Electric Circuits, 9th Edition P27 potx

no energy is stored in the circuit at the instant the switch is closed... Solution a At the instant the switch is moved to position b, the initial voltage on the capacitor is zero.. The

no energy is stored in the circuit at the instant

the switch is closed

Next we observe v o (0 + ) = 120 V, which is

consistent with the fact that i o (0) = 0

Now we observe the solutions for ^ and

i 2 are consistent with the solution for v 0 by

observing

v a = 3 — + 6 —

dt dt

= 360*T5' - 240<T5'

= 120«?-* V, t > 0+,

or

dt dt

= 720e~5' - 600e_5/

= 120e"5' V, t > 0+

The final values of i\ and /2 can be checked

using flux linkages The flux linking the 3 H coil

(Aj) must be equal to the flux linking the 15 H

coil (A2), because

v„ =

Now

and

rfAj

~dt dki dt'

A, = 3ij + 6/2 Wb-turns

A2 = 6i] + 15/2 Wb-turns

Regardless of which expression we use, we obtain

A, = A2 = 24 - 24<T5' Wb-turns

Note the solution for At or A2 is consistent with

the solution for v D

The final value of the flux linking either coil 1 or coil 2 is 24 Wb-turns, that is,

At(oo) = A2(oo) = 24 Wb-turns

The final value of iy is

^(oo) = 24 A

and the final value of i 2 is

/2(oo) = - 8 A

The consistency between these final values

for jj and i 2 and the final value of the flux link-age can be seen from the expressions:

A^oo) = 3/2(00) + 6/2(oo)

= 3(24) + 6 ( - 8 ) = 24 Wb-turns,

A2(oo) = 6^(00) + 15/2(oo)

= 6(24) + 15(-8) = 24 Wb-turns

It is worth noting that the final values of ij and /2 can only be checked via flux linkage

because at t — 00 the two coils are ideal short

circuits The division of current between ideal short circuits cannot be found from Ohm's law

NOTE: Assess your understanding of this material by using the general solution method to solve Chapter

Problems 7.68 and 7.69

7.5 Sequential Switching Whenever switching occurs more than once in a circuit, we have sequential switching For example, a single, two-position switch may be switched back

and forth, or multiple switches may be opened or closed in sequence The time reference for all switchings cannot be f = 0 We determine the volt-ages and currents generated by a switching sequence by using the tech-niques described previously in this chapter We derive the expressions for

v(t) and i(t) for a given position of the switch or switches and then use

these solutions to determine the initial conditions for the next position of the switch or switches

With sequential switching problems, a premium is placed on obtaining

the initial value x(t 0 ) Recall that anything but inductive currents and

capacitive voltages can change instantaneously at the time of switching Thus solving first for inductive currents and capacitive voltages is even more pertinent in sequential switching problems Drawing the circuit that pertains to each time interval in such a problem is often helpful in the solution process

Examples 7.11 and 7.12 illustrate the analysis techniques for circuits

with sequential switching The first is a natural response problem with two

switching times, and the second is a step response problem

Example 7.11 Analyzing an RL Circuit that has Sequential Switching

The two switches in the circuit shown in Fig 7.31

have been closed for a long time At t - 0, switch 1

is opened Then, 35 ms later, switch 2 is opened

a) Find i L {t) for 0 < t < 35 ms

b) Find i L for t a 35 ms

c) What percentage of the initial energy stored in

the 150 mH inductor is dissipated in the 18 ft

resistor?

d) Repeat (c) for the 3 17 resistor

e) Repeat (c) for the 6 D, resistor

t = 0 t = 35 ms

4H

KVr^fr^—

60 V £12 (1 £6(1 <M 150 mH $18(1

Figure 7.31 • The circuit for Example 7.11

Solution

a) For t < 0 both switches are closed, causing the

150 mH inductor to short-circuit the 18 D,

resis-tor The equivalent circuit is shown in Fig 7.32 We

determine the initial current in the inductor by

solving for ii£0~) in the circuit shown in Fig 7.32

After making several source transformations, we

find i L (0~) to be 6 A For 0 < t < 35 ms, switch 1

is open (switch 2 is closed), which disconnects the

60 V voltage source and the 4 H and 12 £l

resis-tors from the circuit The inductor is no longer

behaving as a short circuit (because the dc source

is no longer in the circuit), so the 18 O resistor is

no longer short-circuited The equivalent circuit is

shown in Fig 7.33 Note that the equivalent

resist-ance across the terminals of the inductor is the

parallel combination of 9 O and 18 0 , or 6 i\

The time constant of the circuit is (150/6) X 10~3,

or 25 ms Therefore the expression for i L is

i L = 6e- A{]l A, 0 < t < 35 ms

Figure 7.32 • The circuit shown in Fig 7.31, for t < 0

Figure 7.33 • The circuit shown in Fig 7.31, for 0 < t ^ 3 5 ms

b) When t = 35 ms, the value of the inductor

current is

i L = 6e~ u = 1.48 A

Thus, when switch 2 is opened, the circuit reduces to the one shown in Fig 7.34, and the time constant changes to (150/9) x 1 0_\ or

16.67 ms.The expression for i L becomes

i,= 1.486>-6(,('-a()35>A, t > 3 5 ms

Note that the exponential function is shifted in time by 35 ms

3 (

*'/

6(1 v L ] 150 mH

_ ? |iL{0.035)s 1.48 A

Figure 7.34 • The circuit shown in Fig 7.31, for t > 35 ms

c) The 18 n resistor is in the circuit only during the first 35 ms of the switching sequence During this interval, the voltage across the resistor is

v, = 0.154(6<?"40 ')

dt

= -36e~40' V, 0 < t < 35 ms

The power dissipated in the 18 ft resistor is

p = - ^ = 72<r80' W, 0 < t < 35 ms

18

Hence the energy dissipated is

/.().035

W = 72<T80/ hlV/

.7(1

72 -SO/ 0.035

0

- 8 0

= 0.9(1 - ef2-8)

= 845.27 rnJ

Tlie initial energy stored in the 150 mH inductor is

Wi = j(0.15)(36) = 2.7 J = 2700 mj

Therefore (845.27/2700) x 100, or 31.31% of

the initial energy stored in the 150 mH inductor

is dissipated in the 18 ft resistor

d) For 0 < i < 35 ms, the voltage across the 3 ft

resistor is

*>m »L (3)

40/

Therefore the energy dissipated in the 3 ft

resis-tor in the first 35 ms is

.().035

WMl

"I44e -SO/

dt

Jo 3

= 0.6(1 - e' 2 *)

= 563.51 mJ

For t > 35 ms, the current in the 3 O resistor is

ha = <L = (6e-^)e~W-^ A

Hence the energy dissipated in the 3 ft resistor for

t > 35 ms is

t % i = I iiti X3dt

/().035

f 3(36)e-2 *e- ia *- tt,B5 >A

.7().035

108e - - 8 x

,-120(/-0.035)

1 0 8 _•) o _ ,

— , » - 54.73mJ

The total energy dissipated in the 3 ("1 resistor is

w3(l(total) = 563.51 + 54.73

= 618.24 mJ

The percentage of the initial energy stored is

618.24

2700 X 100 = 22.90%

e) Because the 6 ft resistor is in series with the 3 12 resistor, the energy dissipated and the percent-age of the initial energy stored will be twice that

of the 3 ft resistor:

w6 n(total) = 1236.48 mJ, and the percentage of the initial energy stored is 45.80% We check these calculations by observ-ing that

1236.48 4- 618.24 + 845.27 = 2699.99 mJ and

31.31 + 22.90 + 45.80 = 100.01%

The small discrepancies in the summations are the result of roundoff errors

Example 7.12 Analyzing an RC Circuit that has Sequential Switching

The uncharged capacitor in the circuit shown in

Fig 7.35 is initially switched to terminal a of the

three-position switch At t — 0, the switch is moved

to position b, where it remains for 15 ms After the

15 ms delay, the switch is moved to position c, where

it remains indefinitely

a) Derive the numerical expression for the voltage

across the capacitor

b) Plot the capacitor voltage versus time

c) When will the voltage on the capacitor equal

200 V?

Solution

a) At the instant the switch is moved to position b,

the initial voltage on the capacitor is zero If the

switch were to remain in position b, the capacitor

would eventually charge to 400 V The time

con-stant of the circuit when the switch is in position b

is 10 ms Therefore we can use Eq 7.59 with

t () = 0 to write the expression for the capacitor

voltage:

v = 400 + (0 - 400)e - 1 0 0 /

= (400 - 400e" m") V, 0 =s t < 15 ms

Note that, because the switch remains in

posi-tion b for only 15 ms, this expression is valid only

for the time interval from 0 to 15 ms After the

switch has been in this position for 15 ms, the

voltage on the capacitor will be

y(15ms) = 400 - 400e~15 = 310.75 V

Therefore, when the switch is moved to position c,

the initial voltage on the capacitor is 310.75 V

With the switch in position c, the final value of

the capacitor voltage is zero, and the time

con-stant is 5 ms Again, we use Eq 7.59 to write the

expression for the capacitor voltage:

v = 0 + (310.75 - ())e-200(/-o.oi5)

= 310.75e- ('- >V, 15ms s i /

400 V ( ) L v(t)^:0AfjLF

Figure 7.35 A The circuit for Example 7.12

In writing the expression for ?;, we recognized that r<) = 15 ms and that this expression is valid

only for t ^ 15 ms

b) Figure 7.36 shows the plot of v versus t

c) The plot in Fig 7.36 reveals that the capacitor voltage will equal 200 V at two different times: once in the interval between 0 and 15 ms and once after 15 ms We find the first time by solving the expression

200 = 400 - 400<T10,,\

which yields t\ = 6.93 ms We find the second

time by solving the expression

200 = 310.756>-2(,0(';-°-,),5)

In this case, u = 17.20 ms

v = 4()0-4()0t'" ,m/

v = 3U).75e 2m ' {m5)

t (ms)

Figure 7.36 • The capacitor voltage for Example 7.12

^ A S S E S S M E N T P R O B L E M S

Objective 3—Know how to analyze circuits with sequential switching

7.7 In the circuit shown, switch 1 has been closed

and switch 2 has been open for a long time At

t = 0, switch 1 is opened Then 10 ms later,

switch 2 is closed Find

a) v c (t) for 0 < f < 0.01 s,

b) v c (t) for t > 0.01 s,

c) the total energy dissipated in the 25 kft

resistor, and

d) the total energy dissipated in the 100 kO

resistor

( U 60kn r = 1 0 m

" )10mAf40kft 25kft£lAtF

Answer: (a) 80e~40/ V;

(b) 53.63e-5°('-a 0 1W;

(c) 2.91 mJ;

(d) 0.29 mJ

NOTE: Also try Chapter Problems 7.71 and 7.78

7.8 Switch a in the circuit shown has been open for

a long time, and switch b has been closed for a

long time Switch a is closed at t = 0 and, after

remaining closed for 1 s, is opened again

Switch b is opened simultaneously, and both switches remain open indefinitely Determine

the expression for the inductor current i that is valid when ( a ) 0 s f < h and (b) t > 1 s

Answer: (a) (3 - 3e _a5 ') A, 0 < f < 1 s;

(b) (-4.8 + 5.98tf~ l - 25( '~ 1>) A, t > 1 s

7.6 Unbounded Response

A circuit response may grow, rather than decay, exponentially with time

This type of response, called an unbounded response, is possible if the

cir-cuit contains dependent sources In that case, the Thevenin equivalent resistance with respect to the terminals of either an inductor or a capacitor may be negative This negative resistance generates a negative time con-stant, and the resulting currents and voltages increase without limit In an actual circuit, the response eventually reaches a limiting value when a component breaks down or goes into a saturation state, prohibiting fur-ther increases in voltage or current

When we consider unbounded responses, the concept of a final value

is confusing Hence, rather than using the step response solution given in

Eq 7.59, we derive the differential equation that describes the circuit con-taining the negative resistance and then solve it using the separation of variables technique Example 7.13 presents an exponentially growing response in terms of the voltage across a capacitor

Example 7.13 Finding the Unbounded Response in an RC Circuit

a) When the switch is closed in the circuit shown in

Fig 7.37, the voltage on the capacitor is 10 V

Find the expression for v a for t > 0

b) Assume that the capacitor short-circuits when

its terminal voltage reaches 150 V How many

milliseconds elapse before the capacitor

short-circuits?

20 kn

Figure 7.37 • The circuit for Example 7.13

Solution

a) To find the Thevenin equivalent resistance with

respect to the capacitor terminals, we use the

test-source method described in Chapter 4 Figure 7.38

shows the resulting circuit, where v r is the test

voltage and i T is the test current For Vj expressed

in volts, we obtain

ir = TT: ~ 7 ( - - ) + —- mA

' 10 W 20

Solving for the ratio Vj/ir yields the Thevenin

resistance:

Km = — = - 5 k n

i T

With this Thevenin resistance, we can simplify

the circuit shown in Fig 7.37 to the one shown in

Fig 7.39

'T

Figure 7.38 • The test-source method used to find i?T h

- 5 k O

Figure 7.39 A A simplification of the circuit shown in

Fig 7.37

For t S: 0, the differential equation describing

the circuit shown in Fig 7.39 is

(5 X 10- 6)—^ - - r X 1()-J = 0

dt 5

Dividing by the coefficient of the first derivative yields

dv 0

dt 4(h>„ = 0

We now use the separation of variables technique

to find v ( ,(t):

v () (t) = lOe40 ' V, t>0

b) v a = 150 V when e m = 15 Therefore, 40r = In 15,

and t = 67.70 ms

NOTE: Assess your understanding of this material by trying Chapter Problems 7.85 and 7.87

The fact that interconnected circuit elements may lead to

ever-increasing currents and voltages is important to engineers If such

inter-connections are unintended, the resulting circuit may experience

unexpected, and potentially dangerous, component failures

7.7 The Integrating Amplifier

Recall from the introduction to Chapter 5 that one reason for our interest in

the operational amplifier is its use as an integrating amplifier We are now

ready to analyze an integrating-amplifier circuit, which is shown in Fig 7.40

The purpose of such a circuit is to generate an output voltage proportional

to the integral of the input voltage In Fig 7.40, we added the branch

cur-rents if and / , along with the node voltages v and v , to aid our analysis Figure 7.40 • An integrating amplifier

» i

K„

-2f,

Figure 7.41 • An input voltage signal

Figure 7.42 • The output voltage of an integrating

amplifier

We assume that the operational amplifier is ideal Thus we take advantage of the constraints

Because v„ = 0,

if + i s = 0,

v n = v p

i = ^

-l ^^~dt

Hence, from Eqs 7.61,7.63, and 7.64,

dt ~ R s C f v s

(7.61) (7.62)

(7.63)

(7.64)

(7.65)

Multiplying both sides of Eq 7.65 by a differential time dt and then

inte-grating from f() to t generates the equation

R<C /•/A, v s dy + v 0 (t () ) (7.66)

In Eq 7.66, t () represents the instant in time when we begin the integration Thus u„(?o) is the value of the output voltage at that time Also, because

v n = vp = 0, v o (t 0 ) is identical to the initial voltage on the feedback

capacitor C/

Equation 7.66 states that the output voltage of an integrating ampli-fier equals the initial value of the voltage on the capacitor plus an inverted

(minus sign), scaled (l/R s Cf) replica of the integral of the input voltage If

no energy is stored in the capacitor when integration commences, Eq 7.66 reduces to

vM = -R S C i V s dy (7.67)

f A,

If v s is a step change in a dc voltage level, the output voltage will vary lin-early with time For example, assume that the input voltage is the rectan-gular voltage pulse shown in Fig 7.41 Assume also that the initial value of

v a (t) is zero at the instant v s steps from 0 to V m A direct application of

Eq 7.66 yields

v„ =

1

V m t + 0, 0 < t < t h (7.68)

When t lies between t\ and 2t u

1

v, t =

-# s Q Jt {-V, n )dy - RSC, V m h

RcCi

2V m

R x c t

t h t L < t < 2t v (7.69)

Figure 7.42 shows a sketch of v ( ,(t) versus t Clearly, the output voltage is

an inverted, scaled replica of the integral of the input voltage

The output voltage is proportional to the integral of the input voltage only if the op amp operates within its linear range, that is, if it doesn't sat-urate Examples 7.14 and 7.15 further illustrate the analysis of the inte-grating amplifier

Example 7.14 Analyzing an Integrating Amplifier

Assume that the numerical values for the signal

voltage shown in Fig 7.41 are V m = 50 mV and

t\ = 1 s This signal voltage is applied to the

integrating-amplifier circuit shown in Fig 7.40 The

circuit parameters of the amplifier are R s = 100 kfl,

Cf = 0.1 ^iF, and Vcc = 6 V The initial voltage on

the capacitor is zero

a) Calculate v a (t)

b) Plot v () (t) versus t

Solution

a) For 0 < t < 1 s,

- 1

" (100 X 103)(0.1 x 10"6)

= St V, 0 < f < 1 s

50 X 10~3f + 0

For 1 < t < 2 s,

v <t = (5r - 10) V

b) Figure 7.43 shows a plot of v„(t) versus r

» „ ( 0 ( V ) *

2 t{t)

Figure 7.43 • The output voltage for Example 7.14

At the instant the switch makes contact with

termi-nal a in the circuit shown in Fig 7.44, the voltage on

the 0.1 /AF capacitor is 5 V The switch remains at

terminal a for 9 ms and then moves instantaneously

to terminal b How many milliseconds after making

contact with terminal b does the operational

ampli-fier saturate?

Figure 7.44 • The circuit for Example 7.15

Solution

The expression for the output voltage during the

time the switch is at terminal a is

1

io-27o

= (-5 + 10000 v

( - 1 0 ) ^

Thus, 9 ms after the switch makes contact with ter-minal a, the output voltage is - 5 + 9, or 4 V

The expression for the output voltage after the switch moves to terminal b is

&dy

10"2 9X

1()-= 4 - 800(t - 9 X 10"3)

= (11.2 - 8000 V

During this time interval, the voltage is decreas-ing, and the operational amplifier eventually

satu-rates at —6 V Therefore we set the expression for v a equal to —6 V to obtain the saturation time t s :

11.2 - 800/, = - 6 ,

or

t s = 21.5 ms

Thus the integrating amplifier saturates 21.5 ms after making contact with terminal b

From the examples, we see that the integrating amplifier can perform the integration function very well, but only within specified limits that avoid saturating the op amp The op amp saturates due to the accumula-tion of charge on the feedback capacitor We can prevent it from saturat-ing by placsaturat-ing a resistor in parallel with the feedback capacitor We examine such a circuit in Chapter 8

Note that we can convert the integrating amplifier to a differentiating

amplifier by interchanging the input resistance R s and the feedback

capac-itor Cf Then

v a = -R S C,

We leave the derivation of Eq 7.70 as an exercise for you The differentiat-ing amplifier is seldom used because in practice it is a source of unwanted

or noisy signals

Finally, we can design both integrating- and differentiating-amplifier circuits by using an inductor instead of a capacitor However, fabricating capacitors for integrated-circuit devices is much easier, so inductors are rarely used in integrating amplifiers

^ A S S E S S M E N T P R O B L E M S

Objective 4—Be able to analyze op amp circuits containing resistors and a single capacitor

7.9 There is no energy stored in the capacitor at

the time the switch in the circuit makes contact

with terminal a The switch remains at position

a for 32 ms and then moves instantaneously to

position b How many milliseconds after

mak-ing contact with terminal a does the op amp

saturate?

7.10 a) When the switch closes in the circuit

shown, there is no energy stored in the capacitor How long does it take to saturate the op amp?

b) Repeat (a) with an initial voltage on the capacitor of 1 V, positive at the upper terminal

40 m

10 kO

/ V W

v„ £ 6.8 kO,

Answer: 262 ms

NOTE: Also try Chapter Problems 7.95 and 7.96

Answer: (a) 1.11 ms;

(b) 1.76 ms

Practical Perspective

A Flashing Light Circuit

We are now ready to analyze the flashing light circuit introduced at the

start of this chapter and shown in Fig 7.45 The lamp in this circuit

starts to conduct whenever the lamp voltage reaches a value V max During

the time the lamp conducts, i t can be modeled as a resistor whose

resist-ance is R L The lamp will continue to conduct until the lamp voltage

drops to the value V^ lin When the lamp is not conducting, i t behaves as

an open circuit

Before we develop the analytical expressions that describe the

behav-ior of the circuit, let us develop a feel for how the circuit works by noting

the following First, when the lamp behaves as an open circuit, the dc

voltage source will charge the capacitor via the resistor R toward a value

of V s volts However, once the lamp voltage reaches V max , i t starts

con-ducting and the capacitor will start to discharge toward the Thevenin

voltage seen from the terminals of the capacitor But once the capacitor

voltage reaches the cutoff voltage of the lamp (V m i n ), the lamp will act as

an open circuit and the capacitor will start to recharge This cycle of

charging and discharging the capacitor is summarized in the sketch shown

in Fig 7.46

In drawing Fig 7.46 we have chosen t = 0 at the instant the capacitor

starts to charge The time t 0 represents the instant the lamp starts to

con-duct, and t c is the end of a complete cycle We should also mention that in

constructing Fig 7.46 we have assumed the circuit has reached the

repeti-tive stage of its operation Our design of the flashing light circuit requires

we develop the equation for Vjjj) as a function of V^ax, Vj^^, V s , R, C, and

R L for the intervals 0 to t 0 and t Q to f c

To begin the analysis, we assume that the circuit has been in operation

for a long time Let t = 0 at the instant when the lamp stops conducting

Thus, at t = 0, the lamp is modeled as an open circuit, and the voltage drop

across the lamp is V^ in , as shown in Fig 7.47

From the circuit, we find

vd°°) = Vs,

t>i,(0) = v min ,

T = RC

R

->VvV

V c: v L

Lamp

Figure 7.45 A A flashing light circuit

V L{t)

y

v max

V •

etc

Figure 7.46 • Lamp voltage versus time for the circuit in Fig 7.45

+

R

4-v L

Figure 7.47 A The flashing light circuit at t = 0,

when the lamp is not conducting

Thus, when the lamp is not conducting,

V L (t) = Vs + (Knin ~ V s )e-'S RC

How long does i t take before the lamp is ready to conduct? We can find this

time by setting the expression for v L (t) equal to V max and solving for t I f

we call this value t 0 , then

t n = RC In V • - V y min v s

V — V

When the lamp begins conducting, i t can be modeled as a resistance R L ,

as seen in Fig 7.48 In order to find the expression for the voltage drop

+

T

R

C^

+

Figure 7.48 • The flashing light circuit at t = t 0

when the lamp is conducting