Electric Circuits, 9th Edition P23 doc

Energy Calculations We conclude our first look at mutual inductance with a discussion of the total energy stored in magnetically coupled coils... During this time interval, the voltage

shown in Fig 6.29 Again, the polarity reference assigned to Vy is based on

the dot convention

1 ( T ^ H + I

>l jLL> i /*TN The total flux linking coil 2 is

Figure 6.29 • The magnetically coupled coils of

Fig 6.28, with coil 2 excited and coil 1 open

<t>2 = 4>22 + <Ai2- (6.46)

The flux 4> 2 and its components (f> 22 and <pn are related to the coil current

h as follows:

(f)2 = 8^2^2*2»

(f> 22 = <3> 22 N 2 i 2 ,

012 =

^12^2/2-(6.47)

(6.48)

(6.49)

Tlie voltages v 2 and Vj are

(6.50)

(6.51)

The coefficient of mutual inductance that relates the voltage induced in coil

1 to the time-varying current in coil 2 is the coefficient of di 2/dt in Eq 6.51:

For nonmagnetic materials, the permeances 0^ 2 a nd ^21 a r e equal, and therefore

Hence for linear circuits with just two magnetically coupled coils, attach-ing subscripts to the coefficient of mutual inductance is not necessary

Mutual Inductance in Terms of Self-Inductance

The value of mutual inductance is a function of the self-inductances We derive this relationship as follows From Eqs 6.42 and 6.50,

L 2 = Np 2 ,

(6.54)

(6.55)

respectively From Eqs 6.54 and 6.55,

— \j2\rZc

6.5 A Closer Look at Mutual Inductance 197

We now use Eq 6.41 and the corresponding expression for ?P 2 t o write

L,L 2 = NM(®n + ^ 2 1 ) ( ¾ + ^12) (6.57)

But for a linear system, 0>2i = ^12, so Eq 6.57 becomes

=«•('*&)('* a )

Replacing the two terms involving permeances by a single constant

expresses Eq 6.58 in a more meaningful form:

* - ( • • & ) ( • • & }

Substituting Eq 6.59 into Eq 6.58 yields

M2 = k2L}L2

or

inductance using coupling coefficient

where the constant k is called the coefficient of coupling According to

Eq 6.59, l/k 2 must be greater than 1, which means that k must be less than 1

In fact, the coefficient of coupling must lie between 0 and 1, or

0 < k < 1 (6.61)

The coefficient of coupling is 0 when the two coils have no common

flux; that is, when <£12 = ¢21 = 0 This condition implies that 8P12 = 0, and

Eq 6.59 indicates that l/k 2 = 001 or k — 0 If there is no flux linkage

between the coils, obviously M is 0

The coefficient of coupling is equal to 1 when c/>n and <f> 22 are O.This

condition implies that all the flux that links coil 1 also links coil 2 In terms

of Eq 6.59, SPJI = ^ 2 2 = 0, which obviously represents an ideal state; in

reality, winding two coils so that they share precisely the same flux is

phys-ically impossible Magnetic materials (such as alloys of iron, cobalt, and

nickel) create a space with high permeance and are used to establish

coef-ficients of coupling that approach unity (We say more about this

impor-tant quality of magnetic materials in Chapter 9.)

NOTE: Assess your understanding of this material by trying Chapter

Problems 6.48 and 6.49

Energy Calculations

We conclude our first look at mutual inductance with a discussion of the

total energy stored in magnetically coupled coils In doing so, we will

confirm two observations made earlier: For linear magnetic coupling,

(1) M n = M 2l = M, and (2) M = kVL^L 2 , where 0 < k < 1

total energy stored in the magnetic fields associated with a pair of linearly

coupled coils We begin by assuming that the currents i { and /2 are zero

and that this zero-current state corresponds to zero energy stored in the

coils Then we let /} increase from zero to some arbitrary value I { and

com-pute the energy stored when i l = / } Because i 2 = 0, the total power

input into the pair of coils is v ^ , and the energy stored is

rW\ eh

j dw = Lx \

Jo /0

i]di[

Now we hold /] constant at I { and increase i 2 from zero to some arbitrary

value /2 During this time interval, the voltage induced in coil 2 by ij is

zero because I x is constant The voltage induced in coil 1 by /2 is M ]2 di 2 /dt

Therefore, the power input to the pair of coils is

d'h

p = I\M U — + i 2 v 2

The total energy stored in the pair of coils when i 2 = f 2 is

/ dw = / IiM ]2di2 + / L2i2di2,

Jw, Jo Jo

or

W = W1 + IXI2M{2 + -L2lh

If we reverse the procedure—that is, if we first increase i 2 from zero to /2

and then increase i { from zero to 1 { — the total energy stored is

W = \h x J\ + \ L 2 1 \ + /,/2A#2i (6.64)

Equations 6.63 and 6.64 express the total energy stored in a pair of

lin-early coupled coils as a function of the coil currents, the self-inductances,

and the mutual inductance Note that the only difference between these

equations is the coefficient of the current product l\l 2 We use Eq 6.63 if

/t is established first and Eq 6.64 if /2 is established first

When the coupling medium is linear, the total energy stored is the

same regardless of the order used to establish / ] and / The reason is that

6.5 A Closer Look at Mutual Inductance 199

in a linear coupling, the resultant magnetic flux depends only on the final

values of/j and /2, not on how the currents reached their final values If the

resultant flux is the same, the stored energy is the same Therefore, for

lin-ear coupling, M\i = M 2\ Also, because 1\ and J2 are arbitrary values of /j

and *"2, respectively, we represent the coil currents by their instantaneous

values /*! and /2 Thus, at any instant of time, the total energy stored in the

coupled coils is

1 , 1 ,

We derived Eq 6.65 by assuming that both coil currents entered

polarity-marked terminals We leave it to you to verify that, if one current

enters a polarity-marked terminal while the other leaves such a terminal,

the algebraic sign of the term Mi{i 2 reverses Thus, in general,

1 , 1 ,

w(t) = —L\i\ + ~L2i2 ± Mi {i2 (6.66) A Energy stored in magnetically-coupled

coils

We use Eq 6.66 to show that M cannot exceed VL]L2 The

magneti-cally coupled coils are passive elements, so the total energy stored can

never be negative If w(t) can never be negative, Eq 6.66 indicates that the

quantity

1 , 1 ,

-L { q + -L 2 ii - Mi } i 2

must be greater than or equal to zero when /j and i 2 are either both

posi-tive or both negaposi-tive The limiting value of M corresponds to setting the

quantity equal to zero:

1 , 1 ,

-Liq + ^ 2 ' 2 - Milk = 0 (6.67)

To find the limiting value of M we add and subtract the term

/1/2VL1L2 to the left-hand side of Eq 6.67 Doing so generates a term that

is a perfect square:

+ /V2 VLJ7 2 - M = 0 (6.68)

The squared term in Eq 6.68 can never be negative, but it can be zero

Therefore w{t) 2t 0 only if

which is another way of saying that

M = kVLxL2 ( O ^ H 1)

We derived Eq 6.69 by assuming that i\ and i 2 are either both positive or

both negative However, we get the same result if i\ and i 2 are of opposite

sign, because in this case we obtain the limiting value of M by selecting the

plus sign in Eq 6.66

NOTE: Assess your understanding of this material by trying Chapter Problems 6.44 and 6.45

Practical Perspective Proximity Switches

At the beginning of this chapter we introduced the capacitive proximity switch There are two forms of this switch The one used in table lamps

is based on a single-electrode switch I t is left to your investigation in Problem 6.52 In the example here, we consider the two-electrode switch used in elevator call buttons

EXAMPLE

The elevator call button is a small cup into which the finger is inserted,

as shown in Fig 6.31 The cup is made of a metal ring electrode and a circular plate electrode that are insulated from each other Sometimes two concentric rings embedded in insulating plastic are used instead The electrodes are covered with an insulating Layer to prevent direct contact with the metal The resulting device can be modeled as a capacitor, as shown in Fig 6.32

(a) (b)

Figure 6.31 • An elevator call button, (a) Front view, (b) Side view

Figure 6.32 • A capacitor model of the two-electrode

proximity switch used in elevator call buttons

c 2 Q

Figure 6.33 A A circuit model of a capacitive

proximity switch activated by finger touch

Unlike most capacitors, the capacitive proximity switch permits you to insert an object, such as a finger, between the electrodes Because your fin-ger is much more conductive than the insulating covering surrounding the eLectrodes, the circuit responds as though another electrode, connected to ground, has been added The result is a three-terminal circuit containing three capacitors, as shown in Fig 6.33

The actual values of the capacitors in Figs 6.32 and 6.33 are in the range of 10 to 50 pF, depending on the exact geometry of the switch, how

Practical Perspective

the finger is inserted, whether the person is wearing gloves, and so forth

For the following problems, assume that all capacitors have the same value

of 25 pF Also assume the elevator call button is placed in the capacitive

equivalent of a voltage-divider circuit, as shown in Fig 6.34

a) Calculate the output voltage with no finger present

b) Calculate the output voltage when a finger touches the button

Solution

a) Begin by redrawing the circuit in Fig 6.34 with the call button replaced

by its capacitive model from Fig 6.32 The resulting circuit is shown in

Fig 6.35 Write the current equation at the single node:

_ d(v - v s ) _ dv

Ci ' + C2— = 0

*M(Z)

Button ( A )

Fixed capacitor' ;25pF v(t)

Figure 6.34 • An elevator call button circuit

Vs(t)(l)

Button ]

Fixed capacitor

:c,

+

• C 2 v(t)

r

-Figure 6.35 A A model of the elevator call button

Rearrange this equation to produce a differential equation for the output Q Tm \i W1-th no finger present,

voltage v(t):

dv

dt

Finally, integrate Eq 6.71 to find the output voltage:

v(t) Ci

c, +c -v s (t) + v(0) (6.72)

The result in Eq 6.72 shows that the series capacitor circuit in Fig 6.35

forms a voltage divider just as the series resistor circuit did in Chapter 3

In both voltage-divider circuits, the output voltage does not depend on

the component values but only on their ratio Here, C { = C2 = 25 pF,

so the capacitor ratio is Ci/C 2 = 1 Thus the output voltage is

The constant term in Eq 6.73 is due to the initial charge on the capacitor

We can assume that v(0) - 0 V, because the circuit that senses the

out-put voltage eliminates the effect of the initial capacitor charge Therefore,

the sensed output voltage is

b) Now we replace the call button of Fig 6.34 with the model of the acti-vated switch in Fig 6.33 The result is shown in Fig 6.36 Again, we cal-culate the currents leaving the output node:

_ d(v - v s) dv „ dv

Q - ^ - — - + C2— + C3— = 0

dt dt • dt Rearranging to write a differential equation for v(t) results in

dv _ C\ dvs

~dt ~ Q + C2 + C3 dt '

Finally, solving the differential equation in Eq 6.76, we see

(6.75)

(6.76)

v(t)

C,

vs(t) + v(0)

Cx + C2 + C3

If Q = C 2 = C3 = 25 pF,

v(t) = 0333vs(t) + v(0)

(6.77)

(6.78)

As before, the sensing circuit eliminates v(Q), so the sensed output

voltage is

Comparing Eqs 6.74 and 6.79, we see that when the button is pushed, the output is one third of the input voltage When the button is not pushed, the output voltage is one half of the input voltage Any drop in output voltage is detected by the elevator's control computer and ulti-mately results in the elevator arriving at the appropriate floor

NOTE: Assess your understanding of this Practical Perspective by trying Chapter Problems 6.51 and 6.53

Button

Fixed capacitor

Figure 6.36 A A model of the elevator call button circuit when

activated by finger touch

Summary

Summary 203

Inductance is a linear circuit parameter that relates the

voltage induced by a time-varying magnetic field to the

current producing the field (See page 176.)

Capacitance is a linear circuit parameter that relates the

current induced by a time-varying electric field to the

voltage producing the field (See page 182.)

Inductors and capacitors are passive elements; they can

store and release energy, but they cannot generate or

dissipate energy (See page 176.)

The instantaneous power at the terminals of an inductor

or capacitor can be positive or negative, depending on

whether energy is being delivered to or extracted from

the element

An inductor:

• does not permit an instantaneous change in its

termi-nal current

• does permit an instantaneous change in its teminal

voltage

• behaves as a short circuit in the presence of a constant

terminal current (See page 188.)

A capacitor:

• does not permit an instantaneous change in its

termi-nal voltage

• does permit an instantaneous change in its terminal

current

• behaves as an open circuit in the presence of a

con-stant terminal voltage (See page 183.)

Equations for voltage, current, power, and energy in

ideal inductors and capacitors are given in Table 6.1

Inductors in series or in parallel can be replaced by an

equivalent inductor Capacitors in series or in parallel

can be replaced by an equivalent capacitor The

equa-tions are summarized in Table 6.2 See Section 6.3 for a

discussion on how to handle the initial conditions for

series and parallel equivalent circuits involving

induc-tors and capaciinduc-tors

TABLE 6.1 Terminal Equations for Ideal Inductors and Capacitors

Inductors

i) = J

-0 u dt

i = if v dr +

J fa

p = vi = Li%

w = I Li 2

Capacitors

v = £ 1 idr + Jfa

1 *- dt

p = vi = Cv'j§

w = \Cv 2

Kh)

v(U))

(V) (A) (W) (J)

(V) (A) (W) (J)

TABLE 6.2 Equations for Series- and Parallel-Connected Inductors and Capacitors

Series-Connected

^ecj = L\ + Li + • • • + L n

J - =-L + _L

c„

Parallel-Connected

J_ = i_ + X +

+ G,

Mutual inductance, M, is the circuit parameter relating

the voltage induced in one circuit to a time-varying cur-rent in another circuit Specifically,

Vl = L l - + M l2

-di] dU

»2 = M 2, - + L2- ,

where V\ and i { are the voltage and current in circuit 1,

and v 2 and i 2 are the voltage and current in circuit 2 For

coils wound on nonmagnetic cores, M12 = M 2\ = M

(See page 190.)

The dot convention establishes the polarity of mutually

induced voltages:

When the reference direction for a current enters

the dotted terminal of a coil, the reference

polar-ity of the voltage that it induces in the other coil

is positive at its dotted terminal

Or, alternatively,

When the reference direction for a current leaves

the dotted terminal of a coil, the reference

polar-ity of the voltage that it induces in the other coil

is negative at its dotted terminal

(See page 190.)

The relationship between the self-inductance of each winding and the mutual inductance between windings is

M = kVLJ72

The coefficient of coupling, k, is a measure of the degree

of magnetic coupling By definition, 0 < k < 1 (See

page 197.)

The energy stored in magnetically coupled coils in a lin-ear medium is related to the coil currents and induc-tances by the relationship

w = -Lxi\ + -L2ij ± Mi{i2

(See page 199.)

Problems

Section 6.1

6.1 The current in the 2.5 mH inductor in Fig P6.1 is

known to be 1 A for t < 0 The inductor voltage for

t ^ 0 is given by the expression

v L (t) = 3e~ 4 ' mV, 0+ < t < 2 s

v L {t) = - 3 e "4 ( / _ 2 )m V , 2 s < t < oo

Sketch v L {t) and i L {t) for 0 < t < oo

Figure P6.1

'7.(0

PSPICE

MULTISIM

6.2 The current in a 50 /xH inductor is known to be

iL = 18te~ 10t A for t > 0

a) Find the voltage across the inductor for t > 0

(Assume the passive sign convention.)

b) Find the power (in microwatts) at the terminals

of the inductor when t = 200 ms

c) Is the inductor absorbing or delivering power at

200 ms?

d) Find the energy (in microjoules) stored in the

inductor at 200 ms

e) Find the maximum energy (in microjoules)

stored in the inductor and the time (in

milli-seconds) when it occurs

6.3 The voltage at the terminals of the 200 /xH inductor

PSPICE in pig P6.3(a) is shown in Fig P6.3(b) The inductor

1M current i is known to be zero for t < 0

a) Derive the expressions for i for f > 0

b) Sketch i versus t for 0 < t < oo

Figure P6.3

vs (mV)

200

/xH-t (ms)

6.4 The triangular current pulse shown in Fig P6.4 is

™±l applied to a 20 mH inductor

MULTISIM

a) Write the expressions that describe /(f) in

the four intervals t < 0, 0 ^ / ^ 5 ms,

5 ms ^ t ^ 10 ms, and t > 10 ms

b) Derive the expressions for the inductor volt-age, power, and energy Use the passive sign convention

Figure P6.4 /(mA)

250

t (ms)

Problems 205

6.5 The current in and the voltage across a 5 H inductor

are known to be zero for t < 0 The voltage across

the inductor is given by the graph in Fig P6.5

for t > 0

a) Derive the expression for the current as a

function of time in the intervals 0 < t < 1 s,

I s < t < 3 s, 3 s < t < 5 s, 5 s < t < 6 s, and

6 s < t < oo

b) For t > 0, what is the current in the inductor

when the voltage is zero?

c) Sketch i versus t for 0 < t < oo

Figure P6.5

v (V)

100 - / v

6.9 a) Find the inductor current in the circuit in

PSP.CE Fig P6.9 if v = - 5 0 sin 250 t V, L = 20 mH,

MULTISIM and/(0) = 10 A

b) Sketch v, i, p s and w versus t In making these

sketches, use the format used in Fig 6.8 Plot over one complete cycle of the voltage waveform c) Describe the subintervals in the time interval

between 0 and Sn ms when power is being

absorbed by the inductor Repeat for the subintervals when power is being delivered by the inductor

Figure P6.9

6.10 The current in a 4 H inductor is

i = 10 A, t < 0;

i = {B\ cos At + B 2 sin 4t)e~ tf2 A, t > 0

The voltage across the inductor (passive sign

con-vention) is 60 V at t = 0 Calculate the power at the terminals of the inductor at t = 1 s State whether

the inductor is absorbing or delivering power

6.11 Evaluate the integral

for Example 6.2 Comment on the significance of the result

6.12 The expressions for voltage, power, and energy

derived in Example 6.5 involved both integration and manipulation of algebraic expressions As an engineer, you cannot accept such results on faith alone That is, you should develop the habit of ask-ing yourself, "Do these results make sense in terms

of the known behavior of the circuit they purport to describe?" With these thoughts in mind, test the expressions of Example 6.5 by performing the fol-lowing checks:

a) Check the expressions to see whether the volt-age is continuous in passing from one time inter-val to the next

b) Check the power expression in each interval

by selecting a time within the interval and see-ing whether it gives the same result as the

cor-responding product of v and i For example,

test at 10 and 30 (JLS

c) Check the energy expression within each interval

by selecting a time within the interval and seeing whether the energy equation gives the same

result as \Cv 2 Use 10 and 30 /xs as test points

6.13 Initially there was no energy stored in the 5 H

inductor in the circuit in Fig P6.13 when it was placed across the terminals of the voltmeter At

6.6 The current in a 20 mH inductor is known to be

i = 40 mA, t < 0;

i = A,e~mm[ + yW-4()(X,0'A, 0

The voltage across the inductor (passive sign

con-vention) is 28 V at t = 0

a) Find the expression for the voltage across the

inductor for t > 0

b) Find the time, greater than zero, when the power

at the terminals of the inductor is zero

6.7 Assume in Problem 6.6 that the value of the voltage

across the inductor at t = 0 is - 6 8 V instead of 28 V

a) Find the numerical expressions for i and v for

t > 0

b) Specify the time intervals when the inductor is

storing energy and the time intervals when the

inductor is delivering energy

c) Show that the total energy extracted from the

inductor is equal to the total energy stored

6.8 The current in a 25 mH inductor is known to be

PSPICE - i o A for / < 0 and - ( 1 0 cos 400r - 5 sin 400/)<T2,)()' A

for t Sr 0 Assume the passive sign convention

a) At what instant of time is the voltage across the

inductor maximum?

b) What is the maximum voltage?