Energy Storage Elements: Capacitors and Inductors … - post-midterm.pdf · Energy Storage...

CHAPTER 6

Energy Storage Elements: Capacitors and Inductors

To this point in our study of electronic circuits, time has not beenimportant. The analysis and designs we have performed so far have beenstatic, and all circuit responses at a given time have depended only onthe circuit inputs at that time. In this chapter, we shall introduce twoimportant passive circuit elements: the capacitor and the inductor.

Capacitors and inductors, which are the electric and magnetic duals ofeach other, differ from resistors in several significant ways. Unlike resistors,which dissipate energy, capacitors and inductors do not dissipate but storeenergy, which can be retrieved at a later time. They are called storage el-ements. Furthermore, their branch variables do not depend algebraicallyupon each other. Rather, their relations involve temporal derivatives andintegrals. Thus, the analysis of circuits containing capacitors and inductorsinvolve differential equations in time.

6.1. Capacitors

A capacitor is a passive element designed to store energy in its electricfield. When a voltage source v is connected to the capacitor, the amountof charge stored, represented by q, is directly proportional to v, i.e.,

q(t) = Cv(t)

where C, the constant of proportionality, is known as the capacitanceof the capacitor. The unit of capacitance is the farad (F) in honor ofMichael Faraday.

• 1 farad = 1 coulomb/volt.

6.1.1. Circuit symbol for capacitor of C farads:

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72 6. ENERGY STORAGE ELEMENTS: CAPACITORS AND INDUCTORS

6.1.2. Since i = dqdt , then the current-voltage relationship of the capac-

itor is

(6.2) i = Cdv

dt.

Note that in (6.2), the capacitance value C is constant (time-invariant) andthat the current i and voltage v are both functions of time (time-varying).So, in fact, the full form of (6.2) is

i(t) = Cd

dtv(t).

Hence, the voltage-current relation is

v(t) =1

C

∫ t

to

i(τ)dτ + v(to)

where v(to) is the voltage across the capacitor at time to. Note that capac-itor voltage depends on the past history of the capacitor current. Hence,the capacitor has memory.

6.1.3. The instantaneous power delivered to the capacitor is

p(t) = i(t)× v(t) =

(Cd

dtv(t)

)v(t).

6.1. CAPACITORS 73

The energy stored in the capacitor is

w(t) =

∫ t

−∞p(τ)dτ =

1

2Cv2(t).

In the above calculation, we assume v(−∞) = 0, because the capacitorwas uncharged at t = −∞.

6.1.4. Capacitors are commercially available in different values andtypes. Typically, capacitors have values in the picofarad (pF) to microfarad(µF) range.

6.1.5. Remarks:

(a) The word capacitor is derived from this element’s capacity to storeenergy in an electric field.

(b) A capacitor is an open circuit to dc.When the voltage across a capacitor is not changing with time

(i.e., dc voltage), its derivative wrt. time is dvdt = 0 and hence the

current through the capacitor is i(t) = C dvdt = C × 0 = 0.

(c) The voltage on a capacitor cannot change abruptly (since i = C dvdt )

A discontinuous change in voltage requires an infinite current,which is physically impossible.

(d) The ideal capacitor does not dissipate energy. It takes power fromthe circuit when storing energy in its field and returns previouslystored energy when delivering power to the circuit.

Example 6.1.6. If a 10 µF is connected to a voltage source with

v(t) = 50 sin 2000t

determine the current through the capacitor.


Example 6.1.7. Determine the voltage across a 2-µF capacitor if thecurrent through it is

i(t) = 6e−3000t mA

Assume that the initial capacitor voltage (at time t = 0) is zero.

Example 6.1.8. Obtain the energy stored in each capacitor in the figurebelow under dc conditions.

6.2. SERIES AND PARALLEL CAPACITORS 75

6.2. Series and Parallel Capacitors

We know from resistive circuits that series-parallel combination is apowerful tool for simplifying circuits. This technique can be extended toseries-parallel connections of capacitors, which are sometimes encountered.We desire to replace these capacitors by a single equivalent capacitor Ceq.

6.2.1. The equivalent capacitance of N parallel-connected capacitorsis the sum of the individual capacitance.

Ceq = C1 + C2 + · · ·+ CN

The equivalent capacitance of N series-connected capacitors is the thereciprocal of the sum of the reciprocals of the individual capacitances.

1

Ceq=

1

C1+

1

C2+ · · ·+ 1

CN

Example 6.2.2. Find the Ceq.


6.3. Inductors

An inductor is a passive element designed to store energy in its magneticfield. Inductors find numerous applications in electronic and power sys-tems. They are used in power supplies, transformers, radios, TVs, radars,and electric motors.

6.3.1. Circuit symbol of inductor:

6.3.2. If a current is allowed to pass through an inductor, the voltageacross the inductor is directly proportional to the time rate of change ofthe current, i.e.,

v(t) = Ld

dti(t),

where L is the constant of proportionality called the inductance of theinductor. The unit of inductance is henry (H), named in honor of JosephHenry.

• 1 henry equals 1 volt-second per ampere.

6.3. INDUCTORS 77

6.3.3. By integration, the current-voltage relation is

i(t) =1

L

∫ t

to

v(τ) dτ + i(to),

where i(to) is the current at time to.

6.3.4. The instantaneous power delivered to the inductor is

p(t) = v(t)× i(t) =

(Ld

dti(t)

)i(t)

The energy stored is

w(t) =

∫ t

−∞p(τ) dτ =

1

2Li2(t).

6.3.5. Like capacitors, commercially available inductors come in differ-ent values and types. Typical practical inductors have inductance valuesranging from a few microhenrys (µH), as in communication systems, totens of henrys (H) as in power systems.

6.3.6. Remarks:

(a) An inductor acts like a short circuit to dc.The voltage across an inductor is zero when the current is con-

stant.(b) The current through an inductor cannot change instantaneously.

This opposition to the change in current is an important propertyof the inductor. A discontinuous change in the current through aninductor requires an infinite voltage, which is not physically possible.

(c) The ideal inductor does not dissipate energy. The energy stored init can be retrieved at a later time. The inductor takes power fromthe circuit when storing energy and delivers power to the circuitwhen returning previously stored energy.


Example 6.3.7. If the current through a 1-mH inductor is i(t) =20 cos 100t mA, find the terminal voltage and the energy stored.

Example 6.3.8. Find the current through a 5-H inductor if the voltageacross it is

v(t) =

30t2, t > 0

0, t < 0.

In addition, find the energy stored within 0 < t < 5 s.

6.3. INDUCTORS 79

Example 6.3.9. The terminal voltage of a 2-mH inductor is v = 10(1−t)V. Find the current flowing through it at t = 4 s and the energy stored init within 0 < t < 4 s. Assume i(0) = 2 A.

Example 6.3.10. Determine vC , iL and the energy stored in the capac-itor and inductor in the following circuit under dc conditions.


Example 6.3.11. Determine vC , iL and the energy stored in the capac-itor and inductor in the following circuit under dc conditions.

6.4. SERIES AND PARALLEL INDUCTORS 81

6.4. Series and Parallel Inductors

The equivalent inductance of N series-connected inductors is the sumof the individual inductances, i.e.,

Leq = L1 + L2 + · · ·+ LN

The equivalent inductance of N parallel inductors is the reciprocal ofthe sum of the reciprocals of the individual inductances, i.e.,

1

Leq=

1

L1+

1

L2+ · · ·+ 1

LN

Note that

(a) inductors in series are combined in exactly the same way as resistorsin series and

(b) inductors in parallel are combined in the same way as resistors inparallel.


CHAPTER 6 Capacitors and Inductors 217

where

1

Leq= 1

L1+ 1

L2+ 1

L3+ · · · + 1

LN

(6.30)

The initial current i(t0) through Leq at t = t0 is expected by KCL to bethe sum of the inductor currents at t0. Thus, according to Eq. (6.29),

i(t0) = i1(t0) + i2(t0) + · · · + iN (t0)

According to Eq. (6.30),

The equivalent inductance of parallel inductors is the reciprocal of the sum of thereciprocals of the individual inductances.

Note that the inductors in parallel are combined in the same way as resis-tors in parallel.

For two inductors in parallel (N = 2), Eq. (6.30) becomes

1

Leq= 1

L1+ 1

L2or Leq = L1L2

L1 + L2(6.31)

It is appropriate at this point to summarize the most important character-istics of the three basic circuit elements we have studied. The summaryis given in Table 6.1.

TABLE 6.1 Important characteristics of the basic elements.†

Relation Resistor (R) Capacitor (C) Inductor (L)

v-i: v = iR v = 1

C

∫ t

t0

i dt + v(t0) v = Ldi

dt

i-v: i = v/R i = Cdv

dti = 1

L

∫ t

t0

i dt + i(t0)

p or w: p = i2R = v2

Rw = 1

2Cv2 w = 1

2Li2

Series: Req = R1 + R2 Ceq = C1C2

C1 + C2Leq = L1 + L2

Parallel: Req = R1R2

R1 + R2Ceq = C1 + C2 Leq = L1L2

L1 + L2

At dc: Same Open circuit Short circuit

Circuit variablethat cannotchange abruptly: Not applicable v i

†Passive sign convention is assumed.

E X A M P L E 6 . 1 1

Find the equivalent inductance of the circuit shown in Fig. 6.31.

Example 6.4.1. Find the equivalent inductance Leq of the circuit shownbelow.

6.5. APPLICATIONS: INTEGRATORS AND DIFFERENTIATORS 83

6.5. Applications: Integrators and Differentiators

Capacitors and inductors possess the following three special propertiesthat make them very useful in electric circuits:

(a) The capacity to store energy makes them useful as temporary volt-age or current sources. Thus, they can be used for generating alarge amount of current or voltage for a short period of time.

(b) Capacitors oppose any abrupt change in voltage, while inductorsoppose any abrupt change in current. This property makes induc-tors useful for spark or arc suppression and for converting pulsatingdc voltage into relatively smooth dc voltage.

(c) Capacitors and inductors are frequency sensitive. This propertymakes them useful for frequency discrimination.

The first two properties are put to use in dc circuits, while the thirdone is taken advantage of in ac circuits.

In this final part of the chapter, we will consider two applications in-volving capacitors and op amps: integrator and differentiator.


6.5.1. Integrator. An integrator is an op amp circuit whose output isproportional to the integral of the input signal. We obtain an integrator byreplacing the feedback resistor Rf in the inverting amplifier by a capacitor.

This givesd

dtvo(t) = − 1

RCvi(t),

which implies

vo(t) = − 1

RC

∫ t

0vi(τ)dτ + vo(0).

To ensure that vo(0) = 0, it is always necessary to discharge the integratorscapacitor prior to the application of a signal.

In practice, the op amp integrator requires a feedback resistor to reducedc gain and prevent saturation. Care must be taken that the op ampoperates within the linear range so that it does not saturate.

6.5. APPLICATIONS: INTEGRATORS AND DIFFERENTIATORS 85

6.5.2. Differentiator. A differentiator is an op amp circuit whose out-put is proportional to the differentiation of the input signal. We obtain adifferentiator by replacing the input resistor in the inverting amplifier bya capacitor. This gives

vo(t) = −RC d

dtvi(t).

Differentiator circuits are electronically unstable because any electricalnoise within the circuit is exaggerated by the differentiator. For this rea-son, the differentiator circuit above is not as useful and popular as theintegrator. It is seldom used in practice.

CHAPTER 9

Sinusoids and Phasors

We now begins the analysis of circuits in which the voltage or currentsources are time-varying. In this chapter, we are particularly interested insinusoidally time-varying excitation, or simply, excitation by a sinusoid.

9.0.1. Some terminology:

(a) A sinusoid is a signal that has the form of the sine or cosine func-tion.• Turn out that you can express them all under the same notation

using only cosine (or only sine) function. We ill use cosine.(b) A sinusoidal current is referred to as alternating current (AC).(c) Circuits driven by sinusoidal current or voltage sources are called

AC circuits.

9.1. Sinusoids

9.1.1. Consider the sinusoidal signal (in cosine form)

x(t) = Xm cos(ωt+ φ) = Xm cos(2πft+ φ),

whereXm: the amplitude of the sinusoid,ω: the angular frequency in radians/s (or rad/s),φ: the phase.

• First, we consider the case when φ = 0:

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ECS 203 (ME2) - Part 3A Dr.Prapun Suksompong

110 9. SINUSOIDS AND PHASORS

• When φ 6= 0, we shift the graph of Xm cos(ωt) to the left “by φ”.

9.1.2. The period (the time of one complete cycle) of the sinusoid is

T =2π

ω.

The unit of the period is in second if the angular frequency unit is in radianper second.

The frequency f (the number of cycles per second or hertz (Hz)) isthe reciprocal of this quantity, i.e.,

f =1

T.

9.1.3. Standard form for sinusoid: In this class, when you are askedto find the sinusoid representation of a signal, make sure that your answeris in the form

x(t) = Xm cos(ωt+ φ) = Xm cos(2πft+ φ),

where Xm is nonnegative and φ is between −180 and +180.

• When the signal is given in the sine form, it can be converted intoits cosine form via the identity

sin(x) = cos(x− 90).

In particular,

Xm sin(ωt+ φ) = Xm cos(ωt+ φ− 90).

9.2. PHASORS 111

• Xm is always non-negative. We can avoid having the negative signby the following conversion:

− cos(x) = cos(x± 180).

In particular,

−A cos(ωt+ φ) = A cos(2πft+ φ± 180).

Note that usually you do not have the choice between +180 or−180. The one that you need to use is the one that makes φ±180

falls somewhere between −180 and +180.

9.2. Phasors

Sinusoids are easily expressed in terms of phasors, which are more con-venient to work with than sine and cosine functions. The tradoff is thatphasors are complex-valued.

9.2.1. The idea of phasor representation is based on Euler’s identity:

ejφ = cosφ+ j sinφ,

From the identity, we may regard cosφ and sinφ as the real and imaginaryparts of ejφ:

cosφ = Reejφ, sinφ = Im

ejφ,

where Re and Im stand for the real part of and the imaginary part of ejφ.

9.2.2. A phasor is a complex number that represents the amplitude andphase of a sinusoid. Given a sinusoid v(t) = Vm cos(ωt+ φ), then

v(t) = Vm cos(ωt+φ) = ReVme

j(ωt+φ)

= ReVme

jφ · ejωt

= ReVejωt

,

where

V = Vmejφ = Vm∠φ.

V is called the phasor representation of the sinusoid v(t). In otherwords, a phasor is a complex number that represents amplitude and phaseof a sinusoid.

9.2.3. Remarks:

• Whenever a sinusoid is expressed as a phasor, the term ejωt is im-plicitly present. It is therefore important, when dealing with pha-sors, to keep in mind the (angular) frequency ω of the phasor.


• To obtain the sinusoid corresponding to a given phasor V, multiplythe phasor by the time factor ejωt and take the real part.

Equivalently, given a phasor, we obtain the time-domain repre-sentation as the cosine function with the same magnitude as thephasor and the argument as ωt plus the phase of the phasor.• Any complex number z (including any phasor) can be equivalently

represented in three forms.(a) Rectangular form: z = x+ jy.(b) Polar form: z = r∠φ.(c) Exponential form: z = rejφ

where the relations between them are

r =√x2 + y2, φ = tan−1 y

x± 180.

x = r cosφ, y = r sinφ.

Note that for φ, the choice of using +180 or −180 in the formulais determined by the actual quadrant in which the complex numberlies.

• As a complex quantity, a phasor may be expressed in rectangularform, polar form, or exponential form. In this class, we focus onpolar form.

9.2.4. Summary : By suppressing the time factor, we transform thesinusoid from the time domain to the phasor domain. This transformationis summarized as follows:

v(t) = Vm cos(ωt+ φ)⇔ V = Vm∠φ.

Time domain representation ⇔ Phasor domain representation

9.2. PHASORS 113

9.2.5. Standard form for phasor: In this class, when you are askedto find the phasor representation of a signal, make sure that your answeris a complex number in polar form, i.e. r∠φ where r is nonnegative and φis between −180 and +180.

Example 9.2.6. Transform these sinusoids to phasors:

(a) i = 6 cos(50t− 40) A

(b) v = −4 sin(30t+ 50) V

Example 9.2.7. Find the sinusoids represented by these phasors:

(a) I = −3 + j4 A

(b) V = j8e−j20 V

9.2.8. The differences between v(t) and V should be emphasized:

(a) v(t) is the instantaneous or time-domain representation, while V isthe frequency or phasor-domain representation.

(b) v(t) is time dependent, while V is not.(c) v(t) is always real with no complex term, while V is generally com-

plex.

9.2.9. Adding sinusoids of the same frequency is equivalent to addingtheir corresponding phasors. To see this,

A1 cos (ωt+ φ1) + A2 cos (ωt+ φ2) = ReA1e

jωt

+ ReA2e

jωt

= Re

(A1 + A2) ejωt.


9.2.10. Properties involving differentiation and integration:

(a) Differentiating a sinusoid is equivalent to multiplying its corre-sponding phasor by jω. In other words,

dv(t)

dt⇔ jωV.

To see this, suppose v(t) = Vm cos(ωt+ φ). Then,

dv

dt(t) = −ωVm sin(ωt+ φ) = ωVm cos(ωt+ φ− 90 + 180)

= ReωVme

jφej90 · ejωt

= RejωVejωt

Alternatively, express v(t) as

v(t) = ReVme

j(ωt+φ).

Then,d

dtv(t) = Re

Vmjωe

j(ωt+φ).

(b) Integrating a sinusoid is equivalent to dividing its correspondingphasor by jω. In other words,∫

v(t)dt⇔ V

jω.

Example 9.2.11. Find the voltage v(t) in a circuit described by theintergrodifferential equation

2dv

dt+ 5v + 10

∫vdt = 20 cos(5t− 30)

using the phasor approach.

9.3. PHASOR RELATIONSHIPS FOR CIRCUIT ELEMENTS 115

9.3. Phasor relationships for circuit elements

9.3.1. Resistor R: If the current through a resistor R is

i(t) = Im cos(ωt+ φ)⇔ I = Im∠φ,

the voltage across it is given by

v(t) = i(t)R = RIm cos(ωt+ φ).

The phasor of the voltage is

V = RIm∠φ.

Hence,V = IR.

We note that voltage and current are in phase and that the voltage-currentrelation for the resistor in the phasor domain continues to be Ohms law,as in the time domain.


9.3.2. Capacitor C: If the voltage across a capacitor C is

v(t) = Vm cos(ωt+ φ)⇔ V = Vm∠φ,

the current through it is given by

i(t) = Cdv(t)

dt⇔ I = jωCV = ωCVm∠(φ+ 90).

The voltage and current are 90 out of phase. Specifically, the currentleads the voltage by 90.

• Mnemonic: CIVILIn a Capacitive (C) circuit, I leads V. In an inductive (L) circuit,

V leads V.

9.3. PHASOR RELATIONSHIPS FOR CIRCUIT ELEMENTS 117

9.3.3. Inductor L: If the current through an inductor L is

i(t) = Im cos(ωt+ φ)⇔ I = Im∠φ,

the voltage across it is given by

v(t) = Ldi(t)

dt⇔ V = jωLI = ωLIm∠(φ+ 90).

The voltage and current are 90 out of phase. Specifically, the currentlags the voltage by 90.

CHAPTER 9 Sinusoids and Phasors 369

relations for the capacitor; Fig. 9.14 gives the phasor diagram. Table 9.2summarizes the time-domain and phasor-domain representations of thecircuit elements. v

Re

Im

IV

0

f

Figure 9.14 Phasor diagram for the capa-citor; I leads V.

TABLE 9.2 Summary of voltage-currentrelationships.

Element Time domain Frequency domain

R v = Ri V = RI

L v = Ldi

dtV = jωLI

C i = Cdv

dtV = I

jωC

E X A M P L E 9 . 8

The voltage v = 12 cos(60t + 45) is applied to a 0.1-H inductor. Findthe steady-state current through the inductor.

Solution:

For the inductor, V = jωLI, where ω = 60 rad/s and V = 12 45 V.Hence

I = VjωL

= 12 45

j60 × 0.1= 12 45

6 90= 2 − 45 A

Converting this to the time domain,

i(t) = 2 cos(60t − 45) A

P R A C T I C E P R O B L E M 9 . 8

If voltage v = 6 cos(100t−30) is applied to a 50µF capacitor, calculatethe current through the capacitor.

Answer: 30 cos(100t + 60) mA.

9.5 IMPEDANCE AND ADMITTANCEIn the preceding section, we obtained the voltage-current relations for thethree passive elements as

V = RI, V = jωLI, V = IjωC

(9.38)

These equations may be written in terms of the ratio of the phasor voltageto the phasor current as

VI

= R,VI

= jωL,VI

= 1

jωC(9.39)

From these three expressions, we obtain Ohm’s law in phasor form forany type of element as


9.4. Impedance and Admittance

Thus, we obtained the voltage current relations for the three passiveelements as

V = IR, V = jωLI, I = jωCV.

These equations may be written in terms of the ratio of the phasor voltageto the phasor of current as

V

I= R,

V

I= jωL,

V

I=

1

jωC.

From these equations, we obtain Ohm’s law in phasor form for any type ofelement as

Z =V

Ior V = IZ.

Definition 9.4.1. The impedance Z of a circuit is the ratio of thephasor voltage V to the phasor current I, measured in ohms (Ω).

As a complex quantity, the impedance may be expressed in rectangularform as

Z = R + jX = |Z|∠θ,with

|Z| =√R2 +X2, θ = tan−1 X

R, R = |Z| cos θ, X = |Z| sin θ.

R = Re Z is called the resistance and X = Im Z is called the reac-tance.

The reactanceX may be positive or negative. We say that the impedanceis inductive when X is positive or capacitive when X is negative.

Definition 9.4.2. The admittance (Y) is the reciprocal of impedance,measured in Siemens (S). The admittance of an element(or a circuit) is theratio of the phasor current through it to phasor voltage across it, or

Y =1

Z=

I

V.

9.4. IMPEDANCE AND ADMITTANCE 119

9.4.3. Kirchhoff’s laws (KCL and KVL) hold in the phasorform.

To see this, suppose v1, v2, . . . , vn are the voltages around a closed loop,then

v1 + v2 + · · ·+ vn = 0.

If each voltage vi is a sinusoid, i.e.

vi = Vmi cos(ωt+ φi) = ReVie

jωt

with phasor Vi = Vmi∠φi = Vmiejφi, then

Re

(V1 + V2 + · · ·+ Vn) ejωt

= 0,

which must be true for all time t. To satisfy this, we need

V1 + V2 + · · ·+ Vn = 0.

Hence, KVL holds for phasors.Similarly, we can show that KCL holds in the frequency domain, i.e., if

the currents i1, i2, . . . , in be the currents entering or leaving a closed surfaceat time t, then

i1 + i2 + · · ·+ in = 0.

If the currents are sinusoids and I1, I2, . . . , In are their phasor forms, then

I1 + I2 + · · ·+ In = 0.

Major Implication: Since Ohm’s Law and Kirchoff’s Laws hold inphasor domain, all resistance combination, analysis methods (nodaland mesh analysis) and circuit theorems (linearity, superposition, sourcetransformation, and Thevenin’s and Norton’s equivalent circuits) that wehave previously studied for dc circuits apply to ac circuits!!!

Just think of impedance as a complex-valued resistance!!

In addition, our ac circuits can now effortlessly include capacitors andinductors which can be considered as impedances whose values depend onthe frequency ω of the ac sources!!


9.5. Impedance Combinations

Consider N series-connected impedances as shown below.

The same current I flows through the impedances. Applying KVLaround the loop gives

V = V1 + V2 + · · ·+ VN = I(Z1 + Z2 + · · ·+ ZN)

The equivalent impedance at the input terminals is

Zeq =V

I= Z1 + Z2 + · · ·+ ZN .

In particular, if N = 2, the current through the impedance is

I =V

Z1 + Z2.

Because V1 = Z1I and V2 = Z2I,

V1 =Z1

Z1 + Z2V, V2 =

Z2

Z1 + Z2V

which is the voltage-division relationship.

9.5. IMPEDANCE COMBINATIONS 121

Now, consider N parallel-connected impedances as shown below.

The voltage across each impedance is the same. Applying KCL at thetop node gives

I = I1 + I2 + · · ·+ IN = V

(1

Z1+

1

Z2+ · · ·+ 1

ZN

).

The equivalent impedance Zeq can be found from

1

Zeq=

I

V=

1

Z1+

1

Z2+ · · ·+ 1

ZN.

When N = 2,

Zeq =Z1Z2

Z1 + Z2.

BecauseV = IZeq = I1Z1 = I2Z2,

we have

I1 =Z2

Z1 + Z2I, I2 =

Z1

Z1 + Z2I

which is the current-division principle.


Example 9.5.1. Find the input impedance of the circuit below. Assumethat the circuit operates at ω = 50 rad/s.

Example 9.5.2. Determine vo(t) in the circuit below.

CHAPTER 10

Sinusoidal Steady State Analysis

10.1. General Approach

In the previous chapter, we have learned that the steady-state responseof a circuit to sinusoidal inputs can be obtained by using phasors. In thischapter, we present many examples in which nodal analysis, mesh analysis,Thevenin’s theorem, superposition, and source transformations are appliedin analyzing ac circuits.

10.1.1. Steps to analyze ac circuits, using phasor domain:

Step 1. Transform the circuit to the phasor or frequency domain.• Not necessary if the problem is specified in the frequency do-

main.Step 2. Solve the problem using circuit techniques (e.g., nodal analysis,

mesh analysis, Thevenin’s theorem, superposition, or source trans-formations )• The analysis is performed in the same manner as dc circuit

analysis except that complex numbers are involved.Step 3. Transform the resulting phasor back to the time domain.

10.1.2. ac circuits are linear (they are just composed of sources andimpedances)

10.1.3. The superposition theorem applies to ac circuits the sameway it applies to dc circuits. This is the case when all the sources in thecircuit operate at the same frequency. If they are operating at differentfrequency, see Section 10.2.

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124 10. SINUSOIDAL STEADY STATE ANALYSIS

10.1.4. Source transformation:

Vs = ZsIs, Is =Vs

Zs.

10.1.5. Thevenin and Norton Equivalent circuits:

VTh = ZNIN, ZTh = ZN

10.1. GENERAL APPROACH 125

Example 10.1.6. Compute V1 and V2 in the circuit below using nodalanalysis.

Example 10.1.7. Determine current Io in the circuit below using meshanalysis.


Example 10.1.8. Find the Thevenin equivalent at terminals a-b of thecircuit below.

Example 10.1.9. Op Amp AC Circuits: Find the (closed-loop) gainof the circuit below.

2

2. In this experiment, we study the following op amp circuits:

a) current-to-voltage converter

b) voltage-to-current converter

c) integrating amplifier

3. The voltage-to-current and current-to-voltage converters are used in electronic voltmeters

and ammeters, respectively.

The voltage-to-current converter, as shown in Figure 8-2, produces an output current

that depends on the input voltage and the resistor R. In particular, the output current

Iout = Vi/R

independent of the loading resistance RL.

The current-to-voltage converter, as shown in Figure 8-3, produces an output voltage

that depends on the input current and the resistor R. In particular, the output voltage

Vo = -IinR

independent of the size of the loading resistance RL.

Figure 8-2: Voltage-to-current converter. Figure 8-3: Current-to-voltage converter.

4. An integrating amplifier is shown in Figure 8-4a.

Figure 8-4a: Integrating amplifier

R

+

+

vo

-

iC

iin

vi

V+

V-

X

C+ vC -

RRL

V+

V-

ViIout

+

+

RL

R

V+

V-

Iin

+

Vo

-

10.2. CIRCUIT WITH MULTIPLE SOURCES OPERATING AT DIFFERENT FREQUENCIES 127

10.2. Circuit With Multiple Sources Operating At DifferentFrequencies

A special care is needed if the circuit has multiple sources operatingat different frequencies. In which case, one must add the responses dueto the individual frequencies in the time domain. In other words, thesuperposition still works but

(a) We must have a different frequency-domain circuit for each fre-quency.

(b) The total response must be obtained by adding the individual re-sponse in the time domain.

10.2.1. Since the impedance depend on frequency, it is incorrect to tryto add the responses in the phasor or frequency domain. To see this notethat the exponential factor ejωt is implicit in sinusoidal analysis, and thatfactor would change for every angular frequency ω. In particular, although∑

i

Vmi cos(ωt+ φi) =∑i

ReVie

jωt

= Re

(∑i

Vi

)ejωt

,

when we allow ω to be different for each sinusoid, generally∑i

Vmi cos(ωit+ φi) =∑i

ReVie

jωit6= Re

(∑i

Vi

)ejωit

.

Therefore, it does not make sense to add responses at different frequenciesin the phasor domain.

10.2.2. The Thevenin or Norton equivalent circuit (if needed) must bedetermined at each frequency and we have one equivalent circuit for eachfrequency.


Example 10.2.3. Find vo in the circuit below using the superpositiontheorem.

CHAPTER 11

AC Power Analysis

Our effort in ac circuit analysis so far has been focused mainly on cal-culating voltage and current. The major concern in this chapter is poweranalysis.

11.0.4. Power is the most important quantity in electric utilities, elec-tronic and communication systems because such systems involve transmis-sion of power (or energy) from one point to another.

Every industrial and household electrical device (every fan, motor, lamp,pressing iron, TV, personal computer) has a power rating that indicateshow much power the equipment requires; exceeding the power rating cando permanent damage to an appliance.

11.0.5. The most common form of electric power is 50-Hz (Thailand)or 60-Hz (United States) ac power. The choice of ac over dc allowed high-voltage power transmission from the power generating plant to the con-sumer. (DC attenuation is high.)

11.1. Instantaneous Power

Definition 11.1.1. The instantaneous power p(t) absorbed by anelement is the product of the instantaneous voltage v(t) across the elementand the instantaneous current i(t) through it.

Assuming the passive sign convention as shown on Figure 1,

p(t) = v(t)i(t).

The instantaneous power is the power at any instant of time. It is therate at which an element absorbs energy.

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130 11. AC POWER ANALYSIS

Consider the general case of instantaneous power absorbed by an arbi-trary combination of circuit elements under sinusoidal excitation.

Figure 1. Sinusoidal source and passive linear circuit

Let the voltage and current at the terminals of the circuit be

v(t) = Vm cos(ωt+ θv)

andi(t) = Im cos(ωt+ θi)

where Vm and Im are the amplitudes, and θv and θi are the phase of thevoltage and current, respectively. The instantaneous power absorbed bythe circuit is

p(t) = v(t)i(t) = VmIm cos(ωt+ θv) cos(ωt+ θi)

= VmImej(ωt+θv) + e−j(ωt+θv)

2

ej(ωt+θi) + e−j(ωt+θi)

2

= VmIm1

4

(ej(2ωt+θv+θi) + ej(θi−θv) + ej(θv−θi) + e−j(2ωt+θv+θi)

)= VmIm

1

2

(ej(θv−θi) + ej(θi−θv)

2+ej(2ωt+θv+θi) + e−j(2ωt+θv+θi)

2

)= VmIm

1

2(cos (θv − θi) + cos (2ωt+ θv + θi)) .

Alternatively, we can apply the trigonometric identity

cosA cosB =1

2cos(A−B) + cos(A+B)

to directly arrive at the same result which is

p(t) =1

2VmIm cos(θv − θi)︸︷︷︸

constant term

+1

2VmIm cos(2ωt+ θv + θi)︸︷︷︸

time-dependent term

.

11.1. INSTANTANEOUS POWER 131

This shows that the instantaneous power has two parts.

• First Part: a constant or time-independent term. Its value de-pends on the phase difference between the voltage and the current.• Second Part: a sinusoidal function whose frequency is 2ω, which

is twice the angular frequency of the voltage or current.

11.1.2. Consider the sketch of p(t), we observe that

(a) p(t) is periodic and has a period of To = T2 , where T = 2π

ω is theperiod of the voltage and the current

(b) p(t) may become positive for some part(s) of each cycle and negativefor the rest of the cycle.• When p(t) is positive, power is absorbed by the circuit.• When p(t) is negative, power is absorbed by the source.

– In this case, power is transferred from the circuit to thesource.

– This is possible because of the storage elements (capaci-tors and inductors) in the circuit.


11.2. Average Power

The instantaneous power changes with time and is therefore difficult tomeasure. The average power is more convenient to measure.

Definition 11.2.1. The average power is the average of theinstantaneous power over one period.

Thus, the average power is given by

P =1

T

∫ T

0p(t)dt =

1

2VmIm cos(θv − θi).

Since cos(θv − θi) = cos(θi − θv), what is important is the difference inthe phases of the voltage and the current. Note that p(t) is time varyingwhile P does not depend on time.

11.2.2. Using the phasor forms of v(t) and i(t), which are V = Vm∠θvand I = Im∠θi, we obtain

P =1

2VmIm cos(θv − θi) =

1

2ReVI∗.

11.2.3. Two special cases:

Case 1: When θv = θi, the voltage and the current are in phase. This impliesa purely resistive circuit or resistive load R, and

P =1

2VmIm

This shows that a purely resistive circuit (e.g. resistive load(R)) absorbs power all times.

Case 2: When θv − θi = ±90, we have a purely reactive circuit, and

P =1

2VmIm cos(90) = 0

showing that a purely reactive circuit (e.g. a reactive load Lor C) absorbs no average power.

11.2. AVERAGE POWER 133

11.2.4. From Ohm’s law, we have two more formula.

(a) The average power absorbed by an impedance Z when a voltage Vis applied across it is

P =1

2ReVI∗ =

1

2ReVV∗

Z∗ =

1

2|V|2Re 1

Z∗ =

1

2|V|2ReZ

|Z|2

(b) The average power absorbed by an impedance Z when a current Iflows through it is

P =1

2ReVI∗ =

1

2ReIZI∗ =

1

2|I|2ReZ

Example 11.2.5. Calculate the average power absorbed by an impedanceZ = 30− j70 Ω when a voltage V = 120∠0 is applied across it.

Example 11.2.6. A current I = 10∠30 flows through an impedanceZ = 20∠− 22Ω. Find the average power delivered to the impedance.

Example 11.2.7. For the circuit shown below, find the average powersupplied by the source and the average power absorbed by the resistor.


11.3. Maximum Average Power Transfer

11.3.1. Optimal Load Impedance: Before the midterm, we solvedthe problem of maximizing the power delivered by a power-supplying re-sistive network to a load RL. Representing the circuit by its Theveninequivalent, we proved that the maximum power would be delivered to theload if the load resistance is equal to the Thevenin resistance RL = RTh.We now extend that result to ac circuits.

Consider an ac circuit which is connected to a load ZL and is representedby its Thevenin equivalent.

In a rectangular form, the Thevenin impedance ZTh and load impedanceZL are

ZTh = RTh + jXTh

ZL = RL + jXL

The current through the load is

I =VTh

ZL + ZTh,

and the average power delivered to the load is

P =1

2|I|2RL =

1

2

∣∣VTh2∣∣ RL

(RL +RTh)2 + (XL +XTh)

2

Our objective is to adjust the load parameter RL and XL so that P ismaximum. To do this we set ∂P

∂RLand ∂P

∂XLequal to zero.

11.3. MAXIMUM AVERAGE POWER TRANSFER 135

Setting ∂P∂XL

= 0 givesXL = −XTh.

Setting ∂P∂RL

= 0 gives

RL =√R2Th + (XTh +XL)2

Hence, to get the maximum average power transfer, the load impedanceZL must selected so that

XL = −XTh and RL = RTh,

i.e.,ZL = RL + jXL = RTh − jXTh = Z∗Th

That is, for the maximum average power transfer, the loadimpedance ZL must be equal to the complex conjugate of theThevenin impedance ZTh.

When ZL = Z∗Th, the maximum average power is

Pmax =|VTh|28RTh

.

Example 11.3.2. Determine the load impedance ZL that maximizesthe average power drawn from the circuit below. What is the maximumaverage power?


11.3.3. Optimal Purely Resistive Load: In a situation in which theload must be purely real; that is XL must be 0. Then,

P =1

2|I|2RL =

1

2

∣∣V2Th

∣∣ RL

(RL +RTh)2 + (XTh)

2

Setting ∂P∂RL

= 0 gives

RL =√R2Th +X2

Th = |ZTh|.Hence, for maximum average power transfer to a purely resis-

tive load, the load impedance is equal to the magnitude of theThevenin impedance ZTh. In which case, the maximum average poweris

P =1

4

∣∣VTh2∣∣ 1

|ZTh|+RTh

Note that|ZTh|+RTh ≥ RTh +RTh = 2RTh

Hence,1

4

∣∣VTh2∣∣ 1

|ZTh|+RTh≤ 1

8

∣∣VTh2∣∣

RTh.

11.4. EFFECTIVE OR RMS VALUE 137

11.4. Effective or RMS Value

The idea of effective value arises from the need to measure the effec-tiveness of an ac voltage or current source in delivering power to a resistiveload.

Definition 11.4.1. The effective value Ieff of a periodic currenti(t) is the dc current that delivers the same average power to a resistor asthe periodic current.

11.4.2. Consider the following ac and dc circuits,

our objective is to find the current Ieff that will transfer the same powerto the resistor R as the sinusoid current i

The average power absorbed by the resistor in the ac circuit is

P =1

T

∫ T

0i2(t)Rdt =

R

T

∫ T

0i2(t)dt.

Q: Where don’t we have the factor of 12?

The power absorbed by the resistor in the dc circuit is

P = I2effR.


Equating the two expressions and solving for Ieff, we obtain

Ieff =

√1

T

∫ T

0i2(t)dt

11.4.3. Similarly, the effective value of the voltage is found in the sameway as current; that is,

Veff =

√1

T

∫ T

0v2(t)dt

11.4.4. This indicates that the effective value is the square root of themean (or average) of the square of the periodic signal. Thus, the effectivevalue is often know as the root mean square, or rms value for short.We write

Ieff = Irms, Veff = Vrms

11.4.5. Note that the rms value of a constant is the constant itself.

11.4.6. AC Circuit : For a sinusoid x(t) = Xm cos(ωt+ θx), the effec-tive value or rms value is

Xrms =

√1

T

∫ T

0X2m cos2(ωt+ θx)dt =

Xm√2.

In particular, for i(t) = Im cos(ωt+ θi) and v(t) = Vm cos(ωt+ θv), we have

Irms =Im√

2and Vrms =

Vm√2.

and the average power can be written in terms of the rms values as

P =1

2VmIm cos(θv − θi) = VrmsIrms cos(θv − θi) .

11.4.7. Similarly, the average power absorbed by a resistor R can bewritten as

P = IrmsVrms = I2rmsR =

V 2rms

R.

11.5. APPARENT POWER AND POWER FACTOR 139

11.5. Apparent Power and Power Factor

Definition 11.5.1. The apparent power S (in VA) is the product ofthe rms values of voltage and current.

S = VrmsIrms

Hence, the average power P = S cos(θv − θi).Definition 11.5.2. The power factor (pf) is the ratio of the average

power to the apparent power.

pf =P

S= cos(θv − θi).

Hence,

average power P = apparent power S × power factor pf.

The angle θv − θi is called the power factor angle which is equal tothe angle of the load impedance if V = Vm∠θv is the voltage across theload and I = Im∠θi is the current through it. This is evident from the factthat

Z =V

I=Vm∠θvIm∠θi

=VmIm

∠(θv − θi).

Alternately, define Vrms = V√2

= Vrms∠Θv and Irms = I√2

= Irms∠Θi.

The impedance can then be written as

Z =V

I=

Vrms

Irms=VrmsIrms

∠(θv − θi).

The power factor is the cosine of the phase difference betweenthe voltage and current. It is also the cosine of the angle of theload impedance.

11.5.3. The value of the power factor pf ranges between 0 and 1.

• For a purely resistive load, the voltage and current are in phase, sothat θv − θi = 0 and pf =1. This implies that the average power isequal to the apparent power .• For a purely reactive load, θv − θi = ±90. Hence, pf = 0. In this

case the average power is zero.• In between these two extreme cases, pf is said to be leading or

lagging. Leading power factor means that current leads voltagewhich implies a capacitive load. Lagging power factor meansthat current lags voltage, implying an inductive load.


11.6. Complex Power

The complex power (in VA) S is the product of the rms voltage phasorand the complex conjugate of the rms current phasor. S is a complexquantity whose real part is the real or average power P and imaginarypart is the reactive power Q.

Complex power: S

S = P + jQ =1

2VI∗ = VrmsI

∗rms = VrmsIrms∠(θv − θi) = I2

rmsZ =V 2rms

Z∗.

Note that all previously studied quantities can be derived from the com-plex power. That is,

The apparent power S is the magnitude of the complex power S, i.e.,

S = |S| = VrmsIrms =√P 2 +Q2.

The real or average power P is

P = Re S = S cos(θv − θi) = I2rmsR.

The reactive power Q is

Q = Im S = S sin(θv − θi) = I2rmsX.

The power factor pf is

pf =P

S= cos(θv − θi) = cos(phase of S).

CHAPTER 7

First-Order Circuits

In Chapter 6, we have studied the relationships between current andvoltage of capacitors/inductors. These elements can be used to store energyand release energy when needed. In this chapter, we will see how thevoltage or current behaves during the charging/discharging of these storageelements.

First, we will examine two types of simple circuits with a storage ele-ment:

(a) A circuit with a resistor and one capacitor (called an RC circuit);and

(b) A circuit with a resistor and an inductor (called an RL circuit).

These circuits may look simple but they find continual applications inelectronics, communications, and control systems.

7.0.1. Applying Kirchhoff’s laws to the RC and RL circuits producefirst order differential equations. Hence, the circuits are collectivelyknown as first-order circuits.

7.0.2. There are two ways to excite the circuits.

(a) By initial conditions of the storage elements in the circuit.• Also known as source-free circuits• Assume that energy is initially stored in the capacitive or in-

ductive element.• This is the discharging process.

(b) By using independent sources• This is the charging process• For this chapter, we will consider independent dc sources.

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88 7. FIRST-ORDER CIRCUITS

Before we start our circuit analysis, it is helpful to consider one math-ematical fact which we will use through out this chapter:

7.0.3. The solution of the first-order differential equation

d

dtx(t) = ax+ b

is given by

(7.3) x(t) = ea(t−t0)(x(t0) +

b

a

)− b

a.

In addition, if a < 0, the solution is given by

(7.4) x(t) = ea(t−t0) (x(t0)− x(∞)) + x(∞).

where x(∞) = limt→∞ x(t).The results above take a bit of effort to proof. However, if you have

MATLAB, you can get 7.3 using one line of code:

dsolve(’Dx = a*x+b’,’x(t0) = x0’, ’t’)

When a < 0, from (7.3), x(∞) = −b/a. We can then get (7.4) by substi-tuting b/a in (7.3) with x(∞).

7.0.4. It turns out that you only have two kinds of plots to worry aboutwhen dealing with (7.4):

7.1. SOURCE-FREE RC CIRCUITS 89

7.1. Source-Free RC Circuits

A source-free RC circuit occurs when its dc source is suddenly dis-connected. The energy already stored in the capacitor is released to theresistor.

Consider a series combination of a resistor an initially charged capacitor.We assume that at time t = 0, the initial voltage is v(0) = Vo. (Hence, theinitial stored energy is w(0) = 1

2CV2o .)

Applying KCL, we get

iC + iR = 0

Cdv

dt+v

R= 0

dv

dt+

v

CR= 0

dv

dt=

(− 1

CR

)v

Hence,

v(t) = Voe− tRC = Voe

− tτ ,

where τ = RC.This shows that the voltage response of RC circuit is an exponential

decay of the initial voltage.

• Since the response is due to the initial energy stored and the physicalcharacteristics of the circuit and not due to some external voltageor current source, it is called the natural response of the circuit.


7.1.1. Remarks:

(a) τ = RC is the time constant which is the time required for theresponse to decay to 36.8 percent of its initial value. (e ≈ 2.718 andhence 1/e ≈ 0.368.)

(b) The smaller the time constant, the more rapidly the voltage de-creases, that is, the faster the response.

7.1. SOURCE-FREE RC CIRCUITS 91

(c) The voltage v(t) is less than 1 percent of V0 after 5τ (five timeconstants). Thus, it is customary to assume that the capacitor isfully discharged (or charged) after five time constants. In otherwords, it takes 5τ for the circuit to reach its final state or steadystate when no changes take place with time.

(d) iR(t) = v(t)R = Vo

R e−tτ .

(e) pR(t) = viR = V 2o

R e−2tτ

(f) wR(t) =∫ t

0 p(µ)dµ = 12CV

2o (1− e−2t

τ ) = wC(0)− wC(t)

• Notice that as t → ∞, wR → 12CV

20 , which is the same as

wC(0), the energy initially stored in the capacitor.• The energy that was initially stored in the capacitor is eventu-

ally dissipated in the resistor.

7.1.2. In summary: The key in working with a source-free RC circuitis to determine:

1. The initial voltage v(0) = V0 across the capacitor.2. The time constant τ (= RC)


7.1.3. There are basically three ways that we can extend what we havefound.

(a) The initial condition v(t0) can be given at non-zero t0. This hasalready been taken care of by (7.4).

(b) There can be more than one resistors. In which case, you first sim-plify the circuit by finding the equivalent resistance at the capacitorterminals. See Example 7.1.4.

(c) The initial value of the voltage v(t0) might not be given but condi-tion(s) of the circuit before time t0 is given instead. In which case,you may find v(t−0 ) by considering the long-term behavior of the cir-cuit before t0. In such situation, you should review two propertiesof capacitors:

(i) For DC, capacitor becomes open circuit.(ii) Voltage across the capacitor cannot change instantaneously

See Example 7.1.5.

Example 7.1.4. Let vC(0) = 15 V. Determine vC(t), vx(t), and ix(t) fort > 0.

Remark: The answer is the same if we replace t > 0 in the abovequestion by t ≥ 0.

7.2. SOURCE-FREE RL CIRCUITS 93

In Example 7.1.4, you may ask how can we have the initial voltage of15 V across the capacitor in the first place. The next example suggest away to get the initial voltage.

Example 7.1.5. The switch in the circuit below has been closed for along time, and it is opened at t = 0. Find v(t) for t ≥ 0. Calculate theinitial energy stored in the capacitor.


7.2. Source-Free RL Circuits

Consider the series connection of a resistor and an inductor. Assumethat the inductor has an initial current Io or i(0) = Io.(hence, the energystored in the inductor is w(0) = 1

2LI2o .)

Applying KVL, we get

vL + vR = 0

Ldi

dt+Ri = 0

di

dt+R

Li = 0

Hence,

i(t) = Ioe−RtL = Ioe

−tτ .

Note that:

(a) τ = LR is the time constant which is the time required for the re-

sponse to decay to 36.8 percent of its initial value.(b) vR(t) = iR = IoRe

−tτ .

(c) p(t) = vRi = I2oRe

−2tτ .

(d) wR(t) =∫ t

0 p(µ)dµ = 12LI

2o (1− e−2t

τ ) = wL(0)− wL(t) .

7.2. SOURCE-FREE RL CIRCUITS 95

7.2.1. In summary: The key in working with a source free RL circuitis to determine:

(a) The initial current i(0) = Io.(b) The time constant τ (= L

R).• In general, R is the Thevenin equivalent resistance at the ter-

minal of the inductor.(We take out the inductor L and findR = RTH at its terminals.)

Example 7.2.2. The switch in the circuit below has been closed for along time. At t = 0, the switch is opened. Calculate i(t) for t > 0.


Example 7.2.3. Determine i, io and vo for all t in the circuit. Assumethat the switch was closed for a long time.

7.3. UNIT STEP FUNCTION 97

7.3. Unit step function

We use the step function to represent an abrupt change in voltage orcurrent, like the changes that occur in the circuits of control systems anddigital computers.

The unit step function u(t) is 0 for negative values of t and 1 for positivevalues of t. In mathematical terms,

u(t) =

0, t < 01, t > 0

For example, here is a voltage source of V0u(t)

Similarly, here is a current source of I0u(t)


We can also use the step functions to represent the gate function (pulse)which may be regarded as a step function that switches on at one value oft and switches off at another value of t. The gate function below switcheson at t = 2 s and switches off at t = 5 s.

It consists of the sum of two unit step functions:

v(t) = 10u(t− 2)− 10u(t− 5).

When the dc source of an RC circuit is suddenly applied (i.e., thishappens when the capacitor is being charged), the voltage or current sourcecan be modeled as a step function, and the response is known as a stepresponse.

7.4. STEP RESPONSE OF AN RC CIRCUIT 99

7.4. Step Response of an RC Circuit

7.4.1. Consider an RC circuit with voltage step input below:

The voltage across the capacitor is v. We assume an initial voltageV0 on the capacitor. Since the voltage of a capacitor cannot change in-stantaneously v(0−) = v(0+) = V0, where v(0−) is the voltage across thecapacitor just before switching and v(0+) is its voltage immediately afterswitching.

For t > 0, applying KCL, we have

Cdv

dt+v − VsR

= 0

dv

dt+

v

RC=

VsRC

dv

dt=

(− 1

RC

)v +

VsRC

The solution isv(t) = Vs + (Vo − Vs)e

−tτ , t ≥ 0.


7.4.2. In general, the step response of an RC circuit whose switchchanges position at time t = 0 can be written as

v(t) = v(∞) + (v(0)− v(∞))e−tτ , t > 0.

where v(0) is the initial voltage and v(∞) is the final or steady-state value.We obtain

(a) v(0) from the given circuit for t < 0 and(b) v(∞) and τ from the circuit for t > 0.

7.4.3. If the switch changes position at time t = to instead of at t = 0,the response becomes

v(t) = v(∞) + (v(to)− v(∞))e−(t−to)

τ , t > t0

where v(to) is the initial value at t = to.

7.4.4. The key in finding the step response of an RC circuit is to de-termine three values:

1. The initial capacitor voltage v(0) or V (t0).2. The final capacitor voltage v(∞).3. The time constant τ .


Example 7.4.5. The switch in the circuit below has been in position Afor a long time. At t = 0, the switch moves to B. Determine v(t) for t > 0and calculate its value at t = 1 s and 4 s.

Example 7.4.6. Find v(t) for t > 0 in the circuit below. Assume theswitch has been open for a long time and is closed at t = 0. Numericallyevaluate v(t) at t = 0.5.


Example 7.4.7. An electronic flash unit provides a common exampleof an RC circuit. This application exploits the ability of the capacitorto oppose any abrupt change in voltage. A simplified circuit is shownbelow. It consists essentially of a high-voltage dc supply, a current-limitinglarge resistor R1, and a capacitor C in parallel with the flashlamp of lowresistance R2.

• When the switch is in position 1, the capacitor charges slowly dueto the large time constant (τ = R1C). The capacitor voltage risesgradually from zero to Vs , while its current decreases gradually fromI1 = Vs/R1 to zero.• With the switch in position 2, the capacitor voltage is discharged.

The low resistance R2 of the photolamp permits a high dischargecurrent with peak I2 = Vs/R2 in a short duration.

This simple RC circuit provides a short-duration, high current pulse.Such a circuit also finds applications in electric spot welding and the radartransmitter tube.


Example 7.4.8. An RC circuit can be used to provide various timedelays. Consider the circuit below. It basically consists of an RC circuitwith the capacitor connected in parallel with a neon lamp. The voltagesource can provide enough voltage to fire the lamp.

• When the switch is closed, the capacitor voltage increases graduallytoward 110 V at a rate determined by the circuit’s time constant,(R1 +R2)C. The lamp will act as an open circuit and not emit lightuntil the voltage across it exceeds a particular level, say 70 V.• When the voltage level is reached, the lamp fires (goes on), and

the capacitor discharges through it. Due to the low resistance ofthe lamp when on, the capacitor voltage drops fast and the lampturns off. The lamp acts again as an open circuit and the capacitorrecharges.• By adjusting R2, we can introduce either short or long time delays

into the circuit and make the lamp fire, recharge, and fire repeatedlyevery time constant τ = (R1 +R2)C, because it takes a time periodτ to get the capacitor voltage high enough to fire or low enough toturn off.• The warning blinkers commonly found on road construction sites

are one example of the usefulness of such an RC delay circuit.


7.5. Step Response of an RL Circuit

Here, our goal is to find the inductor current i as the circuit response.

7.5.1. In general, the step response of an RL circuit whose switchchanges position at time t = 0 can be written as

i(t) = i(∞) + (i(0)− i(∞))e−tτ , t > 0.

where i(0) is the initial current and i(∞) is the final or steady-state value.If the switch changes position at time t = to instead of at t = 0, theresponse becomes

i(t) = i(∞) + (i(to)− i(∞))e−(t−to)

τ , t > t0

where v(to) is the initial inductor current value at time t = to.

Example 7.5.2. Find i(t) in the circuit below for t > 0. Assume thatthe switch has been closed for a long time.

7.5. STEP RESPONSE OF AN RL CIRCUIT 105

Example 7.5.3. The switch in the circuit below has been closed for along time. It opens at t = 0. Find i(t) for t > 0.

Example 7.5.4. At t = 0, switch 1 in the circuit below is closed, andswitch 2 is closed 4 s later. Find i(t) for t > 0.

CHAPTER 8

Second-Order Circuits

In this chapter, we will consider circuits containing two storage ele-ments.1 They are known as second-order circuits because their re-sponses are described by differential equations that contain second deriva-tives.

Definition 8.0.5. A second-order circuit is characterized by asecond-order differential equation. It consists of resistors and theequivalent of two energy storage elements.

8.0.6. Second-order differential equations that we will focus on will havethe following general form:

(8.5)d2

dt2x(t) + c1

d

dtx(t) + c2x(t) = f(t).

In this chapter2 f(t) will be some constant C. Hence, (8.5) becomes

(8.6)d2

dt2x(t) + c1

d

dtx(t) + c2x(t) = C.

The solution to (8.6) has two components: the transient response xt(t) andthe steady-state response xss(t); that is;

(8.7) x(t) = xt(t) + xss(t).

(a) The transient response xt(t) is the component of the total re-sponse that dies out with time. To get this, we find the the solutionof (8.6) when C = 0. This part of the solution will have two con-stants which can be determined by the initial conditions.

(b) It turns out that for (8.6), the steady-state response xss(t) isgiven by C/c2. However, we will use that fact that as t → ∞, thetransient response xt(t) will die out and hence we can use for findxss(t) from x(∞).

1A second-order circuit may have two storage elements of different type or the same type (providedelements of the same type cannot be represented by an equivalent single element).

2In Chapter 9 and 10, we work with (8.5) with sinusoidal f(t). Moreover, we shall limit our interestto the steady-state part of the solution.

103

Din

Text Box


104 8. SECOND-ORDER CIRCUITS

8.1. Finding Initial and Final Values

This section is explicitly devoted to the subtleties of getting v(0), i(0),dv(0)dt , di(0)

dt , i(∞) and v(∞). Unless otherwise stated in this chapter, vdenotes capacitor voltage, while i is inductor current. We will write vCand iL when we want to be explicit.

8.1.1. There are two key points to keep in mind in determining theinitial conditions.

(a) v and i are defined strictly according to the passive sign convention.(b) There is no jump in the capacitor’s voltage and the inductor’s cur-

rent:(i) the capacitor voltage is always continuous so that

vC(t+0 ) = vC(t0) = vC(t−0 ),

and(ii) the inductor current is always continuous so that

iL(t+0 ) = iL(t0) = iL(t−0 ),

where t = t−0 denotes the time just before the (switching) event andt = t+0 is the time just after the (switching) event, assuming thatthe (switching) event takes place at time t = t0.

Example 8.1.2. The switch in the figure below has been closed for along time. It is open at t = 0. Find i(0+), v(0+), d

dti(0+), d

dtv(0+), i(∞),and v(∞).

8.2. GENERAL STEP RESPONSE OF SECOND-ORDER CIRCUITS 105

8.2. General Step Response of Second-order circuits

8.2.1. Given a second-order circuit, we determine its step responsex(t) (which may be capacitor-voltage or inductor-current) by taking thefollowing four steps.

Step 1: Determine the initial conditions x(0) and dx(0)dt and the final value

x(∞).Step 2: Find the natural response xn(t) by turning off independent sources

and applying KCL and KVL. Once a second-order differential equa-tion is obtained, we determine its characteristic roots. As a generalrule, you will confront with a differential equation of the form

d2

dt2x(t) + 2α

d

dtx(t) + ω2

0x(t) = 0

whose general form of the characteristic equation is

s2 + 2αs+ ω2o = 0.

The two roots of the characteristic equation are

s1 = −α +√α2 − ω2

0, s2 = −α−√α2 − ω2

0.

We can infer that there are three types of solutions:(i) If α > ωo, we have the overdamped case,

xn(t) = A1es1t + A2e

s2t.

(ii) If α = ωo, we have the critically damped case,

xn(t) = (A2 + A1t)e−αt.

(iii) If α < ωo, we have the underdamped case,

xn(t) = e−αt(B1 cosωdt+B2 sinωdt),

where ωd =√ω2

0 − α2.Depending on whether the response is overdamped (distinctroots), critically damped (repeated roots), or underdamped(complex conjugated roots), we obtain xn(t) with two unknownconstants.

Step 3: We obtain the force response as

xf(t) = x(∞).

106 8. SECOND-ORDER CIRCUITS

Step 4: The total response is now found as the sum of the natural responseand the forced response.

x(t) = xn(t) + xf(t).

We finally determine the constants associated with the natural

response by imposing the initial conditions x(0) and dx(0)dt , deter-

mined in Step 1.

Example 8.2.2. Find the complete response v for t > 0 in the circuit.

Vs 12:= ω 1:=

R1 4:= L 1:=

C12

:= R2 2:=

t0 0:= t1 2:=

Given

v1 t( )tv0 t( )d

d=

tv1 t( )d

d1

C R2⋅

R1L

+⎛⎜⎝

⎞⎟⎠

v1 t( )⋅+1

L C⋅1

R1R2

+⎛⎜⎝

⎞⎟⎠

⋅⎡⎢⎣

⎤⎥⎦

v0 t( )⋅+1

L C⋅Vs⋅=

Initial Conditions: v0 t0( ) 12= v1 t0( ) 12−=

v0

v1⎛⎜⎝

⎞⎟⎠

Odesolvev0

v1⎛⎜⎝

⎞⎟⎠

t, t1, ⎡⎢⎣

⎤⎥⎦

:=

v t( ) v0 t( ):=

0 0.5 1 1.5 20

5

10

v t( )

4 12 e 2− t⋅⋅+ 4 e 3− t⋅⋅−

t

Vs 12:= ω 1:=

R1 4:= L 1:=

C12

:= R2 2:=

vs t( ) Vs cos ω t⋅( )⋅:=t0 0:= t1 30:=

Given

v1 t( )tv0 t( )d

d=

tv1 t( )d

d1

C R2⋅

R1L

+⎛⎜⎝

⎞⎟⎠

v1 t( )⋅+1

L C⋅1

R1R2

+⎛⎜⎝

⎞⎟⎠

⋅⎡⎢⎣

⎤⎥⎦

v0 t( )⋅+1

L C⋅vs t( )⋅=

Initial Conditions: v0 t0( ) 12= v1 t0( ) 12−= (assumed)

v0

v1⎛⎜⎝

⎞⎟⎠

Odesolvev0

v1⎛⎜⎝

⎞⎟⎠

t, t1, ⎡⎢⎣

⎤⎥⎦

:=

v t( ) v0 t( ):=

0 10 20 305−

0

5

10

v t( )

4 12 e 2− t⋅⋅+ 4 e 3− t⋅⋅−

t

AC Analysis

R1 4:=

L 1:=

C12

:=

R2 2:=

parallel x y, ( )x y⋅

x y+:= ZC

1j ω⋅ C⋅

:= ZL j ω⋅ L⋅:=

gparallel R2 ZC, ( )

parallel R2 ZC, ( ) R1+ ZL+:= g

15

15

⎛⎜⎝

⎞⎟⎠

i⋅−→

g225

→ arg g( )180π

⋅ 45−=

0 10 20 305−

0

5

10

v t( )

g Vs⋅ cos ω t⋅ arg g( )+( )⋅

t

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