Chapter 6, Solution 1Chapter 6, Solution 18. For the capacitors in parallel 1 = 15 + 5 + 40 = 60 µF Ceq Hence 10 1 60 1 30 1 20 1 C 1 eq = + + = Ceq = 10 µF Chapter 6, Solution 19.

Chapter 6, Solution 1.

( =+−== −− t3t3 e6e25dtdvCi ) 10(1 - 3t)e-3t A

p = vi = 10(1-3t)e-3t ⋅ 2t e-3t = 20t(1 - 3t)e-6t W


2211 )120)(40(

21Cv

21w ==

w2 = 221 )80)(40(

21

2=Cv1

( ) =−=−=∆ 22

21 8012020www 160 kW Chapter 6, Solution 3.

i = C =−

= −

516028010x40

dtdv 3 480 mA


)0(vidtC1v

t

o+= ∫

∫ +1tdt4sin621

=

= 1 - 0.75 cos 4t Chapter 6, Solution 5.

v = ∫ +t

o)0(vidt

C1

For 0 < t < 1, i = 4t,

∫−=t

o6 t410x201v dt + 0 = 100t2 kV

v(1) = 100 kV

For 1 < t < 2, i = 8 - 4t,

∫ +−= −

t

16 )1(vdt)t48(10x201v

= 100 (4t - t2 - 3) + 100 kV

Thus v (t) =

<<−−

<<

2t1,kV)2tt4(1001t0,kVt100

2

2


610x30dtdvCi −== x slope of the waveform.

For example, for 0 < t < 2,

310x210

dtdv

−=

i = mA15010x210x10x30

dtdv

36 == −

−C

Thus the current i is sketched below.

t (msec)

150

12 10 2

8

6

4

-150

i(t) (mA) Chapter 6, Solution 7.

∫∫ +=+= −−

t

o

33o 10dt10tx4

10x501)t(vidt

C1v

= =+1050t2 2

0.04k2 + 10 V


(a) tt BCeACedtdvC 600100 600100 −− −−==i (1)

BABCACi 656001002)0( −−=→−−== (2)

BAvv +=→= −+ 50)0()0( (3) Solving (2) and (3) leads to A=61, B=-11

(b) J 5250010421)0(

21 32 === − xxxCvEnergy

(c ) From (1),

A 4.264.241041160010461100 60010060031003 tttt eeexxxexxxi −−−−−− −−=−−= Chapter 6, Solution 9.

v(t) = ( ) ( )∫ −− +=+−t

o

tt Vet120dte1621

1

v(2) = 12(2 + e-2) = 25.62 V

p = iv = 12 (t + e-t) 6 (1-e-t) = 72(t-e-2t)

p(2) = 72(2-e-4) = 142.68 W Chapter 6, Solution 10

dtdvx

dtdvCi 3102 −==

<<<<<<

=s4t316t,-64

s 3t116,s10,16

µµµtt

v

<<<<

<<=

s4t3,16x10-s 3t10,

s10,1016

6

6

µµµtx

dtdv

<<<<

<<=

s4t3kA, 32-s 3t10,

s10,kA 32)(

µµµt

ti


v = ∫ +t

o)0(vidt

C1

For 0 < t < 1,

∫ == −−

t

o

36 t10dt10x40

10x41v kV

v(1) = 10 kV For 1 < t < 2,

kV10)1(vvdtC1v

t

1=+= ∫

For 2 < t < 3,

∫ +−= −−

t

2

36 )2(vdt)10x40(

10x41v

= -10t + 30kV Thus

v(t) =

<<+−<<<<⋅

3t2,kV30t102t1,kV101t0,kVt10


π−π== − 4sin)(4(60x10x3dtdvCi 3 t)

= - 0.7e π sin 4πt A P = vi = 60(-0.72)π cos 4π t sin 4π t = -21.6π sin 8π t W

W = ∫ ∫ t dt ππ−=t

o81

o8sin6.21pdt

= πππ 8

86. cos21 8/1

o = -5.4J


Under dc conditions, the circuit becomes that shown below:

i250 Ω

20 Ω

+ − 60V

+

v1

−

i1

30 Ω

10 Ω

+

v2

−

i2 = 0, i1 = 60/(30+10+20) = 1A v1 = 30i2 = 30V, v2 = 60-20i1 = 40V

Thus, v1 = 30V, v2 = 40V Chapter 6, Solution 14. (a) Ceq = 4C = 120 mF

(b) 304

C4

C1

eq

== Ceq = 7.5 mF

Chapter 6, Solution 15. In parallel, as in Fig. (a), v1 = v2 = 100

C2+

v2

−

C1 + − 100V

+

v2

−

C2

+ − v1

C1 +

v1

−

+ −

100V (b)(a)

w20 = == − 262 100x10x20x21Cv

21 0.1J

w30 = =− 26 100x10x30x21 0.15J

(b) When they are connected in series as in Fig. (b):

,60100x5030V

CCC

v21

21 ==

+= v2 = 40

w20 = =− 26 60x10x30x21 36 mJ

w30 = =− 26 4010x302

xx1 24 mJ

Chapter 6, Solution 16

F 203080

8014 µ=→=+

+= CCCxCeq

Chapter 6, Solution 17. (a) 4F in series with 12F = 4 x 12/(16) = 3F 3F in parallel with 6F and 3F = 3+6+3 = 12F 4F in series with 12F = 3F i.e. Ceq = 3F

(b) Ceq = 5 + [6 || (4 + 2)] = 5 + (6 || 6) = 5 + 3 = 8F (c) 3F in series with 6F = (3 x 6)/9 = 6F

131

61

21

C1

eq

=++=

Ceq = 1F


For the capacitors in parallel = 15 + 5 + 40 = 60 µF 1

eqC

Hence 101

601

301

201

C1

eq

=++=

Ceq = 10 µF Chapter 6, Solution 19. We combine 10-, 20-, and 30- µ F capacitors in parallel to get 60µ F. The 60 -µ F capacitor in series with another 60- µ F capacitor gives 30 µ F. 30 + 50 = 80µ F, 80 + 40 = 120 µ F The circuit is reduced to that shown below.

12 120

12 80

120-µ F capacitor in series with 80µ F gives (80x120)/200 = 48 48 + 12 = 60 60-µ F capacitor in series with 12µ F gives (60x12)/72 = 10µ F Chapter 6, Solution 20.

3 in series with 6 = 6x3/(9) = 2 2 in parallel with 2 = 4 4 in series with 4 = (4x4)/8 = 2 The circuit is reduced to that shown below:

20

1 6

2

8

6 in parallel with 2 = 8 8 in series with 8 = 4 4 in parallel with 1 = 5 5 in series with 20 = (5x20)/25 = 4 Thus Ceq = 4 mF

Chapter 6, Solution 21. 4µF in series with 12µF = (4x12)/16 = 3µF 3µF in parallel with 3µF = 6µF 6µF in series with 6µF = 3µF 3µF in parallel with 2µF = 5µF 5µF in series with 5µF = 2.5µF

Hence Ceq = 2.5µF Chapter 6, Solution 22. Combining the capacitors in parallel, we obtain the equivalent circuit shown below:

a b

40 µF 60 µF 30 µF

20 µF Combining the capacitors in series gives C , where 1

eq

101

301

201

601

C11eq

=++= C = 10µF 1eq

Thus

Ceq = 10 + 40 = 50 µF


(a) 3µF is in series with 6µF 3x6/(9) = 2µF v4µF = 1/2 x 120 = 60V v2µF = 60V

v6µF = =(3+

)6036

20V

v3µF = 60 - 20 = 40V

(b) Hence w = 1/2 Cv2 w4µF = 1/2 x 4 x 10-6 x 3600 = 7.2mJ w2µF = 1/2 x 2 x 10-6 x 3600 = 3.6mJ w6µF = 1/2 x 6 x 10-6 x 400 = 1.2mJ w3µF = 1/2 x 3 x 10-6 x 1600 = 2.4mJ


20µF is series with 80µF = 20x80/(100) = 16µF

14µF is parallel with 16µF = 30µF (a) v30µF = 90V

v60µF = 30V v14µF = 60V

v20µF = =+

60x8020

80 48V

v80µF = 60 - 48 = 12V

(b) Since w = 2Cv21

w30µF = 1/2 x 30 x 10-6 x 8100 = 121.5mJ w60µF = 1/2 x 60 x 10-6 x 900 = 27mJ w14µF = 1/2 x 14 x 10-6 x 3600 = 25.2mJ w20µF = 1/2 x 20 x 10-6 x (48)2 = 23.04mJ w80µF = 1/2 x 80 x 10-6 x 144 = 5.76mJ


(a) For the capacitors in series,

Q1 = Q2 C1v1 = C2v2 1

2

2

1

CC

vv

=

vs = v1 + v2 = 21

2122

1

2 vC

CCvvCC +

=+ s21

12 v

CCC+

=v

Similarly, s21

21 v

CCCv+

=

(b) For capacitors in parallel

v1 = v2 = 2

2

1

1

CQ

CQ

=

Qs = Q1 + Q2 = 22

2122

2

1 QC

CCQQCC +

=+

or

Q2 = 21

2

CCC+

s21

11 Q

CCCQ+

=

i = dtdQ s

21

11 i

CCC+

=i , s21

22 i

CCC+

=i


(a) Ceq = C1 + C2 + C3 = 35µF (b) Q1 = C1v = 5 x 150µC = 0.75mC

Q2 = C2v = 10 x 150µC = 1.5mC Q3 = C3v = 20 x 150 = 3mC

(c) w = J150x35x21

222

eq µ=vC1 = 393.8mJ


(a) 207

201

101

51

C1

C1

C1

C1

321eq

=++=++=

Ceq = =µF720 2.857µF

(b) Since the capacitors are in series,

Q1 = Q2 = Q3 = Q = Ceqv = =µV200x720 0.5714mV

(c) w = =µ= J200x720x

21

222

eq vC1 57.143mJ

Chapter 6, Solution 28. We may treat this like a resistive circuit and apply delta-wye transformation, except that R is replaced by 1/C.

Ca

Cb

Cc

50 µF 20 µF

301

401

301

301

101

401

101

C1

a

+

+

=

= 102

401

101

40=++

3

Ca = 5µF

302

101

12001

3001

4001

C1

6

=++

=

Cb = 15µF

154

401

12001

3001

4001

C1

c

=++

=

Cc = 3.75µF Cb in parallel with 50µF = 50 + 15 = 65µF Cc in series with 20µF = 23.75µF

65µF in series with 23.75µF = F39.1775.88

75.23x65µ=

17.39µF in parallel with Ca = 17.39 + 5 = 22.39µF Hence Ceq = 22.39µF Chapter 6, Solution 29. (a) C in series with C = C/(2) C/2 in parallel with C = 3C/2

2C3 in series with C =

5C3

2C5

2C3Cx=

5C3 in parallel with C = C + =

5C3 1.6 C

(b) 2C

Ceq 2C

C1

C21

C21

C1

eq

=+=

Ceq = C


vo = ∫ +t

o)0(iidt

C1

For 0 < t < 1, i = 60t mA,

kVt100tdt6010x3

10v 2t

o6

3

o =+= ∫−

−

vo(1) = 10kV For 1< t < 2, i = 120 - 60t mA,

vo = ∫ +−−

− t

1 o6

3

)1(vdt)t60120(10x3

10 t + = [40t – 10t2 kV10] 1

= 40t – 10t2 - 20

<<−−

<<=

2t1,kV20t10t401t0,kVt10

)t(v2

2

o


<<+−<<<<

=5t3,t10503t1,mA201t0,tmA20

)t(is

Ceq = 4 + 6 = 10µF

)0(vidtC1v

t

oeq

+= ∫

For 0 < t < 1,

∫−

−

=t

o6

3

t2010x10

10v dt + 0 = t2 kV

For 1 < t < 3,

∫ +−=+=t

1

3

kV1)1t(2)1(vdt201010v

kV1t2 −= For 3 < t < 5,

∫ +−=t

3

3

)3(vdt)5t(101010v

kV11t5tkV55t 2t3

2 +−=++−=

<<+−

<<−<<

=

5t3,kV11t5t3t1,kV1t21t0,kVt

)t(v2

2

dtdv10x6

dtdvCi 6

11−==

=

<<−<<<<

5t3,mA30123t1,mA121t0,tmA12

dtdv10x4

dtdvCi 6

21−==

<<−<<<<

=5t3,mA20t83t1,mA81t0,tmA8


(a) Ceq = (12x60)/72 = 10 µ F

13001250501250)0(301012

10 2

00

21

26

3

1 +−=+−=+= −−−−

−

∫ tt

ttt eevdtex

v

23025020250)0(301060

10 2

00

22

26

3

2 −=+=+= −−−−

−

∫ tt

ttt eevdtex

v

(b) At t=0.5s,

03.138230250,15.84013001250 12

11 −=−==+−= −− evev

J 235.4)15.840(101221 26

12 == − xxxw Fµ

J 1905.0)03.138(102021 26

20 =−= − xxxw Fµ

J 381.0)03.138(104021 26

40 =−= − xxxFµw


Because this is a totally capacitive circuit, we can combine all the capacitors using the property that capacitors in parallel can be combined by just adding their values and we combine capacitors in series by adding their reciprocals. 3F + 2F = 5F 1/5 +1/5 = 2/5 or 2.5F The voltage will divide equally across the two 5F capacitors. Therefore, we get: VTh = 7.5 V, CTh = 2.5 F


i = 6e-t/2

2/t3 e21)6(10x10

dtdiL −−

==v

= -30e-t/2 mV v(3) = -300e-3/2 mV = -0.9487 mV p = vi = -180e-t mW

p(3) = -180e-3 mW = -0.8 mW


dtdiLv = ==

∆∆=

−

)2/(6.010x60

t/iVL

3

200 mH


V)t2sin)(2)(12(10x41

dtdiLv 3 −== −

= - 6 sin 2t mV p = vi = -72 sin 2t cos 2t mW But 2 sin A cos A = sin 2A

p = -36 sin 4t mW Chapter 6, Solution 37.

t100cos)100(4x10x12dtdiLv 3−==

= 4.8 cos 100t V p = vi = 4.8 x 4 sin 100t cos 100t = 9.6 sin 200t

w = t200sin6.9pdtt

o

200/11

o∫ ∫=

Jt200cos200

6.9 200/11o−=

= =−π− mJ)1(cos48 96 mJ Chapter 6, Solution 38.

( )dtte2e10x40dtdiLv t2t23 −−− −==

= 240 0t,mVe)t1( t2 >− −


)0(iidtL1i

dtdiLv t

0 +∫=→=

1dt)4t2t3(10x200

1i t0

23 +∫ ++= −

1)t4tt(5t

023 +++=

i(t) = 5t3 + 5t2 + 20t + 1 A Chapter 6, Solution 40

dtdix

dtdiLv 31020 −==

<<+<<

<<=

ms 4t310t,40-ms 3t110t,-20

ms 10,10 tti

<<<<<<

=ms 4t3,10x10ms 3t1,10x10-

ms 10,1010

3

3

3 tx

dtdi

<<<<<<

=ms 4t3V, 200ms 3t1V, 200-

ms 10,V 200 tv

which is sketched below. v(t) V 200 0 1 2 3 4 t(ms) -200


( )∫∫ +−

=+= −t

o

t2t

03.0dt2120

21)0(ivdt

L1i

= Aetet tto

t 745103021 22 .. −+=+

+ −−10

At t = ls, i = 10 - 4.7 + 5e-2 = 5.977 A

L21w = i2 = 35.72J


∫ ∫ −=+=t

o

t

o1dt)t(v

51)0(ivdt

L1i

For 0 < t < 1, ∫ −=−=t

01t21dt

510i A

For 1 < t < 2, i = 0 + i(1) = 1A

For 2 < t < 3, i = ∫ +=+ 1t2)2(idt1051 2

t

= 2t - 3 A For 3 < t < 4, i = 0 + i(3) = 3 A

For 4 < t < 5, i = ∫ +=+t

4

t4 3t2)4(idt10

51

= 2t - 5 A

Thus,

2 1 , 0 11 , 1 2

( ) 2 3 , 2 33 , 3 42 5, 4

t A tA t

i t t A tA tt t

− <

5

<< <

= − < < < < − < <


w = L )(Li21)t(Li

21idt

2t

−∞−=∫ ∞−

( ) 010x60x10x80x21 33 −= −−

= 144 µJ Chapter 6, Solution 44.

( )∫ ∫ −+=+=t

ot

t

oo 1dt)t2cos104(51tivdt

L1i

= 0.8t + sin 2t -1


i(t) = ∫ +t

o)0(i)t(v

L1

For 0 < t < 1, v = 5t

∫−=t

o3 t510x101i dt + 0

= 0.25t2 kA

For 1 < t < 2, v = -10 + 5t

∫ ++−= −

t

13 )1(idt)t510(10x101i

∫= ( +−t

1kA25.0dt)1t5.0

= 1 - t + 0.25t2 kA

<<+−

<<=

2t1,kAt25.0t11t0,kAt25.0

)t(i2

2

Chapter 6, Solution 46. Under dc conditions, the circuit is as shown below: 2 Ω

+

vC

−3 A

iL

4 Ω

By current division,

=+

= )3(24

4iL 2A, vc = 0V

L21w L = =

= 22

L )2(21

21i 1J

C21w c = == )v)(2(

21v2

c 0J

Chapter 6, Solution 47. Under dc conditions, the circuit is equivalent to that shown below: R

+ vC −

5 A

iL

2 Ω

,2R

10)5(2R

2iL +=

+=

2RR10Riv Lc +

==

2

262

cc )2R(R100x10x80Cv

21w

+== −

232

1L )2R(100x10x2Li

21w

+== −

If wc = wL,

2

3

2

26

)2R(100x10x2

)2Rx(R100x10x80

+=

−− 80 x 10-3R2 = 2

R = 5Ω

Chapter 6, Solution 48. Under dc conditions, the circuit is as shown below:

+

vC2

−

+ −

+

vC1

−

iL1

6 Ω

4 Ω

iL2 30V

=+

==64

30ii2L1L 3A

==

1L1C i6v 18V

=2Cv 0V


(a) ( ) =+=++= 36544165Leq 7H (b) ( ) ==+= 41266112Leq 3H (c) ( ) ==+= 446324Leq 2H


( )63124510Leq ++=

= 10 + 5||(3 + 2) = 10 + 2.5 = 12.5 mH


101

301

201

601

L1

=++= L = 10 mH

( )45

35x10102510Leq =+=

= 7.778 mH

Chapter 6, Solution 52. 3//2//6 = 1H, 4//12 = 3H After the parallel combinations, the circuit becomes that shown below. 3H a 1H 1 H b Lab = (3+1)//1 = (4x1)/5 = 0.8 H


[ ])48(6)128(58106Leq +++++= 416)44(816 +=++= Leq = 20 mH


( )126010)39(4Leq +++= 34)40(124 +=++= Leq = 7H Chapter 6, Solution 55.

(a) L//L = 0.5L, L + L = 2L

LLLLLxLLLLLeq 4.1

5.025.025.0//2 =

++=+=

(b) L//L = 0.5L, L//L + L//L = L Leq = L//L = 0.5L


3L

L31LLL ==

Hence the given circuit is equivalent to that shown below: L

L/3 L

L/3

=+

=

+=

L35L

L35Lx

L32LLLeq L

85


Let dtdiLeqv = (1)

221 vdtdi4vvv +=+= (2)

i = i1 + i2 i2 = i – i1 (3)

3

vdtdi

ordtdi

3v 2112 == (4)

and

0dtdi5

dtdi2v 2

2 =++−

dtdi5

dtdi2v 2

2 += (5)

Incorporating (3) and (4) into (5),

3

v5dtdi7

dtdi5

dtdi5

dtdi2v 21

2 −=−+=

dtdi7

351v2 =

+

dtdi

835v2 =

Substituting this into (2) gives

dtdi

835

dtdi4v +=

dtdi

867

=

Comparing this with (1),

==867Leq 8.375H


===dtdi3

dtdiLv 3 x slope of i(t).

Thus v is sketched below:

6

t (s)

7 5 1 4 3 2

-6

v(t) (V) 6 Chapter 6, Solution 59.

(a) ( )dtdiLLv 21s +=

21

s

LLv

dtdi

+=

,dtdiL11 =v

dtdiL22 =v

,vLL

Lv s

21

11 += s

21

2L v

LLL

v+

=

(b) dtdi

Ldtdi

Lvv 22

112i ===

21s iii +=

( )

21

21

21

21s

LLLLv

Lv

Lv

dtdi

dtdi

dtdi +

=+=+=

∫∫ =+

== dtdtdi

LLLL

L1vdt

L1i s

21

21

111 s

21

2 iLL

L+

=+

== ∫∫ dtdtdi

LLLL

L1vdt

L1i s

21

21

222 s

21

1 iLL

L+


8

155//3 ==eqL

( ) tteqo ee

dtd

dtdiLv 22 154

815 −− −===

∫∫ −−− +=+=−+=+=t

tt

ttt

ooo eeeidttvLIi

0

2

0

22

0

A 5.15.05.12)15(512)0()(


(a) is = i1 + i2 i )0(i)0(i)0( 21s += 6 i)0(i4 2+= 2(0) = 2mA (b) Using current division:

( ) ==+

= − t2s1 e64.0i

203020i 2.4e-2t mA

=−= 1s2 iii 3.6e-2t mA

(c) mH1250

20x302030 ==

( ) === −−− 3t231 10xe6

dtd10x10

dtdiLv -120e-2t µV

( ) === −−− 3t232 10xe6

dtd10x12

dtdiLv -144e-2t µV

(d) ( )6t43mH10 10xe3610x30x

21w −−−=

Je8.021t

t4 µ=

−=

= 24.36nJ

( ) 2/1t6t43

mH30 10xe76.510x30x21w =

−−−=

= 11.693nJ

( ) 2/1t6t43

mH20 10xe96.1210x20x21w =

−−−=

= 17.54 nJ


(a) mH 4080

60202560//2025 =+=+=xLeq

∫∫ +−−=+=+=→= −−−

− ttt

eqeq ieidte

xidttv

Li

dtdiLv

0

333

3

)0()1(1.0)0(121040

10)0()(1

Using current division,

iiiii41,

43

8060

21 ===

01333.0)0(01.0)0(75.0)0(43)0(1 −=→−=→= iiii

mA 67.2125e- A )08667.01.0(41 3t-3

2 +=+−= − tei

mA 33.367.2125)0(2 −=+−=i

(b) mA 6575e- A )08667.01.0(43 3t-3

1 +=+−= − tei

mA 67.2125e- -3t2 +=i

Chapter 6, Solution 63. We apply superposition principle and let

21 vvvo += where v1 and v2 are due to i1 and i2 respectively.

<<−<<

===63,2

30,22 11

1 tt

dtdi

dtdiLv

<<−<<<<

===64,4

42,020,4

2 222

ttt

dtdi

dtdiLv

v1 v2 2 4 0 3 6 t 0 2 4 6 t -2 -4 Adding v1 and v2 gives vo, which is shown below. vo(t) V 6 2 0 2 3 4 6 t (s) -2 -6 Chapter 6, Solution 64.

(a) When the switch is in position A, i=-6 =i(0) When the switch is in position B,

8/1/,34/12)( ====∞ RLi τ

A 93)]()0([)()( 8/ tt eeiiiti −− −=∞−+∞= ι (b) -12 + 4i(0) + v=0, i.e. v=12 – 4i(0) = 36 V (c) At steady state, the inductor becomes a short circuit so that v= 0 V


(a) === 22115 )4(x5x

21iL

21w 40 W

=−= 220 )2)(20(

21w 40 W

(b) w = w5 + w20 = 80 W

(c) ( )( )3t20011 10xe502005

dtdvLi −−−==

= -50e-200tA

( )3t20022 10xe50)200(20

dtdvLi −−−==

= -200e-200tA

( )3t20022 10xe50)200(20

dtdvLi −−−==

= -200e-200t A (d) i = i1 + i2 = -250e-200t A

Chapter 6, Solution 66. mH60243640601620Leq =+=++=

dtdiLv =

∫ +=t

o)0(ivdt

L1i

∫= 121 dt + 0 mA −

t

o3 t4sin10x60

=−= tot4cos50i 50(1 - cos 4t) mA

mH244060 =

mV)t4cos1)(50(dtd10x24

dtdiLv 3 −== −

= 4.8 sin 4t mV


∫−= viRC1vo dt, RC = 50 x 103 x 0.04 x 10-6 = 2 x 10-3

∫−

= t50sin10210v

3

o dt

vo = 100 cos 50t mV Chapter 6, Solution 68.

∫−= viRC1vo dt + v(0), RC = 50 x 103 x 100 x 10-6 = 5

vo = ∫ −=+−t

ot20dt10

51

The op amp will saturate at vo = 12± -12 = -2t t = 6s

Chapter 6, Solution 69. RC = 4 x 106 x 1 x 10-6 = 4

∫ ∫−=−= dtv41dtv

RC1v iio

For 0 < t < 1, vi = 20, ∫ =−=t

oo dt2041v -5t mV

For 1 < t < 2, vi = 10, ∫ −−−=+−=t

1o 5)1t(5.2)1(vdt1041v

= -2.5t - 2.5mV

For 2 < t < 4, vi = - 20, ∫ −−=++=t

2o 5.7)2t(5)2(vdt2041v

= 5t - 17.5 mV

For 4 < t < 5m, vi = -10, 5.2)4t(5.2)4(vdt1041v

t

4o +−=+= ∫

= 2.5t - 7.5 mV

For 5 < t < 6, vi = 20, ∫ +−−=+−=t

5o 5)5t(5)5(vdt2041v

= - 5t + 30 mV

Thus vo(t) is as shown below:

2 5

6

5

751 432

5

0 Chapter 6, Solution 70.

One possibility is as follows:

50RC

=1

Let R = 100 kΩ, F2.010x100x50

13 µ==C

Chapter 6, Solution 71. By combining a summer with an integrator, we have the circuit below:

∫ ∫ ∫−−−= dtvCR

1dtvCR

1dtvCR

1v 22

22

11

o

−+

For the given problem, C = 2µF, R1C = 1 R1 = 1/(C) = 1006/(2) = 500 kΩ R2C = 1/(4) R2 = 1/(4C) = 500kΩ/(4) = 125 kΩ R3C = 1/(10) R3 = 1/(10C) = 50 kΩ


The output of the first op amp is

∫−= i1 vRC1v dt = ∫ −=−

t

o63 2t100idt

10x2x10x101

−

= - 50t

∫−= io vRC1v dt = ∫ −−

t

o63 dt)t50(10x5.0x10x20

1−

= 2500t2 At t = 1.5ms, == −62

o 10x)5.1(2500v 5.625 mV Chapter 6, Solution 73.

Consider the op amp as shown below: Let va = vb = v

At node a, R

vvR

v0 o−=

− 2v - vo = 0 (1)

v +

vo

−

b + − vi C

R

v R

a

R

−+

R

At node b, dtdvC

Rvv

Rvv oi +

−=

−

dtdvRCvv2v oi +−= (2)

Combining (1) and (2),

dt

dv2

RCvvv oooi +−=

or

∫= io vRC2v dt

showing that the circuit is a noninverting integrator. Chapter 6, Solution 74.

RC = 0.01 x 20 x 10-3 sec

secmdtdv2.0

dtdvRCv i

o −=−=

<<−<<<<−

=4t3,V23t1,V21t0,V2

vo

Thus vo(t) is as sketched below:

3

t (ms)

2 1

-2

vo(t) (V) 2


,dt

dvRCv i0 −= 5.210x10x10x250RC 63 == −

=−= )t12(dtd5.2vo -30 mV


,dt

dvRCv io −= RC = 50 x 103 x 10 x 10-6 = 0.5

<<<<−

==5t5,55t0,10

dtdv5.0v i

o

The input is sketched in Fig. (a), while the output is sketched in Fig. (b).

t (ms)

5

5

0 10

(b)

15

-10

t (ms)

vi(t) (V)

5

5

0 10

(a)

15

vo(t) (V) Chapter 6, Solution 77. i = iR + iC

( )oF

0i v0dtdC

Rv0

R0v

−+−

=−

110x10CR 66F == −

Hence

+−=

dtdv

vv ooi

Thus vi is obtained from vo as shown below: –dvo(t)/dt – vo(t) (V)

4

-4

t (ms)

1

4

0 2 3

vi(t) (V)

3

8

2 1

t (ms)

-8

4 -4

4

-4

t (ms)

1

4

0 2 3


ooo

2

vdtdv2

t2sin10dtvd

−−=

Thus, by combining integrators with a summer, we obtain the appropriate analog computer as shown below:


We can write the equation as

)(4)( tytfdtdy

−=

which is implemented by the circuit below. 1 V t=0 C R R R R/4 R dy/dt - - -

+ -y + + R dy/dt

f(t)

R

t = 0

-dvo/dt

dvo/dt

d2vo/dt2

d2vo/dt2

-sin2t

2vo

C C

R

R/10

R

R

R

R/2

R

R

− +

−+

− +

−+

− +

+ − sin2t

−+

vo

R


From the given circuit,

dt

dvk200k1000v

k5000k1000)t(f

dtvd o

o2o

2

ΩΩ

−ΩΩ

−=

or

)t(fv2dt

dv5

dtvd

oo

2o

2

=++


We can write the equation as

)(252

2

tfvdtvd

−−=

which is implemented by the circuit below. C C R R - R R/5 d2v/dt2 + - -dv/dt + v - + d2v/dt2 R/2 f(t)

Chapter 6, Solution 82 The circuit consists of a summer, an inverter, and an integrator. Such circuit is shown below. 10R R R R - + - vo + R C=1/(2R) R - + + vs - Chapter 6, Solution 83.

Since two 10µF capacitors in series gives 5µF, rated at 600V, it requires 8 groups in parallel with each group consisting of two capacitors in series, as shown below:

+

600

−

Answer: 8 groups in parallel with each group made up of 2 capacitors in series.


tqI∆∆

=∆ ∆I x ∆t = ∆q

∆q = 0.6 x 4 x 10-6

= 2.4µC

62.4 10 150

(36 20)q xC nv

−∆= = =∆ −

F

Chapter 6, Solution 85. It is evident that differentiating i will give a waveform similar to v. Hence,

dtdiLv =

<<−

<<=

2t1,t481t0,t4

i

<<−

<<==

2t1,L41t0,L4

dtdiLv

But,

<<−

<<=

2t1,mV51t0,mV5

v

Thus, 4L = 5 x 10-3 L = 1.25 mH in a 1.25 mH inductor Chapter 6, Solution 86. (a) For the series-connected capacitor

Cs = 8C

C1....

CC

=+++

111

For the parallel-connected strings,

=µ=== F3

1000x108C10

C10C sseq 1250µF

(b) vT = 8 x 100V = 800V

( ) 262Teq )800(10x1250

21vC

21w −==

= 400J

Chapter 6, Solution 1Chapter 6, Solution 18. For the capacitors in parallel 1 = 15 + 5 + 40 = 60 µF Ceq Hence 10 1 60 1 30 1 20 1 C 1 eq = + + = Ceq = 10 µF Chapter 6, Solution 19.

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