2

at t=0+.

50 V i

K

1H

20 BA

1 FFigure 1.42: Example

Solution:

Before connecting to position 2 switch was atposition 1 at t=0- under steady state conditioncapacitor charges with voltage of v(0−) = 50 = v(0+)and after that it acts as an open circuit which is asshown in Figure 1.43 (a)

50 V i

S 20

(0 ) 50VCv 1

(a)

( )i t1H

20 2

(b)1 F


At t = 0- , inductor is in open circuit and capacitoris after fully charging it is also in open circuit state.That is

i(0−) = 0 and also i(0+) = 0 (1.61)

When switch is at position 2, and at t=0+ the circuitdiagram is as shown in Figure 1.43 (b)


Ri+ Ldi

dt+

1

C

∫idt = 0

Ri+ Ldi

dt+ vc(t)dt = 0

At t = 0+ and vc(0+) = 50

Ri(0+) + Ldi

dt(0+) + vc(0+) = 0

20× 0 + 1di

dt(0+) + 50 = 0

di

dt(0+) =

−50

1= −50 A/sec


Rdi

dt+ L

d2i

dt2+

i

C= 0 (1.62)


Rdi

dt(0+) + L

d2i

dt2(0+) +

i(0+)

C= 0

20× (−50) + 1d2i

dt2(0+) +

0

C= 0

d2i

dt2(0+) =

1000

1= 1000A/sec2

Q DEC-2015 6b) In the circuit shown in Figure ?? thesteady state is reached with switch K is open. Theswitch K is closed at t=0. Solve for i1, i2,

di1dt ,

di2dt at

t=0+.



1iK

1H

20 2i50 V

10 10 1 F


Solution:

When switch is opened and when steady state isreached capacitor acts as open circuit and inductoracts as short circuit which is as shown in Figure1.45(a).

1i20 2i 1i

1H2i

(a) (b)

+-33.3 V

10 10

10

10 10 100 V100 V +-1 F


i1(0−) =V

20 + 10=

100

30= 3.33A (1.63)

i2(0−) = 0A (1.64)

Voltage across capacitor is voltage across R2

vc(0−) = i1(0−)× 10 = 3.33× 10 = 33.33V

When switch is closed at t=0 20Ω is short circuited.Inductor acts as current source with a value of 3.33 A

and capacitor acts as voltage source with a value of33.33 V which is as shown in Figure 1.45(b).

i2(0+) =100− 33.33

10= 6.667

V = 10i1 + Ldi1dt

V = 10i1(0+) + Ldi1(0+)

dt

1di1(0+)

dt= 100− 10× 3.33

di1(0+)

dt= 66.7A/sec

For the capacitor branch

V = 10i2 +1

C

∫i2dt

V = 10i2(0+) +1

C

∫i2(0+)dt

Differentiating we get

= 10di2(0+)

dt+

1

Ci2(0+)

di2(0+)

dt= − 1

10× Ci2(0+)

= − 1

10× 1× 10−6× 6.667 = 0.667× 106 A/sec


Documents

Initial-Conditions€¦ · and evaluation of condition at t= 1are known as nal conditions. The following are the objectives of studying the behavior of the circuit for Initial-Conditions: