Electric Circuits, 9th Edition P39 pptx

Average power is sometimes called real power, because it describes the power in a circuit that is transformed from electric to nonelectric energy.. The average power associated with s

9.67 The op amp in the circuit seen in Fig P9.67 is ideal

PSPICE Find the steady-state expression for v (>(t) when

- = 2 c o s l ( / Y V

Figure P9.67

too kn

40 kO

9.68 The op amp in the circuit in Fig P9.68 is ideal

MULTISIM a) ^n <^ ^e s t e ady-state expression for v 0 (t)

b) How large can the amplitude of v g be before the

amplifier saturates?

Figure P9.68

vg = 25 cos 50,000* V

9.69 The sinusoidal voltage source in the circuit shown in

PSPICE pig P9.69 is generating the voltage v„ = 4 cos 200r V

If the op amp is ideal, what is the steady-state

expres-sion for v 0 (t)1

Figure P9.69

10 kO

20 kQ 20 kH

-f V W

9.70 The 250 nF capacitor in the circuit seen in Fig P9.69

PSPICE is replaced with a variable capacitor The capacitor

MULTISIM -s acjjUSTec] Uxitil the output voltage leads the input

voltage by 135°

a) Find the value of C in microfarads

b) Write the steady-state expression for v () (t) when

C has the value found in (a)

9.71 The operational amplifier in the circuit shown in

PSPICE Fig P9.71 is ideal The voltage of the ideal

sinu-MULTISIM • , 1 i r » i r\f\* t /

soida 1 source is v g = 30 cos 10°t V

a) How small can C a be before the steady-state output voltage no longer has a pure sinusoidal waveform?

b) For the value of C 0 found in (a), write the

steady-state expression for v a

Figure P9.71

10 nF

loo a

loo a

9.72 a) Find the input impedance Za b for the circuit in

Fig P9.72 Express Za b as a function of Z and K where K = (R 2 /R\)

b) If Z is a pure capacitive element, what is the capacitance seen looking into the terminals a,b?

Figure P9.72

9.73 For the circuit in Fig P9.73 suppose

vt = 20 cos(2000f - 36.87°) V v2 = 10cos(5000/ + 16.26°) V

a) What circuit analysis technique must be used to

find the steady-state expression for v v (t)1 b) Find the steady-state expression for v C)(t)

Problems 357

Figure P9.73

l m H / Y Y Y V

100 |xF

If

10 a

9.74 For the circuit in Fig P9.61, suppose

v.d = 5 cos 80,000/ V

vb = - 2 5 cos 320,000/ V

b) Find the coefficient of coupling

c) Find the energy stored in the magnetically

cou-pled coils at t = 1007T /xs and t = 200-7T ^ s

Figure P9.77

30 a

a) What circuit analysis technique must be used to

find the steady-state expression for j„(r)?

b) Find t h e steady-state expression for /0(/)?

9.78 For t h e circuit in Fig P9.78, find t h e Thevenin equivalent with respect t o t h e terminals c,d

Section 9.10

9.75 A series combination of a 300 O resistor a n d a

100 m H inductor is connected to a sinusoidal

volt-age source by a linear transformer T h e source is

operating at a frequency of 1 k r a d / s A t this

fre-quency, t h e internal i m p e d a n c e of t h e source is

100 + /13.74 CI T h e r m s voltage at t h e terminals of

the source is 50 V w h e n it is n o t loaded T h e p a r a m

-eters of t h e linear transformer a r e R\ = 41.68 O ,

L { = 180 m H , R 2 = 500 ft, L 2 = 500 m H , a n d

M = 270 m H

a) What is the value of the impedance reflected

into the primary?

b) W h a t is t h e value of t h e i m p e d a n c e seen from

the terminals of t h e practical source?

9.76 T h e sinusoidal voltage source in t h e circuit seen in

PSPICE Fig P9.76 is operating at a frequency of 200 krad/s

LTISIM ^ e c o ef f }cje t lt 0f coupling is adjusted until t h e

p e a k amplitude of i x is m a x i m u m

a) What is the value of kl

b) W h a t is the p e a k a m p l i t u d e of /j if

v g = 560 cos(2 X 1 0 ¾ V ?

Figure P9.76

150 n so a loo a 2oo a

•—vw—i

12.5 nF

9.77 a) Find t h e steady-state expressions for t h e

cur-rents ig a n d i L in t h e circuit in Fig P9.77 w h e n

PSPICE

MU LTISIM

vg = 70 cos 5000/ V

Figure P9.78

425/0°

45 a

-WV-V (rms)

9.79 T h e value of k in t h e circuit in Fig P9.79 is adjusted

so that Za b is purely resistive when a> = 4 k r a d / s Find Z ab

Figure P9.79

a»-20 a

^VW-12.5 m H !8mH

5 a

-WW

12.5 /JLF

Section 9.11

9.80 At first glance, it may appear from Eq 9.69 that an

inductive load could make the reactance seen

look-ing into the primary terminals (i.e., X ah ) look

capac-itive Intuitively, we know this is impossible Show that X)b can never be negative if X L is an inductive reactance

9.81 a) Show that t h e i m p e d a n c e seen looking into t h e

terminals a,b in the circuit in Fig P9.81 o n t h e next p a g e is given by the expression

<ab

b) Show that if the polarity terminals of either o n e

of the coils is reversed,

-ab

Figure P9.81

a «

Z,

/V,

A'\ Z,

9.82 a) Show that the i m p e d a n c e seen looking into the

terminals a,b in the circuit in Fig P9.82 is given

by the expression

Zab

-Z L

1 +

N-,

b) Show that if the polarity terminal of either o n e

of the coils is reversed that

'ab

1 - ^

Figure P9.82

/V,

Z a b"

b *

AT,:

infinity T h e amplitude and phase angle of the

source voltage are held constant as R x varies

Figure P9.84

>\ = V,n c o s U) '

^wv

9.83 Find the i m p e d a n c e Za b in the circuit in Fig P9.83 if

Z L = 8 0 / 6 0 ' H

R,

9.85 The parameters in the circuit shown in Fig 9.53 are

/?, = 0.1 il,o)L x = 0.8 ft,fl 2 = 24 il,(oL 2 = 32 ft,

a n d V L = 240 + / 0 V

a) Calculate the phasor voltage V s b) Connect a capacitor in parallel with the inductor, hold V L constant, and adjust the capacitor until the magnitude of I is a minimum What is the capacitive reactance? What is the value of V v ? c) Find the value of the capacitive reactance that

k e e p s the m a g n i t u d e of I as small as possible

a n d that at the same time m a k e s

lYvl = |V/J = 240 V

9.86 a) For the circuit shown in Fig P9.86, c o m p u t e Vv

and V/

b) Construct a p h a s o r diagram showing the rela-tionship between V s , V/, and the load voltage of

2 4 0 / 0 ° V

c) R e p e a t p a r t s (a) a n d (b), given that the load voltage remains constant at 240 / 0 ° V , w h e n a

capacitive r e a c t a n c e of - 5 Cl is connected

across the load terminals

Figure P9.86

+ Vj_

+ 0.1 Q~ /0.8 Q +

v , 240/0° v i s a

Figure P9.83

a«

1/6 n -pft;f;

b «

-8:1

Ideal

• 10:1

Ideal

Z

Section 9.12

9.84 Show by using a p h a s o r diagram what h a p p e n s t o

PSPICE the m a g n i t u d e and phase angle of the voltage v„ in

MULTISIM t h e c i r c u i t i n F i g p 9 8 4 a s R ^ j s y a r i c d f r o m z e r o t Q

Sections 9.1-9.12 9.87 You may have the opportunity as an engineering

graduate to serve as an expert witness in lawsuits involving either personal injury or property damage

A s an example of the type of problem on which you may be asked to give an opinion, consider the follow-ing event A t the end of a day of fieldwork, a farmer returns to his farmstead, checks his hog confinement building, and finds to his dismay that the hogs are

Problems 359

dead The problem is traced to a blown fuse that

caused a 240 V fan motor to stop The loss of

ventila-tion led to the suffocaventila-tion of the livestock The

inter-rupted fuse is located in the main switch that

connects the farmstead to the electrical service

Before the insurance company settles the claim, it

wants to know if the electric circuit supplying the

farmstead functioned properly The lawyers for the

insurance company are puzzled because the farmer's

wife, who was in the house on the day of the accident

convalescing from minor surgery, was able to watch

TV during the afternoon Furthermore, when she

went to the kitchen to start preparing the evening

meal, the electric clock indicated the correct time The

lawyers have hired you to explain (1) why the electric

clock in the kitchen and the television set in the living

room continued to operate after the fuse in the main

switch blew and (2) why the second fuse in the main

switch didn't blow after the fan motor stalled After

ascertaining the loads on the three-wire

distribu-tion circuit prior to the interrupdistribu-tion of fuse A, you

are able to construct the circuit model shown in

Fig P9.87 The impedances of the line conductors

and the neutral conductor are assumed negligible

a) Calculate the branch currents It, I2, I3, I4, I5,

and I6 prior to the interruption of fuse A

b) Calculate the branch currents after the

interrup-tion of fuse A Assume the stalled fan motor

behaves as a short circuit

c) Explain why the clock and television set were

not affected by the momentary short circuit that

interrupted fuse A

d) Assume the fan motor is equipped with a

ther-mal cutout designed to interrupt the motor

cir-cuit if the motor current becomes excessive

Would you expect the thermal cutout to

oper-ate? Explain

e) Explain why fuse B is not interrupted when the

fan motor stalls

Figure P9.87

Fuse A (100 A)

120

V 'FQ

Momentary '

short

circuit X

interrupts

fuse A

120

V FQ

-*\fi-9.88 a) Calculate the branch currents I]-I<s in the

cir-pRAcncAL cuit in Fie 9.58

b) Find the primary current Ip 9.89 Suppose the 40 ft resistance in the distribution cir-pRAcncAL cuit in Fie 9.58 is replaced bv a 20 ft resistance

a) Recalculate the branch current in the 2 (1 resistor, I2

b) Recalculate the primary current, Ip c) On the basis of your answers, is it desirable

to have the resistance of the two 120 V loads

be equal?

9.90 A residential wiring circuit is shown in Fig P9.90 In

PRACTICAL this model, the resistor Ri, is used to model a 250 V

PERSPECTIVE

appliance (such as an electric range), and the

resis-tors R] and R 2 are used to model 125 V appliances (such as a lamp, toaster, and iron) The branches carrying ^ and I2 are modeling what electricians refer to as the hot conductors in the circuit, and the

branch carrying \„ is modeling the neutral

conduc-tor Our purpose in analyzing the circuit is to show the importance of the neutral conductor in the sat-isfactory operation of the circuit You are to choose the method for analyzing the circuit

a) Show that l n is zero if R^ = R 2

b) Show that V! = V2 if Ri = R 2

c) Open the neutral branch and calculate Vi and V2

if R } = 40 ft, R2 = 400 ft, and R 3 = 8 ft

d) Close the neutral branch and repeat (c)

e) On the basis of your calculations, explain why the neutral conductor is never fused in such a manner that it could open while the hot conduc-tors are energized

Figure P9.90

+ •

14./0°

kV-

-VAr-• + 0.02x2 /0.02 n 125/0° V V, £ /?,

-A<W

Ideal

• + 0.03X1 125/0° V

/0.03 (1

m m

«-— L + K,fV3

0.02 a /0.02 a

—-WV 1 - ^ 0 ^

•-9.91 a) Find the primary current Ip for (c) and (d) in P™ nw„r PRACTICAL Problem 9.90

K m m o a n PERSPECTIVE

Fuse B( 100 A)

b) Do your answers make sense in terms of known circuit behavior?

CHAPTER

C H A P T E R C O N T E N T S

10.1 Instantaneous Power p 362

10.2 Average and Reactive Power p 363

10.3 The rms Value and Power

Calculations p 368

10.4 Complex Power p 370

10.5 Power Calculations p 371

10.6 Maximum Power Transfer p 375

^ C H A P T E R O B J E C T I V E S

1 Understand the following ac power concepts,

their relationships to one another, and how to

calculate them in a circuit:

Instantaneous power;

Average (real) power;

Reactive power;

Complex power; and

Power factor

Understand the condition for maximum real

power delivered to a load in an ac circuit and be

able to calculate the load impedance required to

deliver maximum real power to the load

Be able to calculate all forms of ac power in

ac circuits with linear transformers and in

ac circuits with ideal transformers

360

Sinusoidal Steady-State Power Calculations

Power engineering has evolved into one of the important

sub-disciplines within electrical engineering The range of problems dealing with the delivery of energy to do work is considerable, from determining the power rating within which an appliance operates safely and efficiently, to designing the vast array of gen-erators, transformers, and wires that provide electric energy to household and industrial consumers

Nearly all electric energy is supplied in the form of sinusoidal voltages and currents Thus, after our Chapter 9 discussion of sinusoidal circuits, this is the logical place to consider sinusoidal steady-state power calculations We are primarily interested in the average power delivered to or supplied from a pair of termi-nals as a result of sinusoidal voltages and currents Other meas-ures, such as reactive power, complex power, and apparent power, will also be presented The concept of the rms value of a sinusoid, briefly introduced in Chapter 9, is particularly pertinent

to power calculations

We begin and end this chapter with two concepts that should

be very familiar to you from previous chapters: the basic equa-tion for power (Secequa-tion 10.1) and maximum power transfer (Section 10.6) In between, we discuss the general processes for analyzing power, which will be familiar from your studies in Chapters 1 and 4, although some additional mathematical tech-niques are required here to deal with sinusoidal, rather than dc, signals

w -_.:

Practical Perspective

Heating Appliances

In Chapter 9 we calculated the steady-state voltages and

cur-rents in electric circuits driven by sinusoidal sources In this

chapter we consider power in such circuits The techniques we

develop are useful for analyzing many of the electrical devices

we encounter daily, because sinusoidal sources are the

pre-dominant means of providing electric power in our homes,

schools, and businesses

One common class of electrical devices is heaters, which

transform electric energy into thermal energy Examples include

electric stoves and ovens, toasters, irons, electric water

heaters, space heaters, electric clothes dryers, and hair dryers

One of the critical design concerns in a heater is power

con-sumption Power is important for two reasons: The more power

a heater uses, the more it costs to operate, and the more heat

i t can produce

Many electric heaters have different power settings corre-sponding to the amount of heat the device supplies You may wonder just how these settings result in different amounts of heat output The Practical Perspective example at the end of this chapter examines the design of a handheld hair dryer with three operating settings (see the accompanying figure) You will see how the design provides for three different power levels, which correspond to three different levels of heat output

Heater tube

Fan and motor

Hot air

361

10.1 Instantaneous Power

l

—*"

+

V

Figure 10.1 A The black box representation of a circuit

used for calculating power

We begin our investigation of sinusoidal power calculations with the

familiar circuit in Fig 10.1 Here, v and /' are steady-state sinusoidal signals

Using the passive sign convention, the power at any instant of time is

VI (10.1)

This is instantaneous power Remember that if the reference direction of the current is in the direction of the voltage rise, Eq 10.1 must be written with a minus sign Instantaneous power is measured in watts when the voltage is in volts and the current is in amperes First, we write expressions

for v and i;

v = V„, cos (cot + 0j,),

i — I„, cos {ait -I- 0,),

(10.2)

(10.3)

where 0,, is the voltage phase angle, and 0-, is the current phase angle

We are operating in the sinusoidal steady state, so we may choose any convenient reference for zero time Engineers designing systems that transfer large blocks of power have found it convenient to use a zero time corresponding to the instant the current is passing through a positive max-imum This reference system requires a shift of both the voltage and cur-rent by 0,- Thus Eqs 10.2 and 10.3 become

v = Vm cos (ait + 0,, - 0,),

i = 1,,, cos cot

(10.4) (10.5)

When we substitute Eqs 10.4 and 10.5 into Eq 10.1, the expression for the instantaneous power becomes

p = VmIm cos {cot + 0 V - 0j) cos cot (10.6)

We could use Eq 10.6 directly to find the average power; however, by sim-ply apsim-plying a couple of trigonometric identities, we can put Eq 10.6 into

a much more informative form

We begin with the trigonometric identity1

1

cos a cos /3 = — cos (a /3) + - c o s ( a + /3)

to expand Eq 10.6; letting a = cot + 0,, — 0, and fS = cot gives

p = — — cos (6 V - 0,) + —r— cos {loot + 0„ - 0,-) (10.7)

Now use the trigonometric identity

cos (a + /3) = cos a cos /3 — sin a sin (3

See entry 8 in Appendix F

to expand the second term on the right-hand side of Eq 10.7, which gives

p = — — cos (6V - 6-) H — cos (0,, - 0,) cos 2(ot

V I

Figure 10.2 depicts a representative relationship among v, i, and p,

based on the assumptions 0.,, - 60° and 6-, = 0° You can see that the

fre-quency of the instantaneous power is twice the frefre-quency of the voltage or

current This observation also follows directly from the second two terms

on the right-hand side of Eq 10.8 Therefore, the instantaneous power

goes through two complete cycles for every cycle of either the voltage or

the current Also note that the instantaneous power may be negative for a

portion of each cycle, even if the network between the terminals is passive

In a completely passive network, negative power implies that energy

stored in the inductors or capacitors is now being extracted The fact that

the instantaneous power varies with time in the sinusoidal steady-state

operation of a circuit explains why some motor-driven appliances (such as

refrigerators) experience vibration and require resilient motor mountings

to prevent excessive vibration

We are now ready to use Eq 10.8 to find the average power at the

ter-minals of the circuit represented by Fig 10.1 and, at the same time,

intro-duce the concept of reactive power

Figure 10.2 • Instantaneous power, voltage, and current versus vt for

steady-state sinusoidal operation

10.2 Average and Reactive Power

We begin by noting that Eq 10.8 has three terms, which we can rewrite as

follows:

p = P + Pcos2wt - Qs'm2a)t, (10.9)

where

P is called the average power, and Q is called the reactive power Average

power is sometimes called real power, because it describes the power in a

circuit that is transformed from electric to nonelectric energy Although

the two terms are interchangeable, we primarily use the term average power in this text

It is easy to see why P is called the average power The average power

associated with sinusoidal signals is the average of the instantaneous power over one period, or, in equation form,

where T is the period of the sinusoidal function The limits on Eq 10.12 imply that we can initiate the integration process at any convenient time t {)

but that we must terminate the integration exactly one period later (We

could integrate over nT periods, where n is an integer, provided we multi-ply the integral by \fnT.)

We could find the average power by substituting Eq 10.9 directly into

Eq 10.12 and then performing the integration But note that the average value of/? is given by the first term on the right-hand side of Eq 10.9,

because the integral of both cos 2cot and sin 2eot over one period is zero

Thus the average power is given in Eq 10.10

We can develop a better understanding of all the terms in Eq 10.9 and the relationships among them by examining the power in circuits that are purely resistive, purely inductive, or purely capacitive

0.01 0.015 Time (s)

0.025

Figure 10.3 • Instantaneous real power and average

power for a purely resistive circuit

Power for Purely Resistive Circuits

If the circuit between the terminals is purely resistive, the voltage and

cur-rent are in phase, which means that $ v = 0, Equation 10.9 then reduces to

The instantaneous power expressed in Eq 10.13 is referred to as the

instantaneous real power Figure 10.3 shows a graph of Eq 10.13 for a

representative purely resistive circuit, assuming co = 377 rad/s By

defini-tion, the average power, P, is the average of/; over one period Thus it is

easy to see just by looking at the graph that P = 1 for this circuit Note

from Eq 10.13 that the instantaneous real power can never be negative, which is also shown in Fig 10.3 In other words, power cannot be extracted from a purely resistive network Rather, all the electric energy is dissi-pated in the form of thermal energy

Power for Purely Inductive Circuits

If the circuit between the terminals is purely inductive, the voltage and current are out of phase by precisely 90° In particular, the current lags the

voltage by 90° (that is, 6-, = $ v - 9 0 ' ) ; therefore 6,, - 6-, = +90° The

expression for the instantaneous power then reduces to

10,2 Average and Reactive Power 365

In a purely inductive circuit, the average power is zero Therefore no

transformation of energy from electric to nonelectric form takes place

The instantaneous power at the terminals in a purely inductive circuit is

continually exchanged between the circuit and the source driving the

cir-cuit, at a frequency of 2co In other words, when p is positive, energy is

being stored in the magnetic fields associated with the inductive elements,

and when p is negative, energy is being extracted from the magnetic fields

A measure of the power associated with purely inductive circuits is

the reactive power Q.The name reactive power comes from the

character-ization of an inductor as a reactive element; its impedance is purely

reac-tive Note that average power P and reactive power Q carry the same

dimension.To distinguish between average and reactive power, we use the

units watt (W) for average power and var (volt-amp reactive, or VAR) for

reactive power Figure 10.4 plots the instantaneous power for a

represen-tative purely inductive circuit, assuming u> = 311 rad/s and Q = 1 VAR

Power for Purely Capacitive Circuits

If the circuit between the terminals is purely capacitive, the voltage and

current are precisely 90° out of phase In this case, the current leads the

voltage by 90° (that is, B t = 6V + 90°); thus, 0V - 0,- = - 9 0 ° The

expres-sion for the instantaneous power then becomes

Again, the average power is zero, so there is no transformation of energy

from electric to nonelectric form In a purely capacitive circuit, the power

is continually exchanged between the source driving the circuit and the

electric field associated with the capacitive elements Figure 10.5 plots the

instantaneous power for a representative purely capacitive circuit,

assum-ing (o = 377 rad/s and Q = - 1 VAR

Note that the decision to use the current as the reference leads to Q

being positive for inductors (that is, $ v — 0,: = 90° and negative for

capac-itors (that is, 6 V - 0, = - 9 0 ° Power engineers recognize this difference in

the algebraic sign of Q by saying that inductors demand (or absorb)

mag-netizing vars, and capacitors furnish (or deliver) magmag-netizing vars.We say

more about this convention later

o

Q, Q (VAR)

§ 0 0.005 0.01 0.015 0.02 0.025

| Time (s)

Figure 10.4 • Instantaneous real power, average

power, and reactive power for a purely inductive circuit

a 0 0.005 0.01 0.015 0.02 0.025

Time (s)

Figure 10.5 • Instantaneous real power and average

power for a purely capacitive circuit

The Power Factor

The angle 6 V - 0, plays a role in the computation of both average and

reactive power and is referred to as the power factor angle The cosine of

this angle is called the power factor, abbreviated pf, and the sine of this

angle is called the reactive factor, abbreviated rf Thus

rf = sin (0,, - 0/) (10.17) Knowing the value of the power factor does not tell you the value of the

power factor angle, because cos (0,, - 0() = cos (0, - 0„) To completely

describe this angle, we use the descriptive phrases lagging power factor and

leading power factor Lagging power factor implies that current lags

volt-age—hence an inductive load Leading power factor implies that current

leads voltage—hence a capacitive load Both the power factor and the

reac-tive factor are convenient quantities to use in describing electrical loads

Example 10.1 illustrates the interpretation of P and Q on the basis of

a numerical calculation