Rizzoni5eSM_CH05

G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 5

Chapter 5: Transient Analysis – Instructor Notes

Chapter 5 was reorganized in the 4th Edition, inresponse to a number of suggestions forwarded by users. The chapter is mostly unchanged in this 5th Edition. The chapter begins with a brief description of what is meant by Transient Analysis in Section 5.1; basic techniques for writing differential equations for dynamic circuits are described in Section 5.2, while Section 5.3 focuses on DC steady-state solutions and on initial and final conditions. The analogy between electrical and thermal systems that was introduced in Chapter 3 is now extended to energy storage elements and transient response (Make The Connection: Thermal Capacitance, p. 218; Make The Connection: Thermal System Dynamics, p. 219). Section 5.4 introduces first order transients; a Focus on Methodology box: First Order Transient Response (p. 229) clearly outlines the methodology that is followed in the analysis of first order circuits; this methodology is then motivated and explained, and is applied to eight examples, including four examples focusing on engineering applications (5.8 - Charging a camera flash; 5.9 and 5.11 dc motor transients; and 5.12, transient response of supercapacitor bank). The box Focus on Measurements: Coaxial Cable Pulse Response (pp. 230-232) illustrates an important transient analysis computation (this problem was suggested many years ago by a Nuclear Engineering colleague). The analogy between electrical and thermal systems is carried further in the sidebars Make The Connection: First-Order Thermal System, p. 232-233;); similarly, the analogy between hydraulic and electrical circuits, begun in Chapter 2 and continued in Chapter 4, is continued here (Make The Connection: Hydraulic Tank, pp. 228-229). Section 5.5 covers second order transients, and, as was done for first order transients, begins with two important boxes: Focus on Methodology: Roots of Second-Order System (p. 254) and Focus on Methodology: Second Order Transient Response (pp. 258-259). These boxes clearly outline the methodology that is followed in the analysis of second order circuits; the motivation and explanations in this section are accompanied by five very detailed examples in which the methodology is applied step by step. The last of these examples takes a look at an automotive ignition circuit (with many thanks to my friend John Auzins, formerly of Delco Electronics, for suggesting a simple but realistic ignition circuit). The analogy between electrical and mechanical systems is explored in Make The Connection: Automotive Suspension, pp. 253-254 and pp. 259-260. The homework problems are divided into four sections, and contain a variety of problems ranging from very basic to the fairly advanced. The focus is on mastering the solution methods illustrated in the chapter text and examples. The homework problems in this chapter are mostly mathematical exercises aimed at mastery of the techniques. The 5th Edition of this book includes 7 new problems, increasing the end-of-chapter problem count from 74 to 81.

Learning Objectives for Chapter 5 1. Write differential equations for circuits containing inductors and capacitors. 2. Determine the DC steady state solution of circuits containing inductors and capacitors. 3. Write the differential equation of first order circuits in standard form and determine the complete

solution of first order circuits excited by switched DC sources. 4. Write the differential equation of second order circuits in standard form and determine the

complete solution of second order circuits excited by switched DC sources. 5. Understand analogies between electrical circuits and hydraulic, thermal and mechanical systems.

5.1 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.


Section 5.2: Writing Differential Equations for Circuits Containing Inductors and Capacitors

Problem 5.1

Solution: Known quantities:

.3,6,6,12,9.0 321 Ω=Ω=Ω=== kRkRkRVVmHL s

Find: The differential equation for t > 0 (switch open) for the circuit of P5.21.

Analysis: Apply KCL at the top node (nodal analysis) to write the circuit equation. Note that the top node voltage is the inductor voltage, . vL

vL

R1 + R2+ iL +

vL

R3= 0

Next, use the definition of inductor voltage to eliminate the variable from the nodal equation: vL

L

R1 + R2

diL

dt+ iL +

L

R3

diL

dt= 0

diL

dt+

R1 + R2( )R3

L R1 + R2 + R3( )iL = 0

Substituting numerical values, we obtain the following differential equation:

diL

dt+ 2.67 ⋅106iL = 0

Problem 5.2

Solution: Known quantities: V1 =12 V,C = 0.5µF, R1 = 0.68 kΩ,R2 = 1.8 kΩ.

Find: The differential equation for t > 0 (switch closed) for the circuit of P5.23.

Analysis: Apply KCL at the top node (nodal analysis) to write the circuit equation. Note that the top node voltage is the

capacitor voltage, . Cv

01

1

2=

−++

RVv

Rvi CC

C

Next, use the definition of capacitor current to eliminate the variable from the nodal equation: Ci

CdvC

dt+

R1 + R2

R1R2vC =

V1

R1 ⇒

dvC

dt+

R1 + R2

C R1R2( )vC =

V1

CR1


dvC

dt+ 4052( )vC − 35292 = 0



Problem 5.3

Solution: Known quantities: V1 =12 V, R1 = 0.68 kΩ,R2 = 2.2 kΩ,R3 = 1.8 kΩ,C = 0.47 µF.


Analysis: Apply KCL at the two node (nodal analysis) to write the circuit equation. Note that the node #1 voltage is the capacitor voltage, . vC

For node #1: iC +vC − v2

R2= 0

For node #2: v2 − vC

R2+

v2

R3+

v2 − V1

R1= 0

Solving the system:

vC =R3

R1 + R3V1 −

R1R2 + R1R3 + R2R3

R1 + R3iC

v2 =R3

R1 + R3V1 −

R1R3

R1 + R3iC

Next, use the definition of capacitor current to eliminate the variable from the nodal equation: CiC R1R2 + R1R3 + R2R3( )

R1 + R3

dvC

dt+ vC =

R3

R1 + R3V1

dvC

dt+

R1 + R3

C R1R2 + R1R3 + R2R3( )vC =

R3

C R1R2 + R1R3 + R2R3( )V1


dvC

dt+ 790( )vC − 6876 = 0

Problem 5.4

Solution: Known quantities: VS2 =13V,L = 170 mH , R2 = 4.3kΩ,R3 = 29 kΩ.


Analysis: Applying KVL we obtain: R2 + R3( )iL + vL + VS2 = 0

Next, using the definition of inductor voltage to eliminate the variable from the nodal equation: Lv

R2 + R3( )iL + LdiL

dt+ VS2 = 0

diL

dt+

R2 + R3( )L

iL +VS2

L= 0


diL

dt+1.96 ⋅105iL + 76.5 = 0



Problem 5.5

Solution: Known quantities: I0 = 17 mA,C = 0.55µF, R1 = 7 kΩ,R2 = 3.3kΩ.

Find: The differential equation for t > 0 for the circuit of P5.32.

Analysis: Using the definition of capacitor current:

CdvC

dt= I0 ⇒

dvC

dt=

I0

C


dvC

dt− 30909 = 0

Problem 5.6

Solution: Known quantities: VS1 = VS2 =11 V , C = 70 nF , , R1 =14 kΩ R2 = 13kΩ ,

. R3 = 14 kΩ


Analysis: Apply KCL at the top node (nodal analysis) to write the circuit equation. v1 − VS2

R1+ iC +

v1

R3= 0

Note that the node voltage is equal to: 1vv1 = R2iC + vC

Substitute the node voltage in the first equation: 1v1

R1+

1

R3

⎛

⎝ ⎜

⎞

⎠ ⎟ vC +

R2

R1+

R2

R3+1

⎛

⎝ ⎜

⎞

⎠ ⎟ iC −

VS2

R1= 0

Next, use the definition of capacitor current to eliminate the variable from the nodal equation: Ci1

R1+

1

R3

⎛

⎝ ⎜

⎞

⎠ ⎟ vC +

R2

R1+

R2

R3+1

⎛

⎝ ⎜

⎞

⎠ ⎟ C

dvC

dt−

VS2

R1= 0


dvC

dt+ 714.3vC − 3929 = 0



Problem 5.7

Solution: Known quantities: VS = 20 V, R1 = 5Ω,R2 = 4 Ω ,R3 = 3Ω,R4 = 6Ω,C1 = 4 F ,C2 = 4 F, IS = 4 A. Find: The differential equation for t > 0 (switch closed) for the circuit of P5.41.

Analysis: Apply KCL at the two node (nodal analysis) to write the circuit equation. Note that the node #1 voltage is equal to the two capacitor voltages, . vC1 = vC 2 = vCFor node #1:

iC1 + iC2 +vC − v2

R2= 0

For node #2: v2 − vC

R2+

v2

R3+

v2

R4− IS = 0

Solving the system:

v2 =R3R4

R3 + R4iC1 + iC 2 − IS( )

vC = − R2 +R3R4

R3 + R4

⎛

⎝ ⎜

⎞

⎠ ⎟ iC1 + iC 2( )+

R3R4

R3 + R4IS

Next, use the definition of capacitor current to eliminate the variables from the nodal equation: Cii

vC + R2 +R3R4

R3 + R4

⎛

⎝ ⎜

⎞

⎠ ⎟ C1 + C2( ) dvC

dt−

R3R4

R3 + R4IS = 0


dvC

dt+

1

48vC −

1

6= 0

Problem 5.8

Solution: Known quantities:

FC µ1= , , . RS =15kΩ R3 = 30 kΩ


Assume: Assume that V, and VS = 9 R1 =10kΩ R2 = 20kΩ .

Analysis: Apply KCL at the top node (nodal analysis) to write the circuit equation.

1. Before the switch opens. Apply KCL at the top node (nodal analysis) to write the circuit equation.

vC − VS

RS+

vC

R1+ iC +

vC

R2+

vC

R3= 0 ⇒

1

RS+

1

R1+

1

R2+

1

R3

⎛

⎝ ⎜

⎞

⎠ ⎟ vC + iC −

VS

RS= 0

Next, use the definition of capacitor current to eliminate the variable from the nodal equation: Ci



1

RS+

1

R1+

1

R2+

1

R3

⎛

⎝ ⎜

⎞

⎠ ⎟ vC + C

dvC

dt−

VS

RS= 0

Substituting numerical value, we obtain the following differential equation:

dvC

dt+ 250vC − 600 = 0

2. After the switch opens. Apply KCL at the top node (nodal analysis) to write the circuit equation.

vC − VS

RS+

vC

R1+ iC = 0 ⇒

1

RS+

1

R1

⎛

⎝ ⎜

⎞

⎠ ⎟ vC + iC −

VS

RS= 0

Next, use the definition of capacitor current to eliminate the variable from the nodal equation: Ci1

RS+

1

R1

⎛

⎝ ⎜

⎞

⎠ ⎟ vC + C

dvC

dt−

VS

RS= 0


dvC

dt+

500

3vC − 600 = 0

Problem 5.9

Solution: Known quantities: Values of the voltage source, of the inductance and of the resistors.


Analysis: Apply KCL at the top node (nodal analysis) to write the circuit equation. v1 −100

10+

v1

5+ iL = 0 ⇒ 0.3v1 + iL −10 = 0

Note that the node voltage is equal to: v1v1 = 2.5iL + vL Substitute the node voltage in the first equation: v1

1.75( )iL + 0.3( )vL −10 = 0

Next, use the definition of inductor voltage to eliminate the variable from the nodal equation: Lv

(0.3) 0.1( ) diL

dt+ 1.75( )iL −10 = 0 ⇒

diL

dt+ 58.33( )iL − 333 = 0



Problem 5.10

Solution: Known quantities: IS = 5A, , , . L1 = 1H L2 = 5H Ω= kR 10


Analysis: Applying KCL at the top node (nodal analysis) to write the circuit equation.

−IS +v1

R+ iL = 0

Note that the node voltage is equal to: 1vv1 = vL1 + vL2

Substitute the node voltage in the first equation: 1vvL1 + vL2

R+ iL − IS = 0

Next, use the definition of inductor voltage to eliminate the variable from the nodal equation: LvL1 + L2( )

R

diL

dt+ iL − IS = 0

Substituting numerical value, we obtain the following differential equation:

03

250003

5000=−+ L

L idt

di



Section 5.3: DC Steady State Solution of Circuits Containing Inductors and Capacitors – Initial and Final Conditions

Problem 5.11

Solution: Known quantities: L = 0.9 mH ,Vs = 12V , R1 = 6kΩ,R2 = 6 kΩ,R3 = 3kΩ.

Find: The initial and final conditions for the circuit of P5.21.

Analysis: Before opening, the switch has been closed for a long time. Thus we have a steady-state condition, and we treat the inductor as a short circuit. The voltages across the resistances R is equal to zero, since it is in parallel to the short

circuit, so all the current flow through the resistor and : 3

1R R2

46K1

6K11211)0(

21=⎟

⎠⎞

⎜⎝⎛ +=⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

RRVi SL mA

After the switch has been opened for a long time, we have again a steady-state condition, and we treat the inductor as a short circuit. When the switch is open, the voltage source is not connected to the circuit. Thus, iL ∞( )= 0 A.

Problem 5.12

Solution: Known quantities: V1 =12 V,C = 0.5µF, R1 = 0.68 kΩ,R2 = 1.8 kΩ.


Analysis: Before closing, the switch has been opened for a long time. Thus we have a steady-state condition, and we treat the capacitor as an open circuit. When the switch is open, the voltage source is not connected to the circuit. Thus,

vC (0) = 0 V. After the switch has been closed for a long time, we have again a steady-state condition, and we treat the capacitor as an open circuit. The voltage across the capacitor is equal to the voltage across the resistance : R2

vC ∞( )=R2

R1 + R2V1 =

1800

1800 + 68012 = 8.71 V.



Problem 5.13

Solution: Known quantities: V1 =12 V, R1 = 0.68 kΩ,R2 = 2.2 kΩ,R3 = 1.8 kΩ,C = 0.47 µF.


Analysis: Before closing, the switch has been opened for a long time. Thus we have a steady-state condition, and we treat the capacitor as an open circuit. When the switch is open, the voltage source is not connected to the circuit. Thus,

vC (0) = 0 V. After the switch has been closed for a long time, we have again a steady-state condition, and we treat the capacitor as an open circuit. Since the current flowing through the resistance is equal to zero, the voltage across the capacitor is equal to the voltage across the resistance :

R2R3

vC ∞( )=R3

R1 + R3V1 =

1800

1800 + 68012 = 8.71 V.

Problem 5.14

5.9

Solution: Known quantities: VS1 =VS2 =13 V,L = 170 mH , R2 = 4.3kΩ,R3 = 29 kΩ.


Analysis: In a steady-state condition we can treat the inductor as a short circuit. Before the switch changes, applying the KVL we obtain:

VS1 − VS2 = R1 + R2( )iL 0( ) ⇒ iL 0( )=VS1 − VS2

R1 + R2= 0 mA.

After the switch has changed for a long time, we have again a steady-state condition, and we treat the inductor as a short circuit. Thus, applying the KVL we have:

iL ∞( )= −VS2

R2 + R3= −0.39 mA.

PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.


Problem 5.15

Solution: Known quantities: I0 = 17 mA,C = 0.55µF, R1 = 7 kΩ,R2 = 3.3kΩ.


Analysis: Before the switch changes, the capacitor is not connected to the circuit, so we don’t have any information about its initial voltage. After the switch has changed, the current source and the capacitor will be in series so the current to the capacitor will be constant at . Therefore, the rate at which charge accumulates on the capacitor will also be constant and, consequently, the voltage across the capacitor will rise at a constant rate, without ever reaching an equilibrium state.

I0

Problem 5.16

Solution: Known quantities: VS1 =17 V , , VS2 =11 V C = 70 nF , R1 =14 kΩ ,

, . R2 = 13kΩ R3 = 14 kΩ


Analysis: In a steady-state condition we can treat the capacitor as an open circuit. Before the switch changes, applying the KVL we have that the voltage across the capacitor is equal to the source voltage : VS1

vC (0) = VS1 = 17 V. After the switch has changed for a long time, we have again a steady-state condition, and we treat the capacitor as an open circuit. Since the current flowing through the resistance is equal to zero, the voltage across the capacitor is equal to the voltage across the resistance :

R2R3

vC ∞( )=R3

R1 + R3VS2 =

14000

14000 +1400011 = 5.5 V.



Problem 5.17

Solution: Known quantities: VS = 20 V, R1 = 5Ω,R2 = 4 Ω ,R3 = 3Ω,R4 = 6Ω,C1 = 4 F ,C2 = 4 F, IS = 4 A.


Analysis: The switch is always open and the switch closes at

. Before closing, the switch has been opened for a long time. Thus we have a steady-state condition, and we tthe capacitors as open circuits. When the switch is open, the current source is not connected to the circuit. Thus,

S1 S2t = 0 S2

reat

vC1(0) = 0 V, vC2 (0) = 0 V.

After the switch has been closed for a long time, we have again a steady-state condition, and we treat the

capacitors as open circuits. The voltages across the capacitors are both equal to the voltage across the resistance : 2S

R3

vC1 ∞( )= vC2 ∞( )= R3 || R4( )IS =6 ⋅ 3

6 + 34 = 8 V.

Problem 5.18

Solution: Known quantities: C = 1µF , , . RS =15kΩ R3 = 30 kΩ


Assume: Assume that V, and VS = 9 R1 =10kΩ R2 = 20kΩ .

Analysis: Before opening, the switch has been closed for a long time. Thus, we have a steady-state condition, and we treat the capacitor as an open circuit. The voltage across the capacitor is equal to the voltage across the resistances ,

, and . Thus, R1

R2 R3

vC (0) =R1 || R2 || R3( )

RS + R1 || R2 || R3( )VS =

10k ||12k

15k +10k ||12k9 = 2.4 V.

After opening, the switch has been opened for a long time. Thus we have a steady-state condition, and we treat the capacitor as an open circuit. The voltage across the capacitor is equal to the voltage across the resistance . Thus, R1

vC ∞( )=R1

R1 + RSVS =

10000

10000 +150009 = 3.6 V.



Problem 5.19

Solution: Known quantities: Values of the voltage source, of the inductance and of the resistors.


Analysis: Before the switch changes, apply KCL at the top node (nodal analysis) to write the following circuit equation. v1 −100

1000+

v1

5+

v1

2.5= 0 ⇒ v1 =

100

601= 0.165 V.

iL (0) =v1

2.5=

40

601= 66 mA.

After the switch has changed, apply KCL at the top node (nodal analysis) to write the following circuit equation. v1 −100

10+

v1

5+

v1

2.5= 0 ⇒ v1 =

100

7= 14.285 V.

iL (∞) =v1

2.5=

40

7= 5.714 A.

Problem 5.20

Solution: Known quantities: IS = 5A, , , L1 = 1H L2 = 5H R = 10kΩ .


Analysis: Before closing, the switch has been opened for a long time. Thus we have a steady-state condition, and we treat the inductors as short circuits. The values of the two resistors are equal so the current flowing through the inductors is:

iL (0) =IS

2= 2.5 A.

After the switch has been closed for a long time, we have again a steady-state condition, and we treat the inductors as short circuits. In this case the resistors are short-circuited and so all the current is flowing through the inductors. A. iL (∞) = IS = 5



Section 5.4: Transient Response of First Order Circuits

Pro

SolKnoCircL =

FinIf thopen

AssiL =

AnaDetesteadsteadcomAt s

3RV

Thuappl

App

AppThe in a

PROeduc

Focus on Methodology

First-order transient response 1. Solve for the steady-state response of the circuit before the switch changes state (t = 0-),

and after the transient has died out (t → ∞). We shall generally refer to these responses as x(0-) and x(∞).

2. Identify the initial condition for the circuit, x(0+), using continuity of capacitor voltages and inductor currents (vC(0+) = vC(0-), iL(0+) = iL(0-)), as illustrated in Section 5.4.

3. Write the differential equation of the circuit for t = 0+, that is, immediately after the switchhas changed position. The variable x(t) in the differential equation will be either a capacitor voltage, vC(t), or an inductor current, iL(t). It is helpful at this time to reduce the circuit to Thévenin or Norton equivalent form, with the energy storage element (capacitor or inductor) treated as the load for the Thévenin (Norton) equivalent circuit. Reduce this equation to standard form (Equation 5.8).

4. Solve for the time constant of the circuit: τ = RTC for capacitive circuits, τ = L/RT for inductive circuits.

5. Write the complete solution for the circuit in the form:

blem 5.21

ution: wn quantities:

uit shown in Figure P5.21, 0.9 mH ,Vs = 12V , R1 = 6kΩ,R2 = 6 kΩ,R3 = 3kΩ.

d: e steady-state conditions exist just before the switch was ed.

umptions: 1.70 mA before the switch is opened at 0=t .

lysis: rmine the steady state current through the inductor at 0<t . If this current is equal to the current specified, y-state conditions did exit; otherwise, opening the switch interrupted a transient in progress. To determine the y-state current before the switch was opened, replace the inductor with an equivalent DC short-circuit and

pute the steady-state current through the short circuit. teady state, the inductor is modeled as a short circuit:

( ) ( ) 0000 3 == −−Ri

s, the short-circuit current through the inductor is simply the current through which can be found by ying Kirchoff’s Laws and Ohm’s Law.

2R

ly KVL: ( ) ( ) mHRViRiV s

RRs 2106

12000 32

223 =×

===+− −−

ly KCL: ( ) ( ) ( ) ( ) ( ) mHiiiii RLRLR 200,0000 232 ===++− −−−−−

actual steady state current through the inductor is larger than the current specified. Therefore, the circuit is not steady state condition just before the switch is opened.

x(t) = x(∞) + x(0) − x(∞)( )e−t /τ

5.13 PRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and ators for course preparation. If you are a student using this Manual, you are using it without permission.


Problem 5.22

Solution: Known quantities: Circuit shown in Figure P5.22, .23,7,17,11,130,35 32121 Ω=Ω=Ω==== kRkRkRFCVVVV SS µ

Find: At the initial current through just after the switch is changed.

t = 0+ R3

5.14

Assumptions: None.

Analysis: To solve this problem, find the steady state voltage across the capacitor before the switch is thrown. Since the voltage across a capacitor cannot change instantaneously, this voltage will also be the capacitor voltage immediately after the switch is thrown. At that instant, the capacitor may be viewed as a DC voltage source. At : Determine the voltage across the capacitor. At steady state, the capacitor is modeled as an open circuit:

t = 0−

( ) ( ) 000 21 == −−RR ii

Apply KVL:

( )

( ) VVVV

VVV

SSC

SCS

950

0000

21

21

−=−=

=−+−+−

−

At : t = 0+

( ) ( )( ) ( )++

−+

=

−==

00

9500

32 RR

CC

ii

VVV

Apply KVL:

( ) ( ) ( )( ) ( ) mA

RRVVi

RiVRiV

CSR

RCRS

167.11023107

9513000

0000

3332

23

33232

=×+×

−=

++

=

=−+−+

+

+++



Problem 5.23

Solution: Known quantities: Circuit shown in Figure P5.23,

.8.1,68.0,5.0,12 211 Ω=Ω=== kRkRFCVV µ

Find: The current through the capacitor just before and just after the switch is closed.

Assumptions: The circuit is in steady-state conditions for t < 0.

Analysis: At , assume steady state conditions exist. If charge is stored on the plates of the capacitor, then there will be energy stored in the electric field of the capacitor and a voltage across the capacitor. This will cause a current flow through which will dissipate energy until no energy is stored in the capacitor at which time the current ceases and the voltage across the capacitor is zero. These are the steady state conditions, i.e.:

t = 0−

2R

( ) ( ) 0000 == −−CC Vi

At , the switch is closed and the transient starts. Continuity requires: += 0t ( ) ( ) 000 == −+

CC VV At this instant, treat the capacitor as a DC voltage source of strength zero i.e. a short-circuit. Therefore, all of the voltage is across the resistor and the resulting current through is the current into the capacitor. V1 R1 R1 Apply KCL: (Sum of the currents out of the top node)

( ) ( ) ( )

( ) mARVi

RVV

RVi

C

CCC

65.171068.0

120

00000

31

1

1

1

2

=×

==

=−

+−

+

+

+++

The CURRENT through the capacitor is NOT continuous but changes from to 17.65 when the switch is closed. The voltage across the capacitor is continuous because the stored energy CANNOT CHANGE INSTANTANEOUSLY.

0 mA



Problem 5.24

Solution: Known quantities: Circuit shown in Figure P5.23, V1 =12 V,C = 150 µF, R1 = 400 mΩ,R2 = 2.2 kΩ.

Find: The current through the capacitor just before and just after the switch is closed.


Analysis: At , assume steady state conditions exist. If charge is stored on the plates of the capacitor, then there will be energy stored in the electric field of the capacitor and a voltage across the capacitor. This will cause a current flow through which will dissipate energy until no energy is stored in the capacitor at which time the current ceases and the voltage across the capacitor is zero. These are the steady state conditions, i.e.:

t = 0−

R2

( ) ( ) 0000 == −−CC Vi

At , the switch is closed and the transient starts. Continuity requires: t = 0+ ( ) ( ) 000 == −+CC VV

At this instant, treat the capacitor as a DC voltage source of strength zero i.e. a short-circuit. Therefore, all of the voltage is across the resistor and the resulting current through is the current into the capacitor. V1 R1 R1Apply KCL: (Sum of the currents out of the top node)

( ) ( ) ( ) ( ) ARV

iR

VVR

Vi C

CCC 30

10400120 0

0000

31

1

1

1

2=

×==⇒=

−+

−+

−+

+++

The CURRENT through the capacitor is NOT continuous but changes from to when the switch is closed. The voltage across the capacitor is continuous because the stored energy CANNOT CHANGE INSTANTANEOUSLY.

0 A30

Problem 5.25

Solution: Known quantities: Circuit shown in Figure P5.21, .3,6,6,9.0,12 321 Ω=Ω=Ω=== kRkRkRmHLVVs

Find: The voltage across just after the switch is open. 3RAssumptions: iL = 1.70 mA before the switch is opened at 0=t .

Analysis: When the switch is opened the voltage source is disconnected from the circuit and plays no role. Since the current through the inductor cannot change instantaneously the current through the inductor at t = 0+ is also 1.70 mA. At this instant, treat the inductor as a DC current source and solve for the voltage across R3 by current division or KCL and Ohm’s Law. Specify the polarity of the voltage across 3R . ( ) ( ) mAii LL 7.100 == −+

Ω=+= kRRReq 1221Apply KCL: (Sum of the currents out of the top node)

( ) ( ) ( ) ( ) ( )

V

RR

iV

RV

iR

V

eq

LR

RL

eq

R 080.4

1031

10121

107.111

00 0

00

0

33

3

3

33

33 −=

×+

×

×=

+−=⇒=++

−++

++

+



Problem 5.26


.100,22,7.0,12 11 mHLkRRVV s =Ω=Ω==

Find: The voltage through the inductor just before and just after the switch is changed.


Analysis: In steady-state the inductor acts like a short-circuit so it has no voltage across it for t < 0. However, its current is non-zero and is equal to the current out of the source VS and through RS. At the instant the switch is changed the current through the inductor is unchanged since the current through an inductor cannot change instantaneously. Also notice that after the switch is changed the current through R1 is always equal to the inductor current and the voltage across R1 is always equal to the inductor voltage. Thus, at t = 0+ the voltage across the inductor must be non-zero. That’s fine since the voltage across an inductor can change instantaneously (or relatively so.) Assume a polarity for the voltage across the inductor. t = 0− : Steady state conditions exist. The inductor can be modeled as a short circuit with: VL 0−( )= 0

Apply KVL;

ARV

i

VRiV

S

SL

LSLS

14.177.0

12)0(

0)0()0(

===

=++−

−

−−

At , the transient commences. Continuity requires:+= 0t ( ) ( )−+ = 00 LL ii Apply KVL:

( ) ( ) ( ) ( ) kVRiVVRi LLLL 1.337102214.1700 000 311 −=××−=−=⇒=+ ++++

Problem 5.27

Solution: Known quantities: Circuit shown in Figure P5.27, V1 =12 V, R1 = 0.68 kΩ,R2 = 2.2 kΩ,R3 = 1.8 kΩ,C = 0.47 µF.

Find: The current through the capacitor at , just after the switch is closed.

t = 0+


Analysis: For t < 0, the switch is open and no power source is connected to the left half of the circuit. In steady state, by definition, the voltage across the capacitor and the current out of it must be constant. However, without a power source to replenish the energy dissipated by the resistors, that constant must be zero. Otherwise, current would flow out of the capacitor, its voltage would drop as it lost charge, and the energy of that charge would be dissipated by the resistors. This process would continue until no net charge remained on the capacitor and its voltage was zero. At steady state, then, the voltage across the capacitor is zero.



At t = 0+, the voltage across the capacitor is still zero since the voltage across a capacitor cannot change instantaneously. At that instant, the capacitor can be treated as a voltage source of strength zero (i.e. a short-circuit.) However, the current through the capacitor can change instantaneously (or relatively so) from 0 to a new value. In this problem it will change as the switch is closed because the voltage source V1 will drive current through R1and the parallel combination of R2 and R3. The current through R2 is the capacitor current.

VC 0+( )= VC 0−( )= 0

Apply KCL:

( ) ( ) ( )

( ) V

RR

RRV

RRR

RV

V

RVV

RV

RV

R

RRR

114.7

8.168.0

2.268.01

12

11110

00000

3

1

2

11

321

1

1

3

1

13

3

3

2

3

=++

=++

=++

=

=−

++−

+

+++

Recall that the voltage across the capacitor (Volts = Joules/Coulomb) represents the energy stored in the electric field between the plates of the capacitor. The electric field is due to the amount of charge stored in the capacitor and it is not possible to instantaneously remove charge from the capacitor’s plates. Therefore, the voltage across the capacitor cannot change instantaneously when the circuit is switched. However, the rate at which charge is removed from the plates of the capacitor (i.e. the capacitor current) can change instantaneously (or relatively so) when the circuit is switched. Note also that these conditions hold only at the instant t = 0+. For t > 0+, the capacitor is gaining charge, all voltages and currents exponentially approach their final or steady state values. Apply KVL:

( ) ( ) ( ) ( ) ( ) ( )mA

RVV

iVRiV CRCRCC 234.3

102.2114.700

0 000032

332 =

×=

−=⇒=++−

++++++

Problem 5.28

Solution: Known quantities: NOTE: Typo in problem statement: should read “For t < 0, …” Circuit shown in Figure P5.22. . 23 ; 7 ; 17 F; 11 V; 130 V; 35 32121 Ω=Ω=Ω==== kRkRkRCVV SS µ

Find: The time constant of the circuit for t > 0.


Analysis: For t > 0, the transient is in progress. The time constant is a relative measure of the rate at which the voltages and currents are changing in the transient phase. The time constant for a single capacitor system is ReqC where Req is the Thévenin equivalent resistance as seen by the capacitor, i.e., with respect to the port or terminals of the capacitor. To calculate Req turn off (set to zero) the ideal independent voltage source and ask the question “What is the net equivalent resistance encountered in going from one terminal of the capacitor to the other through the network?” Then:

msCRkRRR eqeq 0.33010111030 301023107 633332 =×××==Ω=×+×=+= −τ



Problem 5.29

Solution: Known quantities: Circuit shown in Figure P5.29, .29,3.4,7.2,170,13,13 32121 Ω=Ω=Ω==== kRkRkRmHLVVVV SS


5.19


Analysis: For t > 0, the transient is in progress. The time constant is a relative measure of the rate at which the voltages and currents are changing in the transient phase. The time constant for a single inductor system is L/Req where Req is the Thévenin equivalent resistance as seen by the inductor, i.e., with respect to the port or terminals of the inductor. To calculate Req turn off (set to zero) the ideal independent voltage source and ask the question “What is the net equivalent resistance encountered in going from one terminal of the inductor to the other through the network?” Then:

sRL

kRRR

eq

eq

µτ 105.51030.33

10170

30.331029103.4

3

3

3332

=×

×==

Ω=×+×=+=

−

Problem 5.30

Solution: Known quantities: Circuit shown in Figure P5.27, .8.1,2.2,680,47.0,12 3211 Ω=Ω=Ω=== kRkRRFCVV µ



Analysis: For t > 0, the transient is in progress. The time constant is a relative measure of the rate at which the voltages and currents are changing in the transient phase. The time constant for a single capacitor system is ReqC where Req is the Thévenin equivalent resistance as seen by the capacitor, i.e., with respect to the port or terminals of the capacitor. To calculate Req turn off (set to zero) the ideal independent voltage source and ask the question “What is the net equivalent resistance encountered in going from one terminal of the capacitor to the other through the network?” Then:

msCR

kRR

RRRR

eq

eq

266.11047.010694.2

694.21068.0108.11068.0108.1102.2

63

33

333

13

132

=×××==

Ω=×+×

×××+×=

++=

−τ



Problem 5.31

Solution: Known quantities: Circuit shown in Figure P5.21, VS =12 V,L = 0.9 mH , R1 = 6 kΩ,R2 = 6 kΩ,R3 = 3kΩ.


Assumptions: The current through the inductor is before the switch is opened at t = 0. mAiL 70.1=Analysis: For t > 0, the transient is in progress. The time constant is a relative measure of the rate at which the voltages and currents are changing in the transient phase. The time constant for a single inductor system is L/Req where Req is the Thévenin equivalent resistance as seen by the inductor, i.e., with respect to the port or terminals of the inductor. To calculate Req turn off (set to zero) the ideal independent voltage source and ask the question “What is the net equivalent resistance encountered in going from one terminal of the inductor to the other through the network?” Then:

Req =R3(R1 + R2)

R3 + (R1 + R2)=

3×103(6 ×103 + 6 ×103)

3×103 + (6 ×103 + 6 ×103)= 2.400 kΩ

τ =L

Req=

0.9 ×10−3

2.4 ×103= 0.3750 µs

Problem 5.32

Solution: Known quantities: Circuit shown in Figure P5.32, .3.3,7,55.0,17,7)0( 210 Ω=Ω===−=− kRkRFCmAIVVc µ

Find: The voltage across the capacitor for t > 0. )(tVc

Assumptions: Before the switch is thrown the voltage across the capacitor is –7 V.

Analysis: The current source and the capacitor will be in series so the cTherefore, the rate at which charge accumulates on the capacitor will also be constant and, consequently, the voltage across the capacitor will rise at a constant rate. The integral form of the capacitor i-V relationship best expresses this accumulation process. The continuity of the voltage across the capacitor requires:

urrent to the capacitor will be constant at I0.

( ) ( )

( ) ( )tt

tCIVdt

CIV

dtIdttiC

dttiC

tV

mAItiVVVC 0 =+

tC

tC

tC

tCC

C

C

36

3

00

00

0 00

0

1091.3071055.0

10177

|00

))((1)(1)(

17)(70

×+−=×

×+−=

+=∫+=

∫+∫=∫=

==−=

−

−

++

∞−∞−

−



Problem 5.33

Solution: Known quantities: Circuit shown in Figure P5.29, .330,13,7.0,23,20,23 32121 Ω=Ω=Ω==== kRRRmHLVVVV SS

Find: The current through resistor R)(3 tiR 3 for t > 0.

Assumptions: The circuit is in steady-state conditions for 0<t . It is a resistive circuit with one storage element (e.g. inductor)

Analysis:

so an assumed solution is of the form

here is to first find the initial condition at t = 0+ for the inductor and use it to determine the initial .

me steady state conditions exist. At steady state, the inductor is modeled as a short-circuit:

iR3(t) = ISS + (I0 − ISS )e−t /τ

The approachcondition on the current through resistor R3. Second, find the final steady-state condition of the circuit for t > 0To do so, simply apply DC circuit analysis to solve for the current through resistor R3 i.e. replace the inductor with ashort-circuit. Finally, solve for the time constant of the circuit for t > 0. Each of these three results is needed to construct the complete transient solution. At t = 0− : AssuApply KVL:

ARRVV

iVRiRiV

5.21

SSLSLLS 22.0

137.02320)0( 0)0()0(

21

121212 −=

+−

=+−

==+++ −−−

This current is flowing in the direction from the inductor to the switch.

of the rrent through the inductor requires that:

Find ISS at infinity: eno lapsed for steady state conditions to return. In steady state the inductor is modeled

Find τ for stant t one first needs to determine the Thevenin equivalent resistance RTH across the terminals

−

Find I0 at t = 0+ : Continuity cu

AiiIAii LRLL 22.0)0()0( 22.0)0()0( 30 −===−== ++−+

=tAssume that ugh time has eas a short circuit; therefore, the voltage across the inductor is zero. The result is a simple series connection of resistors R2 and R3. The current across R3 is found directly from Ohm’s Law in this case.

0>t : To find the time conof the inductor. To do so, set all independent ideal sources to zero and determine the equivalent resistance "seen" by the inductor, i.e. with respect to the port or terminals of the inductor:

nsRL 10233 ×kRRReq

eq 70.69100.330

0.33010330133

3

32 =×

==Ω=×+=+=−

τ

The complete response can now be written using the solution for transient voltages and currents in first-order

circuits:

Ae

eeiiitit

tt

RRRR

9

9

1070.69

1070.693333

28.0058.0

)058.022.0(058.0))()0(()()(

−

−

×−

×−−+

−=

−−+=∞−+∞= τ

mAVRR

ViI SRSS 6.60

k 013.33020

32

23

≅Ω

=+

==



5.22

Problem 5.34

Known quantities: Circuit shown in Figure P5.34,

Solution:

.70,14,13,14,11,17 32121 nFCkRkRkRVVVV SS =Ω=Ω=Ω===

Find: V (t) for

or 0>t . a)

b) The time f V (t) to change by 98% of its total change is operated.

it is in steady-state conditions for

in voltage after the switch

Assumptions: 0<t . The circu It is a

e storage element (e.g. capacitor) so sum tion is of the form

this ross resistor R3. Second, find the final steady-state

o do so, simply apply DC circuit analysis to solve for the voltage across the pacitor with an open-circuit and solve.) Finally, solve for the time constant of the Thevenin equivalent resistance RTH across the terminals of the capacitor. Each of

sults is needed to construct the complete transient solution.

−)R2 + VC (0−)

resistive circuit with onan as ed solu

VR3(t) = VSS + (V0 − VSS )e−t /τ

Analysis: In general, the approach in problems such as one is to first findand use it to determine the initial condition on the voltage accondition of the circuit for t > 0. T

the initial condition at t = 0+ for the capacitor

resistor R3 (i.e. replace the cacircuit for t > 0 by finding the these three re

a) At t = 0− :

Steady state conditions are specified. At steady state, the capacitor is modeled as an open circuit:

iC (0−) = 0

Apply KVL:

iC (0VS1 =

VC (0−) = VS1 = 17V

At :

ross the capacitor remains the same:

Apply KCL:

t = 0+

The voltage ac

VC (0+) = VC (0−) = 17V

V

RRR

RV

RV

V 0( +

RRR 321++

VVVVV

SC

CS

525.9

10141

10131

10141

101411

101317

111

0(

)

00)0()0()0()()0

333

33

321

1

2

2

2

=

×+

×+

×

×+

×=++

+=

=−−−

+

++++

For

etermine the equivalent resistance as "seen" by the capacitor, ie, with resp ct to the port or terminals of the capacitor. Suppress the independent ideal voltage source:

(

)

0>t :

D e



5.23

= R + R RReq 2 1 3( )= R +R1R3

2 R1 + R3= 13×103 +

14 ×103 ×14 ×103

14 ×103 +14 ×103

= 20.00 kΩ

τ = ReqC = 20 ×103 × 70 ×10−9 = 1.400 ms

At infinity:

Steady state is again established. At steady state the capacitor is once again modeled as an open circuit:

ickly by

=t

iC (∞) = 0

Since the current through the capacitor branch is zero in steady state the voltage across R3 may be found quvoltage division

=14 ×103 +14 ×103

11×14 ×103= 5.500V V (∞) =

R3 + R1

VS2R3

The complete response for is then:

0>t

V (t) = V (∞) + (V (0+) − V (∞))e−

t

τ = 5.5+ (9.525− (5.5))e−

t

1.4×10−3 = 5.5+ 4.025e−

t

1.4×10−3V

b)

The time required for V(t) to reach 98% of its final value is found from

or

mst

eVVtV 025.45.55805.5)025.4(98.0525.998.0)0()( +==−×+=∆+=−

+

Vt

477.5)025.4

5.55805.5ln(104.1

025.4)525.9(5.5)0()(

31

104.113

=−

×−=

−=−=−∞=∆

−

×

+

−

VVV

=V (t) − V (0+)

V (∞) − V (0+)0.98



Problem 5.35


.7.1,37.0,12 Ω=Ω== kRRVV GG

Find: The value of L and . R1

Assumptions: The voltage across the spark plug gap just after the switch is changed is and the voltage will change exponentially with a time constant

VR 23kVτ = 13ms .

Analysis: At : t = 0−

Assume steady state conditions exist. At steady state the inductor is modeled as a short circuit:

VL (0−) = 0

The current through the inductor at this point is given directly by Ohm’s Law:

1

)0(RR

ViG

GL +

=−

At : t = 0+

Continuity of the current through the inductor requires that:

Ω=−×−

××−=−−=

+−=−=

+==

+

++

−+

5170.037.01023

107.112)0(

)0()0(

)0()0(

3

3

1

1

1

GR

G

G

GLR

G

GLL

RV

RVR

RRRV

RiV

RRV

ii

Note that the voltage across the gap VR was written as –23 kV since the current from the inductor flows opposite to the polarity shown for VR; that is, the actual polarity of the voltage across R is opposite that shown. For : t > 0

Determine the Thevenin equivalent resistance as "seen" by the inductor, ie, with respect to the port or terminals of the inductor:

HRRL

RRL

RL

RRR

eq

eq

11.22)107.15170.0(1013)( 331

1

1

=×+××=+=

+==

+=

−τ

τ



Problem 5.36

Solution: Known quantities: Circuit shown in Figure P5.36, when , the relay functions. mAiL 2+≥

.1.3,9.10,12 1 Ω=== kRmHLVVS

Find: 2R so that the relay functions at . st 3.2=

Assumptions: The circuit is in steady-state conditions for 0<t .

Analysis: In this problem the current through the inductor is clearly zero before the switch is thrown. The task is determine the value of the resistance R2 such that the current through the inductor will need 2.3 seconds to rise to 2 mA. Once again, we must find the complete transient solution, this time for the current through the inductor. Assume a solution of the form

iL (t) = i∞ + (i0 − i∞ )e−t /τ

At : t = 0−

The current through the inductor is zero since no source is connected.

iL (0−) = 0 At : += 0t

i0 = iL (0+) = iL (0−) = 0For : 0>t

Determine the Thevenin equivalent resistance as "seen" by the inductor, i.e., with respect to the port or terminals of the inductor:

( )21

2121 RR

RRRRRTH +==

Hence,

t infinity:

tablished. At steady state the inductor is again modeled as a short circuit. Thus, the

lug in the above quantities to the complete solution and set the current through the inductor equal to 2 mA and the

τ =L

RTH=

L R1 + R2( )R1R2

A =tSteady state is again escurrent through R2 is zero and the current through the inductor is given by

mAR

ii SL 87.3

101.3)( 3

1=

×==∞=∞

V 12

Ptime equal to 2.3 s.



( ) seconds16.3or 87.321ln3.2or 11087.3102 /3.233 ≅⎥⎦

⎤⎢⎣⎡ −=

−−×=× −−− τ

ττe

5.26

Solving for R2: ( )

Ω≅−

=+

= mLR

LRRRR

RRL 4.316.3

or 16.31

12

21

21

Problem 5.37

Solution: ntities:

5.37, Known quaCircuit shown in Figure PV1 =12 V, R1 = 400 mΩ,R2 = 2.2 kΩ,C = 150 µF.

Find: ent through the capacitor just before and just after the

dy-state conditions for

The currswitch is closed.

Assumptions:The circuit is in stea t < 0.

sume steady state conditions exist. If charge is stored on the plates of the capacitor, then there will be Analysis: At t = 0− , asener red in the electric field of the capacitor and a voltage across the capacitor. This will cause a current flow through R2 which will dissipate energy until no energy is stored in the capacitor at which time the current ceases and the voltage across the capacitor is zero. Thus, in this circuit, and others like it that contain no connected sources prior to the switch being thrown, the steady state condition is zero currents and zero voltages.

gy sto

( ) ( ) 0000 == −−CC Vi

At , the switch is close ; the transient starts. Continuity requires: t = 0+ d

( ) ( ) 000 == −+CC VV

Since the voltage across the capacitor is zero at this instant, the voltage across the resistor R2 is also zero at this f the instant which implies that no current passes through the resistor at t = 0+. Therefore, at t = 0+, the current out o

source must be the current into the capacitor.

( ) AViC 30121 ==+ R 4.0

01

=

e CURRENT through the capacitor is NOT continuous but changes from to Th 0 30A when the switch is closed. The voltage across the capacitor is continuous because the stored energy CANNOT CHANGE INSTANTANEOUSLY.



Problem 5.38


VS =12 V, RS = 0.24 Ω,R1 = 33kΩ,L = 100 mH .

Find: The voltage across the inductor before and just after the switch is changed.


Analysis: The solution to this problem can be visualized by noting that before the switch is thrown the circuit is presumed to be in a DC steady state. Therefore, before the switch is thrown the inductor may be modeled as a short circuit such that the voltage across it is zero. Moreover, the current through the inductor before the switch is thrown is simply VS/RS. Immediately after the switch is thrown the current through the inductor must be this same value since the current through an inductor cannot change instanteneously. However, after the switch is thrown this current must also be the current drawn through R1 since that resistor and the inductor are then in series. Finally, the voltage across the inductor after the switch is thrown is always equal to the voltage across R1, which is simply the product of iL and R1. Thus, the voltage across the inductor immediately after the switch is thrown must be iL(t=0+) times R1. In quantitative terms, at t = 0-,

vL (0−) = 0

ARVi

S

SL 50

24.012)0( ===−

At t = 0+,

iL (0+) = iL (0−) = 50 A

and

vL (0+ ) = iL (0+)R1 = (50 A)(33kΩ) = 1.65MV

This high side of this voltage is located at the ground terminal due to the direction of current flow through R1. ANSWER: MVV 65.1,0 −



Problem 5.39

Solution: Known quantities: Circuit shown in Figure P5.27, .6,80,4,150,12 3211 Ω=Ω=Ω=== MRMRMRFCVV µ

Find: The time constant of the circuit for . 0>tAssumptions: The circuit is in steady-state conditions for 0<t .

Analysis: At , just after the switch is closed, a transient starts. Since this is a first order circuit (a single independent capacitance), the transient will be exponential with some time constant. That time constant is the product of the capacitance of the capacitor and the Thevenin equivalent resistance seen across the terminals of the capacitor.

+= 0t

To find the Thevenin equivalent resistance, suppress the independent, ideal voltage source which is equivalent to replacing it with a short circuit. Then with respect to the terminals of the capacitor:

Ω=+=+= MRRRRTH 4.82)64(80)( 312 The time constant is simply τ = RTH C = (82.4 MΩ)(150 µF) = 12360 sec = 206 minutes = 3.43 hours+ Notice that this value is independent of the magnitude of the voltage source. ANSWER: hrks 433.3min0.20636.2 ==

Problem 5.40

Solution: Known quantities: Circuit shown in Figure P5.21, VS =12 V,L = 100 mH , R1 = 400 Ω,R2 = 400Ω ,R3 = 600Ω .

Find: The time constant of the circuit for . t > 0

Assumptions: iL = 1.70 mA just before the switch is opened at t = 0.

Analysis: At , just after the switch is opened, a transient starts. Since this is a first order circuit (a single independent capacitance), the transient will be exponential with some time constant. That time constant is the product of the inductance of the inductor and the Thevenin equivalent resistance seen across the terminals of the inductor.

t = 0+

In this problem, the Thevenin equivalent resistance is particularly easy to find since there are no sources connected. With respect to the terminals of the inductor: RTH = (R1 + R2) R3 = 800 600 ≅ 343 Ω

τ =L

RTH=

0.1

343≅ 292 ns

The time constant is simply ANSWER: 291.7 ns



Problem 5.41

Solution: Known quantities: Circuit shown in Figure P5.41, .4,4,4,6,3,4,5,20 214321 AIFCFCRRRRVV SS ===Ω=Ω=Ω=Ω==

Find: a) The capacitor voltage at . )(tVC t = 0+

b) The time constant τ for . t ≥ 0

c) The expression for and sketch the function. VC (t)

d) Find for each of the following values of : VC (t) t0,τ,2τ ,5τ ,10τ .

Assumptions: Switch is always open and switch closes at S1 S2 t = 0.

Analysis: a) Without any power sources connected the steady state voltages are zero due to relentless dissipation of energy in the resistors.

VC (0−) = VC (0+) = 0V When the initial condition on a transient is zero, the general solution for the transient simplifies to

VC (t) = V (∞) 1− e−t /τ( )

b) The two capacitors in parallel can be combined into one 8 F equivalent capacitor. The Thevenin equivalent resistance seen by the 8 F capacitance is found by suppressing the current source (i.e. replacing it with an open circuit) and computing R2 + R3||R4. RTH = 4 + (3 6) = 6 Ω τ = RTH C = (6)(8) = 48 s c) The long-term steady state voltage across the capacitors is found by replacing them with DC open circuits and solving for the voltage across R3. This voltage is found readily by current division

VC (∞) =6Ω

3Ω + 6Ω(4 A)(3Ω) = 8V

Plug into the generalized solution given above to find

)

C

CC

C

0.8)10(;95.7)5(;9.6)2(

;06.5)( ;0)0(

===

==

τττ

VC (t) = 8 1− e−t / 48( ) , t ≥ 0

VC (t) = 0 , t ≤ 0d

VVVVVV

VVVVC τ



Problem 5.42

Solution: Known quantities: Circuit shown in Figure P5.41, VS = 20 V, R1 = 5Ω,R2 = 4 Ω ,R3 = 3Ω,R4 = 6Ω,C1 = 4 F ,C2 = 4 F, IS = 4 A.

Find: a) The capacitor voltage V at . C (t) t = 0+



d) Find for each of the following values of : VC (t) t 0,τ,2τ ,5τ ,10τ .

Assumptions: Switch has been open for a long time and closes at S1 t = 0; conversely, switch has been closed for a long time and opens at .

S2t = 0

Analysis: a) The capacitor voltage immediately after the switch is closed and the switch is opened is equal to that when the switches were the first opened and the second closed respectively. Since the second switch was closed for a long time one can assume that the capacitor voltage had reached a long-term steady state value. This value is found by replacing both capacitors with DC open circuits and solving for the voltage across R

S1 S2

3. This voltage is found readily by current division:

VAVV CC 8)3)(4(63

6)0()0( =ΩΩ+Ω

Ω== −+

b) The Thevenin equivalent resistance seen by the parallel capacitors is R1 || R2 + R3( ).

sCRTH 370)44()

341

51( 1 =+×

++== −τ

c) The generalized solution for the transient is

VC (t) = V (∞) +[V (0+) − V (∞)]e−t /τ

The long-term steady state voltage across the capacitors is found by replacing them with DC open circuits and solving for the voltage across R3. This voltage is found readily by voltage division. Thus,

VC (∞) =R2 + R3

R1 + R2 + R3VS =

4 Ω + 3Ω5Ω + 4 Ω + 3Ω

20V( )=35

3V

Plug in to the generalized solution given above to find

VC (t) = V (∞) + V (0+) − V (∞)[ ]e−t /τ =35

3+ 8 −

35

3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ e−3t / 70 =

35

3−

11

3e−3t / 70 , t ≥ 0

VC (t) = 8 , t ≤ 0

d)

VVVVVV

VVVV

C

CC

CC

666.11)10(; 642.11)5(; 170.11)2(

; 318.10)( ; 8)0(

===

==

τττ

τ



Problem 5.43


Find: a) The capacitor voltage at . )(tVC

+= 0t

b) The expression for and sketch the function. )(tVC

Assumptions: Switch is always open; switch has been closed for a long time, and opens at S2 S1 t = 0. At t = t1 = 3τ ,

switch closes again. 1SAnalysis: The approach here is to find the transient solution in the interval 0 < t < 3 seconds and use that solution to determine the initial condition (the capacitor voltage) for the new transient after the switch S1 closes again.

a) has been closed for a long time and in DC steady state the capacitors can be replaced with open circuits. Thus, by voltage division

1S

VC (0+) = VC (0−) =7

12× 20 = 11.67V

b) As mentioned above, to find the complete transient solution for t > 0 it is necessary to find the capacitor voltage when switch S1 closes at t = 3 seconds. To do so it is first necessary to find the complete transient solution for when the switch is open (i.e. as if the switch never closed again.) The long-term steady state capacitor voltage when the switch is held open is zero. The time constant is simply RTH CEQ = 56 seconds. Thus, the complete transient solution for the first 3 seconds is

VC (t) = V (0+)e−t / 56 = 11.67e−t / 56 , 0 ≤ t ≤ 3 At t = 3 seconds, the capacitor voltage is

VC (t = 3−) = 11.67e−3/ 56 = 11.06 At t=3 seconds the switch S1 closes again. Continuity of voltage across the capacitors still holds so

VC (t = 3+) = VC (t = 3−) = 11.06 With the switch closed the long-term steady state capacitor voltage is the same as that found in part a.

VC (∞) =7

12× 20 = 11.67V

The new time constant is found after suppressing the independent voltage source (i.e. replacing it with a short circuit) and finding the new Thevenin equivalent resistance seen by the capacitors. RTH = (4 + 3) 5 = 2.92 Ω and τ = RTHC = (2.92)(4 + 4) = 23.3 seconds Finally, the transient solution for t > 3 is

VC (t) = 11.67 + 11.06 −11.67[ ]e−t / 23.3 = 11.67 − 0.61e−(t−3)/ 23.3 , t ≥ 3 Notice the use of the shifted time scale (t-3) in the exponent.



Problem 5.44


Find: a) The capacitor voltage at . )(tVC t = 0+



d) Find for each of the following values of : VC (t) t0,τ,2τ ,5τ ,10τ .

Assumptions: Both switches and close at . S1 S2 t = 0

Analysis: a) Without any power sources connected the steady state voltages are zero due to the complete dissipation of all circuit energy by the resistors.

VC (0−) = VC (0+) = 0V When the initial condition on a transient is zero, the general solution for the transient simplifies to

VC (t) = V (∞) 1− e−t /τ( )

b) The two capacitors in parallel can be combined into one 8 F equivalent capacitor. The Thevenin equivalent resistance seen by the 8 F capacitance is found by suppressing the independent sources (i.e. by replacing the current source with an open circuit and the voltage source with a short circuit) and computing R1||(R2 + R3||R4).

RTH = 5 (4 + (3 6))[ ]= 5 6[ ]=30

11≅ 2.73Ω τ = RTH C =

30

118 =

240

11≅ 21.8 s

c) At this point only the long-term steady state capacitor voltage is needed to write down the complete transient solution. In DC steady state the capacitors can be modeled as open circuits. Furthermore, R3||R4 can be replaced with an equivalent resistance. This resistance is in parallel with the independent current source and the two can be replaced with a Thevenin source transformation of an appropriate voltage source in series with the same resistance. Once this replacement is made it is a simple matter of voltage division to determine the capacitor voltage. R3 R4 = 3 6 = 2 Ω The source transformation results in an 8V voltage source in series with this resistance. Then, by voltage division,

VC (∞) = 8 +4 + 2

4 + 2 + 5(20 − 8) =

160

11≅ 14.55 V

Now plug in to the general form of the transient solution to find

VC (t) ≅ 14.5 1− e−11t / 240[ ] , t ≥ 0

d) VC (0) = 0V; VC (τ ) = 9.17V ;

VC (2τ ) = 12.5V; VC (5τ ) = 14.4 V;

VC (10τ ) = 14.5V



Problem 5.45



b) The time constant τ for . 0 ≤ t ≤ 48s

c) The expression for valid for VC (t) 0 ≤ t ≤ 48s .

d) The time constant τ for . t > 48s

e) The expression for valid for . VC (t) t > 48s

f) Plot for all time. VC (t)

Assumptions: Switch opens at ; switch opens at S1 t = 0 S2 t = 48s .

Analysis: The approach here is to find the transient solution in the interval 0 < t < 48 seconds and use that solution to determine the initial condition (the capacitor voltage) for the new transient after the switch S2 opens. a) and have been closed for a long time and in DC steady state the capacitors can be replaced with open circuits. Thus, by node analysis and voltage division

S1 S2

VS − VC (0−)

R1=

VC (0−)R2

R2 R2 + R3 + R4( ) ⇒ VC (0−) =

VS R2 + R3 + R4( )R1 + R2 + R3 + R4( )

VC (0+) = VC (0−) =20 4 + 3+ 6( )5+ 4 + 3+ 6( )

=260

18V ≅ 14.4V

b) The two capacitors in parallel can be combined into one 8 F equivalent capacitor. The Thevenin equivalent resistance seen by the 8 F capacitance is found by suppressing the independent sources (i.e. by replacing the current source with an open circuit) and computing (R2 + R3||R4). RTH = R2 + R3 || R4( )= 4 + 3 || 6( )= 4 + 2 = 6Ω τ = RTH C1 + C2( )= 6 ⋅8 = 48 s

c) As mentioned above, to find the complete transient solution for t > 0 it is necessary to find the capacitor voltage when switch S2 opens at t = 48 s. To do so it is first necessary to find the complete transient solution for when only the switch S1 is open (i.e. as if the switch S2 never opens.). The generalized solution for the transient is

VC (t) = V (∞) +[V (0+) − V (∞)]e−t /τ

The long-term steady state voltage across the capacitors is found by replacing them with DC open circuits and solving for the voltage across R3. This voltage is found readily by current division. Thus,

VAVC 8)3)(4(63

6)( =ΩΩ+Ω

Ω=∞

Substitute into the generalized solution given above to find

VC (t) = V (∞) + V (0+) − V (∞)[ ]e−t /τ = 8 + 14.4 − 8( )e−t / 48 = 8 + 6.4e−t / 48 , 0 ≤ t ≤ 48

At t = 48 s, the capacitor voltage is

VC (t = 48−) = 8 + 6.4e−1 = 10.35V Continuity of voltage across the capacitors still holds so

VC (48+) = VC (48−) = 10.35V d) The two capacitors in parallel can be combined into one 8 F equivalent capacitor. When both the switches are opened, there are no independent sources connected to the circuit. Thus, the Thevenin equivalent resistance seen by the 8 F capacitance is found by computing (R2 + R3). 5.33 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.


RTH = R2 + R3 = 4 + 3 = 7Ω τ = RTH C1 + C2( )= 7 ⋅8 = 56 s

e) The generalized solution for the transient is

VC (t) = V (∞) +[V (t0+) − V (∞)]e−( t−t0 ) /τ

The long-term steady state capacitor voltage after the switch has been opened is zero since no independent sources are connected and all the initial energy in the circuit is eventually dissipated by the resistors. Thus, VC (∞) = 0V Substitute into the generalized solution given above to find

VC (t) = V (48+)e−t /τ = 10.35e−(t−48)/ 56 , t > 48 f) The plot of for all time is shown in the following figure. VC (t)



Problem 5.46



b) The time constant τ for . 0 ≤ t ≤ 96s

c) The expression for valid for VC (t) 0 ≤ t ≤ 96s .

d) The time constant τ for . st 96>e) The expression for valid for . )(tVC t > 96s

f) Plot for all time. VC (t)

Assumptions: Switch opens at ; switch opens at S1 t = 96s S2 t = 0.

Analysis: The approach here is to find the transient solution in the interval 0 < t < 96 seconds and use that solution to determine the initial condition (the capacitor voltage) for the new transient after the switch S1 opens. a) and have been closed for a long time and in DC steady state the capacitors can be replaced with open circuits. Thus, by node analysis and voltage division

S1 S2

( )4322

2

1

)0()0(RRRR

RVRVV CCS

++=

− −− ⇒

( )( )4321

432)0(RRRR

RRRVV SC +++

++=−

( )( ) VVVV CC 4.14

18260

634563420)0()0( ≅=

+++++

== −+

b) The two capacitors in parallel can be combined into one 8 F equivalent capacitor. The Thevenin equivalent resistance seen by the 8 F capacitance is found by suppressing the independent sources (i.e. by replacing the current

source with an open circuit) and computing R1 R2 + R3( ). RTH = R1 R2 + R3( )= 5 4 + 3( )= 2.92Ω τ = RTH C1 + C2( )= 2.92 ⋅8 = 23.3s

c) As mentioned above, to find the complete transient solution for t > 0 it is necessary to find the capacitor voltage when switch S1 opens at t = 96 s. To do so it is first necessary to find the complete transient solution for when only the switch S2 is open (i.e. as if the switch S1 never opens.). The generalized solution for the transient is

VC (t) = V (∞) +[V (0+) − V (∞)]e−t /τ

The long-term steady state voltage across the capacitors is found by replacing them with DC open circuits and solving for the voltage across and . This voltage is found readily by voltage division. Thus, R2 R3

VVVC 67.11)20(345

34)( =Ω+Ω+Ω

Ω+Ω=∞


[ ] ( )960,73.267.11)(

67.114.1467.11)()0()()(3.23/

3.23//

≤≤+=

−+=∞−+∞=−

−−+

tetV

eeVVVtVt

C

ttC

τ

At t = 96 s, the capacitor voltage is

VC (t = 96−) = 11.67 + 2.73e−96 / 23.3 = 11.71V Continuity of voltage across the capacitors still holds so



VVV CC 71.11)96()96( == −+

d) The two capacitors in parallel can be combined into one 8 F equivalent capacitor. When both the switches are opened, there are no independent sources connected to the circuit. Thus, the Thevenin equivalent resistance seen by the 8 F capacitance is found by computing (R2 + R3). RTH = R2 + R3 = 4 + 3 = 7Ω τ = RTH C1 + C2( )= 7 ⋅8 = 56 s

e) The generalized solution for the transient is

VC (t) = V (∞) +[V (t0+) − V (∞)]e−( t−t0 ) /τ

The long-term steady state capacitor voltage after the switch has been opened is zero since no independent sources are connected and all the initial energy in the circuit is eventually dissipated by the resistors. Thus, VC (∞) = 0V Plug in to the generalized solution given above to find

VC (t) = V (96+)e−t /τ = 11.71e−(t−96)/ 56 , t > 96 f) The plot of for all time is shown in the following figure. VC (t)



Problem 5.47

Solution: Known quantities: RS =15kΩ , τ =1.5ms, τ ' =10 ms, R3 = 30 kΩ , C = 1µF .

Find: The value of resistors and . R1 R2

Assumptions: None.

Analysis: Before the switch opens:

Req = RS //R1 //R2 //R3

τ = ReqC = 1.5ms

After the switch opens:

Req

' = RS //R1

τ ' = Req' C = 10 ms

Solving the system of equations we have,

R1 =RSτ '

RSC −τ '=

15000 ⋅ 0.01

15000 ⋅10−6 − 0.01= 30kΩ

R2 =C

τ−

1

RS−

1

R1−

1

R3

⎛

⎝ ⎜

⎞

⎠ ⎟ −1

=10−6

0.0015−

1

15000−

1

30000−

1

30000

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

−1

= 1875Ω

Problem 5.48

Solution: Known quantities: Circuit shown in Figure P5.47, VS = 100V , RS = 4 kΩ, R1 = 2 kΩ,R2 = R3 = 6 kΩ,C = 1µF.

Find: The value of the voltage across the capacitor after t = 2.666 ms.

Assumptions: None.

Analysis: Before opening, the switch has been closed for a long time. Thus we have a steady-state condition, and we treat the capacitor as an open circuit. The voltage across the capacitor is equal to the voltage across the resistance R1. Thus,

VC (t0−) = VC (t0

+) =RS || R1 || R2 || R3

RSVS =

300

13V ≅ 23.077V

After the switch opens, the time constant of the circuit is

Req = RS //R1 =4000

3Ω τ = ReqC =

1

750ms ≅ 1.3ms

the generalized solution for the transient is VC (t) = V (∞) + [V (t0+) − V (∞)]e−( t−t0 ) /τ

The long-term steady state voltage across the capacitors is found by replacing them with DC open circuits and solving for the voltage across R1. This voltage is found readily by voltage division. Thus,

VC (∞) =R1

RS + R1VS =

2000Ω4000Ω + 2000Ω

20V =20

3V ≅ 6.67V




VC (t) = V (∞) + [V (t0+) − V (∞)]e−( t−t0 ) /τ =

20

3+

300

13−

20

3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ e−750(t−t0 ) =

20

3+

640

39e−750(t−t0 )

Finally,

VC (t0 + 2.666ms) =20

3+

640

39e−750(0.002666) = 8.888V

Problem 5.49

Solution: Known quantities: As described in Figure P5.49.

Find: The time at which the current through the inductor is equal to 5 A, and the expression for for . )(tiL 0≥t

Assumptions: None.

Analysis: At : Using the current divider rule:

0<t

( ) mAiL 5.66)5.25

5)(5.2//51000

100(0 =++

=−

At : 0>tUsing the current divider rule:

( ) AiL 71.5)5.25

5)(5.2//510

100( =++

=∞

To find the time constant for the circuit we must find the Thevenin resistance seen by the inductor:

Req = 10 //5+ 2.5 = 5.83Ω τ =L

Req=

0.1

5.83= 17.1ms

Finally, we can write the solution:

iL t( )= iL ∞( )− (iL ∞( )− iL 0( ))e− tτ = 5.71− (5.71− 0.0665)e

− t17.1×10−3

= 5.71− 5.64e− t

17.1×10−3A

Solving the equation we have,

( ) 564.571.5ˆ 3101.17ˆ

=−=−×

− tL eti ⇒ ˆ t = 35.437ms

The plot of for all time is shown in the figure. iL (t)



Problem 5.50


Find: The expression for for . The maximum voltage between the contacts during the 5-ms duration of the switch.

iL (t) 0 ≤ t ≤ 5ms

Assumptions: The mechanical switching action requires 5 ms.

Analysis: a) For t < 0: Using the current divider rule:

( ) mAiL 5.66)5.25

5)(5.2//51000

100(0 =++

=−

For : 0 ≤ t ≤ 5msThe long term steady state inductor current after the switch has been opened is zero since no independent source is connected to the circuit and all the initial energy in the circuit is eventually dissipated by the resistors. Thus,

iL ∞( )= 0 A

To find the time constant for the circuit we must find the Thevenin resistance seen by the inductor: Req = 5+ 2.5 = 7.5Ω

τ =L

Req=

0.1

7.5= 13.33ms

Finally, we can write the solution:

iL t( )= iL 0( )e− tτ = 0.0665( )e

− t13.33×10−3

A 0 ≤ t ≤ 5ms( )

b) The voltage between the contacts during the 5-ms duration of the switching is equal to: Vcont (t) = VS − V5Ω = 100 − VL + 2.5iL( ) where,

VL (t) = LdiL (t)

dt= −

(0.1)(0.0665)

13.33×10−3e

− t13.33×10−3

= −0.5( )e− t

13.33×10−3V

Therefore,

Vcont (t) = 100 − (2.5)(0.0665) − 0.5[ ]e− t

13.33×10−3= 100 + (0.33)e

− t13.33×10−3

V Thus, the maximum voltage between the contacts during the 5-ms duration of the switch is:

VcontMAX = Vcont (t = 0) = 100.33V



Problem 5.51

Solution: Known quantities: As described in Figure P5.51. The switch closes when

the voltage across the capacitor voltage reaches ;

The switch opens when the voltage across the capacitor

voltage reaches

CMv

vMO = 1V . The period of the capacitor

voltage waveform is 200 ms.

Find: The voltage vM

C .

Assumptions: The initial capacitor voltage is and the switch has just opened. 1V

Analysis: With the switch open: VC ∞( )= 10V

τ = RC = 0.15s

VC t( )= VC ∞( )− VC ∞( )− VC 0( )[ ]e− tτ = 10 − (10 −1)e

− t0.15 = 10 − 9e

− t0.15 V

Now we must determine the time when VC (t) = vMC

. Using the expression for the capacitor voltage:

vMC = 10 − 9e

−t00.15 ⇒ e

−t00.15 =

10 − vMC

9 ⇒ t0 = −0.15ln

10 − vMC

9

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

With the switch closed, the capacitor sees the Thevenin equivalent defined by:

msCRkRVVV eqeqCeq 15.0 101010 division) (voltage10110100001010)( 2 ==Ω≅ΩΩ=×≈×

+=∞= − τ

The initial value of this part of the transient is vMC

at t = t0. With these values we can write the expression for the capacitor voltage:

VC t( )= VC ∞( )− VC ∞( )− VC t0( )[ ]e−(t−t0 )τ = 0.01+ vM

C − 0.01( )e−(t−t0 )0.15×10−3

V

The end of one full cycle of the waveform across the 10-Ω resistor occurs when the second transient reaches

vMO = 1V . If we call the time at which this event occurs t1,

then: VC t( )= 1V at t = t1 = 200ms

and so 1 = 0.01+ vMC − 0.01( )e−(t1 −t0 )

0.15×10−3

Graphically, the solution is the intersection between the following two functions:

vMC = 10 − 9e

−t00.15 (blue line)

vMC = 0.01+

1− 0.01( )

e−(0.2−t0 )

0.15×10−3

(red line)

which correspond to vMC = 7.627 V and t0 ≅ 0.1997ms.



Problem 5.52

Solution: Known quantities: As describes in Figure P5.52. At , the switch closes. t = 0

Find: a) for . iL (t) t ≥ 0b) V for . L1(t) t ≥ 0

Assumptions: iL (0) = 0 A .

Analysis: a) In the long-term DC steady state after the switch is closed the inductors may be modeled as short circuits and so all of the current from the source will travel through the inductors. iL ∞( )= 5 A

With the current source suppressed (treated as an open circuit) the Thevenin equivalent resistance seen by the inductors in series and the associated time constant are

Req = 10 kΩ

Leq = L1 + L2 = 6 H

τ =Leq

Req= 0.6 ms

iL (t) = iL (∞) 1− e− t

τ⎡

⎣ ⎢

⎤

⎦ ⎥ = 5 1− e

− t0.6×10−3

⎡

⎣ ⎢

⎤

⎦ ⎥ A

b) The voltage across either of the inductors is derived directly from the differential relationship between current and voltage for an inductor.

VL1(t) = L1

diL (t)

dt= (1)(5)

d

dt(1− e

− t0.6×10−3

) = 51

0.6 ×10−3e

− t0.6×10−3

⎛

⎝ ⎜

⎞

⎠ ⎟ = 8.333e

− t0.6×10−3

kV

Problem 5.53

Solution: Known quantities: As describes in Figure P5.52. At , the switch closes. t = 0

Find: The voltage across the 10-kΩ resistor in parallel with the switch for t ≥ 0.

Assumptions: None.

Analysis: When the switch closes at t = 0, the 10-kΩ resistor is in parallel with a short circuit, so its voltage is equal to zero for all time (t ≥0).



Problem 5.54

Solution: Known quantities: As describes in Figure P5.54.

Find: The heat capacity of the burner, if the burner reaches 90 percent of the desired temperature in 10 seconds.

Analysis: It is a first order dynamic system. We just need to calculate the time constant and related with the value of capacitor.

SSSBurner C.CR 51==τ , SSCSS IRV = ( ) 00 =CV ( ) ( )-t/τSSC - eIRtV 1=

So, , -t/τ-e. 190 = 32.τt

−=− , 325110 .C. S

=

So, F92 .CS =

Problem 5.55


Find: a) Final temperature of the water in the pot. b) Time takes for the water to reach 80 percent of its final temperature.

Assumptions: PS CC <<

Analysis: a) As t→∞, the capacitors become open circuits, and we can compute an equivalent circuit, as shown below.

Voc

RS RL

RpVpss

+

-

+-

RT

VT Cp

+-

In the circuit above V51125175 ..RIV SSOC =×==

Therefore,

V6.585252805.1

5112=

++=

++= .

...R

RRRVV P

PLS

OCPSS

(b) The Thèvenin equivalent seen by the capacitance, Cp, is

shown:

In the circuit above ( ) Ω21 .||RpRRR LST =+=

and the time constant for this circuit is: s96 CRτ pT ==′

To find the 80% time we set and solve for t: -t/τ-e. 180 =s155 t =



Problem 5.56


Find: The range of variable resistor.

Analysis: When , µA100 iL =

100001000100010100

10001000 6 RRv -

C +

=×××+

=

A general expression for the capacitor current is: ( ) ( )-t/τC - e

RRtV 110

1000200001000

××++

+=

At the time the alarm sounds, we have

( )-t/τC - e

RRRv 110

210001000

100001000

××+

+=

+= , -t/τe- R- =⎟

⎠⎞

⎜⎝⎛ + 1

10000021000

The time constant is given by the expression ( )[ ]200001000µF100 ||R +=τ

Substituting the expression for the time constant into the above equation we have one equation in one unknown, R.

This is a transcendental equation, and can be solved by iteration or by graphical analysis. the solution is that R

must be between 33,580 Ω and 53,510 Ω or Ω≤≤Ω 53,510 R 33,580

Problem 5.57


Find: The voltage across the . 1C

Analysis: Since both two resistance and two capacitors are series connected, we can first calculate the equivalent capacity and

equivalent resistance, and then calculate the time constant.

µF3

10µF1550

21

21 ==+

=CC

CCCeq ( ) µs7663

10119 .CRτ eqeq =+==

( ) ( ) A50191

100 . Ω

Vi =+

= ( ) ( ) µS766500 .- t/- t/τ e.eit i == ( ) ( )∫= λλ diC

tVC1

( ) ( )[ ]VµS766

0µS766µS766

1

676676

50µF5

µS76650µF51

.t/

t./t

o

.λ/C

e.- .

e..dλe.tV

−

−−

=

=∫= λ



Problem 5.58

Ω

Ω 1kΩv

S+-

5

5 Ω 20

1

Ω

2.5 Ω

1 Fµ

4k Ω

Solution: Known quantities: As describes in Figure P5.58

Find: a) time constant for 9<t<10 s b) time constant for t>10 s

Analysis: a) With the switch open we must consider the following circuit.

To find the time constant for this circuit we must find the Thèvenin

resistance seen by the capacitor:

( ) Ω5005120555240001000 ||||.RT =++++=

( )( ) ms0055µF15005 . CRτ T =Ω==

b) With the switch closed, the 1-kΩ and 4-kΩ resistors are not included since there is a short circuit across them

with the switch closed. The resulting circuit is shown

below.

Ω

ΩvS

+-

5

5Ω 20

1

Ω 1 F

2.5Ω

µNow the Thèvenin resistance seen by the

capacitor is:

( ) Ω51205552 ||||.RT =++=

and the time constant is,

( )( ) µs5µF15 CRτ T =Ω==



Problem 5.59

Solution: Known quantities: As describes in Figure P5.59

Find: a) Time to wait for 99% recharge b) Energy delivered to flush during 1/30 seconds. c) Energy delivered to flush when not fully recharged.

Analysis: a) ( ) at/τ

BattBattC eVVtV −−=

( )( ) s51µF1500Ω1000 .RCτa ===

( ) aready /τtreadyC etV −−=×= 5.75.75.799.0

( ) s96010log51 ...-t eready ==

b) We must find an expression for the voltage across the flash bulb (R2) after the shutter button has closed. This

is the circuit:

( ) ms51s0015010001 . .C||C Rτ T =≈==

( ) ( ) ( ) ( ) ( )( )

( ) ( )

mV4925.75.7

575711000

15711000

1

ms51

ms51

.tt

.tt

τtt

flashCCCCfalsh

flash

flash

flash

e

e...

etVVVtVtV

−−

−−

−−

+=

⎟⎠⎞

⎜⎝⎛ −

+−

+=

−∞−∞==

( ) ( )2RtVti C

flash =

vflash(t)

iflash(t)

1K

1500 F 1Ωµ 7.5V

+

-

7.5=VC(tflash

)

+

-

( ) ( ) mJ27.424925.75.74925.75.7301

0

ms51ms51301

=∫⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+=∫=

−−+

λλλλλλ

deediVW ..t

tflashc

flash

flash

c) In this case the capacitor has not fully charged but has achieved the value of

( ) V48.65.75.7s3 s513

=−==−

.s

C etV

after the shutter switch closes at t = 3 s.

The voltage becomes: flashV

( ) mV4775.65.7 ms513

.t

falsh etV−

−+=

mJ6.314775.65.74775.65.7301

0

ms51ms51 =∫⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+=

−−λ

λλ

deeW ..



Problem 5.60 Known quantities: As describes in Figure P5.60

Find: Voltage-time curve

Analysis: Applying KCL : vo

R+

1

Lvodt∫ = iS t( )

Differentiating, we have

oo v

LR

dtdv

−=

Solve it, we can have

( ) ( )( )

( ) ( )00 10

00tt

ott

LR

oo etvetvtv −−−−==

region I: t < 0 ( ) 00 =tv

region II: 0< t < 0.8 s

Since the current through an inductor cannot change instantaneously

( ) ( ) ( ) V50mA10000 ====== tv,ti,ti oRL

( ) V5 10toII etv −=

( ) ( ) ( ) ( ) ( ) mA1010 mA,10 1010 etititi eR

tvti tRIISLII

toIIRII

−− −=−===

region III: 0.8 < t < 0.9 sec

( ) ( ) ( ) ( ) V08.008.08.0mA108.0 ======== −−+− tv,ti,titi oRLL

( ) ( ) ( )( ) ( ) V68.0mA,128.08.08.0 −==−==+=−== ++++ tvtititi oIIISIIILRIII

( ) ( )V6 8.010 −−−= toIII etv

region IV: 0.9 < t < 1.3 sec

( ) ( ) ( ) ( )+−−− ====−==−== 9.0mA4.29.0mA4.49.0 V,2.29.0 tit,ititv LLRo

( ) ( ) ( )( ) ( ) V2.09.0mA,4.09.09.09.0 −==−==+=−== ++++ tvtititi oIVSIVLRIV

( ) ( )V2.0 9.010 −−−= toIV etv

region V: 1.3 < t < 1.4 ec s

( ) ( ) ( ) ( )−−−− ======== 3.1mA23.103.1 V,03.1 tit,ititv LLRo

( ) ( ) ( )( ) ( ) V13.1mA,23.13.13.1 −==−==+=−== ++++ tvtititi oVSVLRV

( ) ( )V3.110 −−−= toV etv

region VI: t > 1.4 sec

( ) ( ) ( ) ( )−−−− ====−==−== 4.1mA74.04.1mA74.04.1 V,37.04.1 tit,ititv LLRo

( ) ( ) ( ) ( ) V68.44.1mA,36.94.14.14.1 ====−=== ++++ tvtititi oVILSVIRVI

( ) ( )V68.4 4.110 −−= toV etv



Section 5.5: Transient response of Second-Order Circuits

Focus on Methodology – roots of second order systems Case 1: Real and distinct roots. This case occurs when ζ>1, since the term under the square root is

positive in this case, and the roots are: s1,2 = −ζωn ± ωn ζ 2 −1 . This leads to an

overdamped response. Case 2: Real and repeated roots. This case holds when ζ=1, since the term under the square root is

zero in this case, and s1,2 = −ζωn = −ωn . This leads to a critically damped response.

Case 3: Complex conjugate roots. This case holds when ζ<1, since the term under the square root is

negative in this case, and s1,2 = −ζωn ± jωn 1−ζ 2 . This leads to an underdamped response.

Focus on Methodology

Second-order transient response 1. Solve for the steady-state response of the circuit before the switch changes state (t = 0-), and after

the transient has died out (t → ∞). We shall generally refer to these responses as x(0-) and x(∞).

2. Identify the initial conditions for the circuit, x(0+), and Ý x (0+) using continuity of capacitor voltages and inductor currents (vC(0+) = vC(0-), iL(0+) = iL(0-)), and circuit analysis. This will be illustrated by examples.

3. Write the differential equation of the circuit for t = 0+, that is, immediately after the switch has changed position. The variable x(t) in the differential equation will be either a capacitor voltage, vC(t), or an inductor current, iL(t). Reduce this equation to standard form (Equation 5.9, or 5.48).

4. Solve for the parameters of the second-order circuit: ωn and ζ . 5. Write the complete solution for the circuit in one of the three forms given below, as appropriate:

Overdamped case (ζ > 1):

x(t) = xN (t) + xF (t) = α1e−ζω n +ω n ζ 2 −1⎛

⎝ ⎜ ⎞

⎠ ⎟ t

+α2e−ζω n −ω n ζ 2 −1⎛

⎝ ⎜ ⎞

⎠ ⎟ t

+ x(∞) t ≥ 0

Critically damped case (ζ = 1):

x(t) = xN (t) + xF (t) = α1e−ζω n( )t +α2te −ζω n( )t + x(∞) t ≥ 0

Underdamped case (ζ = 1):

x(t) = xN (t) + xF (t) = α1e−ζω n + jω n 1−ζ 2⎛

⎝ ⎜ ⎞

⎠ ⎟ t

+α2e−ζω n − jω n 1−ζ 2⎛

⎝ ⎜ ⎞

⎠ ⎟ t

+ x(∞) t ≥ 0

6. Apply the initial conditions to solve for the constants α1 and α2.



Problem 5.61


.35.0,17,700,1.1,290,130,9,15 212121 FCmHLRkRRRVVVV SSSS µ==Ω=Ω=Ω=Ω===

Find: The voltage across the capacitor and the current through the inductor and as approaches infinity. RS2 tAssumptions: The circuit is in DC steady-state conditions for 0<t .

Analysis: The conditions that exist at have no effect on the long-term DC steady state conditions at . In the long-term DC steady state the inductor may be modeled as a short circuit and the capacitor as an open circuit. In this case, the inductor short circuits the R

0<t ∞→t

1 branch and the R2 C branch. Thus, the voltage across these branches and the current through them are zero. In other words all of the current produced by the 9V source travels through the inductor in this case.

iL (∞) =VS2

RS2=

9

290= 31.03mA

Of course, this current is also the current traveling through the 290 Ω resistor. iRS2 (∞) = iL (∞) = 31.03mA

And since the voltage across the inductor in the long-term DC steady state is zero (short circuit) 0 + VC (∞) + i R2 (∞)R2 = 0

VC (∞) = 0

Problem 5.62

Solution: Known quantities: Circuit shown in Figure P5.61, VS1 =12 V, VS2 =12 V, RS1 = 50 Ω,RS2 = 50 Ω,R1 = 2.2 kΩ,R2 = 600Ω,L = 7.8 mH ,C = 68 µF.

Find: The voltage across the capacitor and the current through the inductor as approaches infinity. tAssumptions: The circuit is in DC steady-state conditions for 0<t .

Analysis: The conditions that exist at have no effect on the long-term DC steady state conditions at 0<t ∞→t . In the long-term DC steady state the inductor may be modeled as a short circuit and the capacitor as an open circuit. In this case, the inductor short circuits the R1 branch and the R2 C branch. Thus, the voltage across these branches and the current through them are zero. In other words all of the current produced by the 12V source travels through the inductor in this case.

iL (∞) =VS2

RS2=

12

50= 240 mA

Of course, this current is also the current traveling through the 290 Ω resistor. iRS2 (∞) = iL (∞) = 240 mA

And since the voltage across the inductor in the long-term DC steady state is zero (short circuit) 0)( 0)()(0 22 =∞=∞+∞+ CRC VRiV



Problem 5.63

Solution: Known quantities: Circuit shown in Figure P5.63, VS =170 V,RS = 7 kΩ, R1 = 2.3kΩ,R2 = 7 KΩ,L = 30 mH ,C = 130 µF.

Find: The current through the inductor and the voltage across the capacitor and R1 at steady state.

Assumptions: None.

Analysis: As , the circuit will return to DC steady state conditions. In the long-term DC steady state the inductor may be modeled as a short circuit and the capacitor as an open circuit. Therefore, in this case, the current through R

t → ∞2 is

zero and thus the voltage across the capacitor must be equal to the voltage across R1. Furthermore, the current through the inductor and R1 is simply VS/(RS + R1).

iC (∞) = 0 iL (∞) = i R1 (∞) =VS

RS + R1=

170

7000 + 2300≅ 18.3 mA

VR1 (∞)= iL (∞)R1 = 18.28 ×10−3 × 2.3×103 = 42.04 V

and VC (∞)= VR1 (∞) = 42.04 V

Problem 5.64

Solution: Known quantities: Circuit shown in Figure P5.64, VS =12 V,C = 130 µF, R1 = 2.3kΩ,R2 = 7 KΩ,L = 30 mH .

Find: The current through the inductor and the voltage across the capacitor and at steady state. 1R

Assumptions: None.

Analysis: As ∞→t , the circuit will return to DC steady state conditions (practically after about 5 time constants.) In the long-term DC steady state the inductor may be modeled as a short circuit and the capacitor as an open circuit. Therefore, in this case, the voltage across the capacitor must be equal to the voltage across R2. Furthermore, the current through the inductor and R2 is simply VS/(R1 + R2). iC (∞) = 0 VL (∞) = 0

iL (∞)= iR 2 (∞) =VS

R1 + R2=

12

9.3×103= 1.29 mA

VR1 (∞)= iS (∞)R1 = 1.29 ×10−3 × 2.3×103 = 2.968V

And by observation

VC (∞)=VR2(∞)= (1.29 ×10−3)(7 ×103) = 9.03V

All answers are positive indicating that the directions of the currents and polarities of the voltages assumed initially are correct. (You did do that, didn't you?)



Problem 5.65

Solution: Known quantities: Circuit shown in Figure P5.65, VS =12 V,C = 0.5µF, R1 = 31kΩ,R2 = 22 KΩ,L = 0.9 mH .

Find: The current through the inductor and the voltage across the capacitor at steady state.

Assumptions: None.

Analysis: As , the circuit will return to DC steady state conditions (practically after about 5 time constants). In the long-term DC steady state the inductor may be modeled as a short circuit and the capacitor as an open circuit. Therefore, in this case, the voltage across the capacitor must be equal to the voltage across R

t → ∞

2. Furthermore, the current through the inductor and R2 is simply VS/(R1 + R2). iC (∞) = 0 VL (∞) = 0

iL (∞)= iR 2 (∞) =VS

R1 + R2=

12

(31+ 22) ×103≅ 226 µA

VR2(∞)= iR2

(∞)R2 = (226 ×10−3)(22 ×103) ≅ 4.98 V

And by observation VC (∞)=VR2(∞)= 4.98 V

Theoretically, when the switch is OPENED, the current through the inductor must continue to flow, at least momentarily. However, the inductor is in series with an OPEN switch through which current CANNOT flow. What the theory does not predict is that a very large voltage is developed across the gap and this causes an arc with a current (kind of like a teeny, weeny lightning bolt). The energy stored in the magnetic field of the inductor is rapidly dissipated in the arc. The same effect will be important later when discussing transistors as switches.

Problem 5.66

Solution: Known quantities: Circuit shown in Figure P5.66, VS =12 V,C = 3300 µF,R1 = 9.1kΩ,R2 = 4.3kΩ,R1 = 4.3kΩ,L = 16 mH ,.

Find: The initial voltage across just after the switch is changed.

R2

Assumptions: At the circuit is at steady state and the voltage across the capacitor is .

0<t+7V

Analysis: It is important to remember that only the values of the capacitor voltage and the inductor current are guaranteed continuity from immediately before the switch is thrown to immediately afterward. Therefore, to determine the initial voltage across R2 it is necessary to first determine the initial voltage across the capacitor and the initial current through the inductor. Assume that before the switch was thrown DC steady state conditions existed. In DC steady state the inductor may be modeled as a short circuit and the capacitor as an open circuit. The initial voltage across the capacitor is given as +7V. The initial current through the inductor is equal to the current through R3, which is given by Ohm’s Law.



iL (0+ )= iL (0− )=VS

R3=

12

4.3×103= 2.791mA and VC 0+( )= VC 0−( )= 7V

Apply KCL:

VR2 (0+) − VC (0+)

R1+

VR2 (0+)

R2+ iL (0+ ) = 0

VR2 (0+) =

VC (0+)

R1− iL (0+)

1

R1+

1

R2

=(VC (0+ ) − iL (0+)R1)R2

R2 + R1=

(7 − 2.791×10−3 × 9.1×103) × 4.3×103

4.3×103 + 9.1×103= −5.93V

One could also solve for VR2 by superposition.

VR2 (0+) =R2

R1 + R2(7V ) −

R1R2

R1 + R2(2.8mA) = −5.93 V

Problem 5.67

Solution: Known quantities: Circuit shown in Figure P5.67, VS1 =15 V, VS2 = 9 V, RS1 =130 Ω,RS2 = 290 Ω,R1 = 1.1kΩ,R2 = 700Ω,L = 17 mH ,C = 0.35µF.

Find: The current through and the voltage across the inductor

and the capacitor and the current through at RS2 t = 0+ .

Assumptions: The circuit is in DC steady-state conditions for t < 0.

Analysis: Since this was not done in the specifications above, you must note on the circuit the assumed polarities of voltages and directions of currents. At : t = 0−

Assume that steady state conditions exist. At steady state the inductor is modeled as a short circuit and the capacitor as an open circuit. Choose a ground. Note that because the inductor is modeled as a short circuit, there is no voltage drop from the top node to the bottom node and so before the switch there is no current through R1.

VL 0−( )= 0 iC 0−( )= 0

Apply KCL:

0 − VS1

RS1+ iL (0−) +

0

R1+ 0 +

0 − VS2

RS2= 0

iL (0−) =VS1

RS1+

VS2

RS2=

15

130+

9

290= 146.4 mA

Apply KVL: VC 0−( )+ iC 0−( )R2 = 0

VC 0−( )= 0

At : += 0t 5.51 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.


iL 0+( )= iL 0−( )= 146.4 mA

VC 0+( )= VC 0−( )= 0

Apply KVL:

−VL (0+ ) + 0 + iC (0+ )R2 = 0

iC (0+ )=VL (0+ )

R2

Apply KCL:

iL (0+) +VL (0+)

R1+

VL (0+)

R2+

VL (0+) − VS2

RS2= 0

VL (0+) =

VS2

RS2− iL (0+)

1

R1+

1

R2+

1

RS2

=VS2 − iL (0+)RS2

RS2

R1+

RS2

R2+1

=9 −146.4 ×10−3 × 0.29 ×103

0.29

1.1+

0.29

0.7+1

= −19.94 V

iC (0+) =VL (0+ )

R2=

−19.94

0.7 ×103= −28.49 mA

Apply KVL again:

−VL (0+ ) + iRS2 (0+ )RS2 + VS2 = 0

iRS2 (0+ )=VL (0+) − VS2

RS2=

−19.94 − 9

0.29 ×103= −99.79 mA

Problem 5.68

Solution: Known quantities: Circuit shown in Figure P5.67, VS1 =12 V, VS2 =12 V, RS1 = 50 Ω,RS2 = 50 Ω,R1 = 2.2 kΩ,R2 = 600Ω,L = 7.8 mH ,C = 68 µF.

Find: The voltage across the capacitor and the current through the inductor as approaches infinity. tAssumptions: The circuit is in DC steady-state conditions for t < 0.

Analysis: The conditions that exist at have no effect on steady state conditions as t < 0 t → ∞. In the long-term DC steady state the inductor may be modeled as a short circuit and the capacitor as an open circuit. In this case, the inductor short circuits the R1 branch and the R2 C branch. Thus, the voltage across these branches and the current through them are zero. In other words all of the current produced by the 12V source travels through the inductor in this case.

Apply KCL;

iL (∞) + iR1(∞)+ iR2 (∞) +0 − VS2

RS2= 0

iR1(∞) = iR2 (∞) = 0

iL (∞) =VS2

RS2=

12

50= 240 mA

Apply KVL: 0)( 0)()(0 22 =∞=∞+∞+ CRC VRiV



Problem 5.69


Find: An expression for the inductor current for . t ≥ 0

Assumptions: The switch has been closed for a long time. It is suddenly opened at and then reclosed at t = 0 t = 5s

Analysis: For : 0 ≤ t ≤ 5Define clockwise mesh currents, in the lower loop, and 0=t 0=t in the upper loop. Then the mesh equations are: 03)55( 21 =−+ IIs

−3I1 + (3+1

4s)I2 = 0

from which we determine that

(5+ 5s)(3+

1

4s) − 9 = 0

s = 0.242 ± j0.158

Therefore, the inductor current is of the form:

i (t) = e−0.242t[ A cos(0.158t) + B sin(0.159t)]From the initial conditions:

i (0) =6

3= 2 = A

Ldi

dt|t=0= 5

di

dt|t=0= VC (0+) = −10

⇒di

dt|t=0= −2 = −0.242A + 0.158B

Solving the above equations: A = 2 B = −9.59

i (t) = e−0.242t[2 cos(0.158t) − 9.59sin(0.158t)] A for 0 ≤ t ≤ 5s

The solution for capacitor voltage will have the same form.

VC (t) = e−0.242t[ A cos(0.158t) + B sin(0.158t)]V From the initial conditions:

VC (0) = 6 = A

dVC

dt|t=0=

1

CiC (0) = 0⇒− 0.242A + 0.158B = 0

Solving the above equations:

A = 6 B = 9.18

VC (t) = e−0.242t[6 cos(0.158t) + 9.18sin(0.158t)]V for 0 ≤ t ≤ 5s

From the above results:

VC (5) = 3.2807V

i (5) = −1.641 AThese are the initial conditions for the solution after the switch recloses. For : t ≥ 5



The mesh equations are:

(3+ 5s)I1 − 3I2 =6

s+ 5i (5)

−3I1 + (3+1

4s)I2 = −

VC (5)

s

from which we determine that

60s2 + 5s + 3 = 0

s = 0.041± j0.220Therefore, the inductor current is of the form:

i (t) = 2 + e−0.041tA cos[0.220(t − 5)]+ B sin[0.220(t − 5)]From the initial conditions: 2 + A = −1.641⇒ A = −3.641

di

dt|t=5=

VL (5)

5=

6 − 238

5= 0.543

⇒− 0.41A + 0.220B = 0.543

Solving the above equations: A = −3.641 B = 1.77

i (t) = 2 + e−0.041t−3.641cos[0.220(t − 5) +1.77sin[0.220(t − 5)] A for t ≥ 5s

This, together with the previous result, gives the complete solution to the problem.

Problem 5.70


Find: Determine if the circuit is underdamped or overdamped. The capacitor value that results in critical damping.

Assumptions: The circuit initially stores no energy. The switch is closed at 0=t .

Analysis: a) For t ≥ 0: The characteristic polynomial is:

Ls2 + Rs +1

C= 0

The damping ratio is:

12.0102104001

210

8<=

⋅==

−

LCRCζ

The system is underdamped, in fact we have the following complex conjugate roots:

s1,2 = −2 ×104 ± j(9.79 ×104 )

b) The capacitor value that results in critical damping is:

112

==LC

RCζ ⇒ 1100200 =C

C

1100200 22 =C

C ⇒ FFC µ25.01041

6 =×

=



Problem 5.71


Find: a) The capacitor voltage as t approaches infinity b) The capacitor voltage after 20 µs c) The maximum capacitor voltage.

Assumptions: The circuit initially stores no energy. The switch is closed at 0=t .

Analysis: For : 0≥tThe characteristic polynomial is:

0.01s2 + 400s +108 = 0

s = −2 ×104 ± j (9.79 ×104 )Therefore, the solution is of the form:

VC (t) = 10 + e−2×104 t[ A cos(9.79 ×104 t) + B sin(9.79 ×104 t)]From the initial conditions: 10 + A = 0

−2 ×104 A + 9.79 ×104 B = 0

Solving the above equations:

VttetV

BAt

C )]1079.9sin(04.2)1079.9cos(10[10)(

04.21044102 4

×−×−+=

−=−=×−

a) The capacitor voltage as t approaches infinity is: VC (∞) = 10V

b) The capacitor voltage after 20 µs is: VsVC 26.11)20( =µ

c) Graphically, the maximum capacitor voltage is:

VCmax ≅ 15V



Problem 5.72


Find: An expression for the capacitor voltage for . 0≥tAssumptions: The circuit initially stores no energy, the switch is open and the switch is closed. The switch is closed at

S1S2 S1 t = 0

and the switch is opened at . S2 t = 5s

Analysis: This circuit has the same configuration during the interval 0 ≤ t ≤ 5sas the one for Problem 5.39 did for . Therefore, the roots of the characteristic polynomial will be the same as those determined in that problem. They are:

t > 5s

s = −0.041± j0.220Ad the general form of the capacitor voltage is

VC (t) = 6 + e−0.041t[ A cos(0.220t) + B sin(0.220t)]For : 0 ≤ t ≤ 5sThe initial conditions are:

VC (0) = 0⇒ 6 + A = 0

dVC

dt|t=0=

1

CiC (0) = 0

⇒− 0.41A + 0.220B = 0

Solving the above equations: A = −6 B = −0.904

VC (t) = 6 + e−0.041t[−6 cos(0.220t) − 0.904 sin(0.220t)]V for0 ≤ t ≤ 5s

Note that . VC (5) = 3.127V For : t > 5sWe have a simple RC decay:

VC (t) = 3.127e−

t−5

12 V



Problem 5.73

Solution: Known quantities: Circuit shown in Figure P5.73, C = 1.6 nF ; After the switch

is closed at , the capacitor voltage reaches an initial

peak value of when

0=t70V t = 5π

3 µs , a second peak value of

when 53.2V t = 5π µs , and eventually approaches a steady-state of . 50V

Find: The values of R and L .

Assumptions: The circuit is underdamped and the circuit initially stores no energy.

Analysis: Using the given characteristics of the circuit step response and the assumption that the circuit is underdamped, the damping ratio and natural frequency can be determined as follows:

ζ =1

π / ln(a / A)( )2 +1

where a is the overshoot distance and A is the steady-state value

ωn =2π

T 1−ζ 2

where T is the period of oscillation In our case, and a = 70 − 50 = 20 A = 50

ζ =1

π / ln(20 /50)( )2 +1= 0.28

The period of the waveform is

T = 5π −5π3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ×10−6 =

10π3

×10−6

ωn =2π

10π3

×10−6 1− 0.282= 6.25×105

Implying that the characteristic polynomial for the circuit is

s2 + 2ζωns + ωn2

Compare this with the standard form of the characteristic polynomial for a series RLC circuit:

s2 +R

Ls +

1

LC

Matching terms yields L = 1.6 µH , R = 0.56Ω .



Problem 5.74

Solution: Known quantities: Same as P5.73, but the first two peaks occur at 5π µs and 15π µs

Find: Explain how to modify the circuit to meet the requirements.

Assumptions: The capacitor value C cannot be changed.

Analysis: Assuming we wish to retain the same peak amplitudes, we proceed as follows: The new period is

T = 15π ×10−6 − 5π ×10−6 = 10π ×10−6 Using the given characteristics of the circuit step response and the assumption that the circuit is underdamped, the damping ratio and natural frequency can be determined as follows:

ζ =1

π / ln(a / A)( )2 +1

where a is the overshoot distance and A is the steady-state value

ωn =2π

T 1−ζ 2

where T is the period of oscillation In our case, and a = 70 − 50 = 20 A = 50

ζ =1

π / ln(20 /50)( )2 +1= 0.28

ωn =2π

10π ×10−6 1− 0.282= 2.17 ×105

Implying that the characteristic polynomial for the circuit is

s2 + 2ζωns + ωn2

Compare this with the standard form of the characteristic polynomial for a series RLC circuit:

s2 +R

Ls +

1

LC

Matching terms yields L = 13.3µH , R = 1.61Ω .

Note that the frequency for this problem is one-third that of Problem 5.66, the inductance is times that of

Problem 5.66, and the resistance is 3 times that of Problem 5.66.

32



Problem 5.75

Solution: Known quantities: Circuit shown in Figure P5.75, i(0) = 0 A,V (0) = 10V .

Find: i(t) for . t > 0

Assumptions: None.

Analysis: The initial condition for the capacitor voltage is V (0−) = 10V . Applying KCL,

iR + iC + i = 0

where

iR =V

1Ω, iC = 0.5

dV

dt

Therefore,

i + V + 0.5dV

dt= 0

where

V = 2di

dt+ 4i

Thus, d 2i

dt 2+ 4

di

dt+ 5i = 0

Solving the differential equation:

i(t) = k1e(−2+ j )t + k2e(−2− j )t t > 0

i(0) = k1 + k2 = 0

V (0) = 2di (0)

dt+ 4i (0) = − 4 − j2( )k1 − 4 + j2( )k2 = 10

Solving for k1 and k2 and substituting, we have

i(t) = − j5

2e(−2+ j )t + j

5

2e(−2− j )t A for t > 0



Problem 5.76


Find: The maximum value of V .

Assumptions: The circuit is in steady state at . t = 0−

Analysis:

V (0−) = V (0+) = 0 Applying KVL: 48442

2=++ V

dtdV

dtVd

Solving the differential equation: 1222

21 ++= −− tt tekekV

From the initial condition:

V (0) = 0⇒ k1 = −12

iL (0) = CdV (0)

dt⇒ 6 +

k2

4= 3⇒ k2 = −12

V (t) = −12e−2t −12te−2t +12V for t > 0

The maximum value of V is: Vmax = V (∞) = 12V

Problem 5.77


Find: The value of t such that i = 2.5 A .


Analysis: In steady state, the inductors behave as short circuits. Using mesh analysis, we can find the initial conditions.

VVAii 0)0( 5)0()0( === −+−

After the switch is closed, the circuit is modified.

Applying nodal analysis: VV 0)0( =+ 0672

2=++ i

dtdi

dtid


i(t) = k1e−t + k2e−6t for t > 0

i(0) = k1 + k2 = 5

V (0) = −k1 − 6k2 = 0

Solving for the unknown constants, using the initial conditions, we have: 0 6)( 6 >−= −− tforAeeti tt

Therefore, i(t ) = 6e−t − e−6t = 2.5 ⇒ t 1 = 873ms



Problem 5.78


Find: The value of t such that i = 6 A .


Analysis: In steady state, the inductors behave as short circuits. Using mesh analysis, we can find the initial conditions. i(0−) = i (0+) = 12.5A

V (0−) = 0V

After the switch is closed, the circuit is modified. Applying nodal analysis:

V (0+) = −15V

d 2i

dt 2+ 7

di

dt+ 6i = 0


i(t) = k1e−t + k2e−6t for t > 0

i(0) = k1 + k2 = 12.5

V (0) = −k1 − 6k2 = −15Solving for the unknown constants, using the initial conditions, we have:

i(t) = 12e−t + 0.5e−6t A for t > 0Therefore,

i(t ) = 12e−t + 0.5e−6t = 6 ⇒ t 1 = 694ms



Problem 5.79


Find: The value of t such that . V = 7.5V


Analysis: The circuit at has the capacitors replaced by open circuits.

−= 0tBy current division:

i2Ω =3

3+ 5× 20 = 7.5 A

V (0−) = V (0+) = 2i2 = 15V

i (0−) = 0A

After the switch opens, apply KCL:

i (0+) = 0A

i + i1 + i2 = 0

i = CdV

dt=

1

6

dV

dt, i2 =

V

2

i1 = −1

6

dV

dt−

V

2

Applying KVL:

V = 3i1 + V1 = 3(−1

6

dV

dt−

V

2) + V1

V1 = V − 3(−1

6

dV

dt−

V

2)

i1 =1

6

dV1

dt=

1

12

d 2V

dt 2+

5

12

dV

dt

Applying KCL: 1

12

d 2V

dt 2+

5

12dV

dt+

V

2+

1

6

dV

dt= 0

⇒d 2V

dt 2+ 7

dV

dt+ 6V = 0

Solving the differential equation,

V (t) = k1e−t + k2e−6t for t > 0

V (0) = k1 + k2 = 15

i(0) =1

6−k1 − 6k2( )= 0

Solving for the unknown constants, using the initial conditions, we have:

V (t) = 18e−t − 3e−6t V for t > 0Therefore,

V (t ) = 18e−t − 3e−6t = 7.5 ⇒ t 1 = 873ms



Problem 5.80


Find: The maximum value of V and the maximum voltage between the contacts of the switches.

Assumptions: The circuit is in steady state at . L = 3 H. t = 0−


i(0−) = i (0+) =10

5= 2 A

V (0−) = V (0+) = 0V

After the switch is closed: Applying KVL: 1

LVdt

−∞

t∫ +

1

12

dV

dt+

V

3= 0

⇒d 2V

dt 2+ 4

dV

dt+

12

LV = 0

The particular response is zero for because the circuit is source-free. 0>t

L = 3H ⇒ s2 + 4s + 4 = 0

s1 = s2 = −2

V (t) = e−2t (A + Bt) for t > 0From the initial condition:

V (0+) = 0 = e0 (A + B(0))⇒ A = 0

Substitute the solution into the original KCL equation and evaluate at t = 0+ : 1 1

12

d

5.63

3(0) +

dt[e−2 + Bt)] |t=0 +2 = 0

0 +1

12

t (A

[Be−2t − 2Bte−2t )] |t=0 +2 = 0⇒ B = −24

V (t) = −24te−2t V for t > 0

he maximum absolute value of V is:

e

T

V = V (t = 0.5s) = −12

max V ≅ 4.414V

The maximum voltage between the contacts of the switches is:

since the voltage be een the contacts of the switches is a constant. Vswitch

MAX = VS = 10V tw



Problem 5.81


Find: V for . t > 0



V (0−) = 12V

iL (0−) = iL (0+) = 6 A

iC (0−) = 0 A

VC (0−) = VC (0+) = 4 V

VL (0−) = 0V

For : t > 0

iC (0+) = −2A

VL (0+) = 4 V

V (0+) = 8VdV

dt0+( )= −

dVC

dt0+( )= −

iCC

= −−21

4= 8

Using KVL and KCL, we can find the differential equation for the voltage in the resistor:

12 − 2i − 0.8diL

dt= 0

i = iL + iC ⇒ iL = i − iC

iC = CdVC

dt=

1

4

d

dt12 − V( )= −

1

4

dV

dt

12 − 2i − 0.8d

dti − ic( )= 0 ⇒ 12 − 2i − 0.8

di

dt+ 0.8

d

dt−

1

4

dV

dt

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 0

i = V / 2 ⇒ 12 − V − 0.4dV

dt− 0.2

d 2V

dt 2= 0

0.2d 2V2Ω

dt 2+ 0.4

dV2Ωdt

+ V2Ω = 12

⇒d 2V2Ω

dt 2+ 2

dV2Ωdt

+ 5V2Ω = 60

Solving the differential equation: Homogeneous Solution:



V2Ω,h = K1e−1+2 j( )t + K2e −1−2 j( )t t > 0

V (0) = 8V ,dV

dt0( )= 8

K1 + K2 = 8

−1+ 2 j( )K1 − 1+ 2 j( )K2 = 8

K1 = 4 − 4 j

K2 = 4 + 4 j

V2Ω,h = 4 − 4 j( )e −1+2 j( )t + 4 + 4 j( )e −1−2 j( )t t > 0

Particular Solution:

V2Ω,p = −6 + 3 j( )e −1+2 j( )t + −6 − 3 j( )e −1−2 j( )t +12 t > 0

The Complete Solution:

V2Ω = V2Ω,h + V2Ω,p

V2Ω = 4 − 4 j( )e −1+2 j( )t + 4 + 4 j( )e −1−2 j( )t −6 + 3 j( )e −1+2 j( )t + −6 − 3 j( )e −1−2 j( )t +12

V2Ω = −2 − j( )e −1+2 j( )t + −2 + j( )e −1−2 j( )t +12 t > 0


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