CHAPTER 6 - CAPACITORS AND INDUCTORS

List of topics for this chapter :CapacitorsSeries and Parallel CapacitorsInductorsSeries and Parallel InductorsApplications

CAPACITORS

Problem 6.1 For the circuit shown in Figure 6.1, find )t(ic given(a) V)6t2()t(v +=(b) V)tcos(2)t(v =

Figure 6.1

(a) ==+== 7-7-cc 102dt)6t2(d

10dtdv

C)t(i A2.0

(b) )tsin()2)(-10(dt))tcos(2(d

10dtdv

C)t(i 7-7-cc =

==

=)t(ic At)sin(0.2-

Problem 6.2 Find )t(vc as shown in Figure 6.1, given that

(a)

, 2CCC 21 =+= mF

V501021.0

V 3 ==

Now, J5.2)50)(102(21

CV21

w 232 ===

+

v2

+

v1

1 mF 1 mF

t = 0

20 F

C1 10 F C2

Clearly, the energy has gone from 5 J to 2.5 J. What happened to 2.5 J of energy? Well, theswitch cannot close fast enough to keep from having a spark. Thus, 2.5 J of energy must bedissipated in the spark.

EVALUATE the solution and check for accuracy.After the switch closes, the charge of 0.1 C remains the same and the voltage across both isnow the same, 50 V.Our check for accuracy was successful.

Has the problem been solved SATISFACTORILY? If so, present the solution; if not,then return to ALTERNATIVE solutions and continue through the process again.This problem has been solved satisfactorily.

== 21 vv V50

Problem 6.6 Find the equivalent capacitance for the collection of capacitors shown inFigure 6.1.

Figure 6.1

10 in parallel with 10 201010 =+=

20 in series with 20 102020

)20)(20(=

+=

10 in parallel with 10 201010 =+=

Therefore, =eqC F20

Problem 6.7 Given that the equivalent capacitance of the collection of capacitors shown inFigure 6.1 is F30 , find 1C and 2C .

Figure 6.1

10 F 10 F 10 F

20 F

There is an infinite number of solutions.To find one solution, let F10C2 = , the network is similar to the one shown in Problem 6.6.The first two combinations are the same. Hence, we have

10 in parallel with eq1 CC =

or 20C30C10 11 ==+

Therefore, =1C F20 and =2C F10 produce F30Ceq =

Problem 6.8 [6.17] Calculate the equivalent capacitance for the circuit in Figure 6.1.All capacitances are in mF.

Figure 6.1

3 in series with 6 236)3)(6(=

+=

2 in parallel with 2 422 =+=

4 in series with 4 244)4)(4(=

+=

The circuit is reduced to that shown below:

6 21

20

8

3

4

5

15

8

1 26 6

6 in parallel with 2 826 =+=

8 in series with 8 488)8)(8(=

+=

4 in parallel with 1 514 =+=

5 in series with 20 4205

)20)(5(=

+=

Therefore, =eqC mF4

INDUCTORS

Problem 6.9 For the circuit shown in Figure 6.1,

Figure 6.1

calculate )t(vL given that(a) A)6t5()t(i +=(b) A)30tsin(3)t(i +=

(a) =+== dt)6t5(d

5dt)t(di

L)t(vL V25

(b) dt))30tsin(3(d

5dt)t(di

L)t(vL+

==

=)t(vL V)30tcos(15 ++++

i(t) 5 H+

vL(t)

Problem 6.10 For the circuit shown in Figure 6.1, calculate )t(vL given that

Figure 6.1

dt)t(di

L)t(v LL =

(a) For -3t < , == dt)2(d

3)t(vL V0

For 0t3-

For 5t4

(a) =)t(iL

Problem 6.15

Figure 6.1

There is an infinite number of solutions.

However, if H5L2 = , the network is similar to the one shown in Problem 6.13. The first twocombinations are the same. Hence, we have

5 in series with eq1 LL =

or 15L20L5 11 ==+

Therefore, =1L H15 and =2L H5 produce H20Leq =

APPLICATIONS

Problem 6.16 Calculate the voltage across the current source in Figure 6.1 given thatA)5t()t(i += .

Figure 6.1

Carefully DEFINE the problem.Each component is labeled completely. The problem is clear, except for the value of thecapacitor voltage at some point in time.

10 H

5 H

L1

L2

i(t)+

v(t)

10 5 H

1/5 F

PRESENT everything you know about the problem.We know the current as well as the values of the elements. However, we do not know theinitial condition on the voltage across the capacitor. We will solve for the voltage across thecurrent source assuming that the capacitor voltage at 0t = is equal to )0(vC .

Establish a set of ALTERNATIVE solutions and determine the one that promises thegreatest likelihood of success.The three solution techniques that can be used are nodal analysis, mesh analysis, and basiccircuit analysis. Basic circuit analysis will be used to solve this problem.

ATTEMPT a problem solution.)t(v)t(v)t(v)t(v CLR ++=

++= d)(iC1dt )t(diL)t(Ri)t(v)0(vd)5(5)1)(5()5t)(10()t(v C

t

0+++++=

)0(vt52t)5(550t10)t(v C

2

+

++++=

)0(vt25t5.2550t10)t(v C2 +++++=)0(vV]55t35t5.2[)t(v C2 +++=

EVALUATE the solution and check for accuracy.The current through each element is the same. The voltage across each element wasdetermined while attempting a problem solution.

For the resistor, 5t10

50t1010

)t(v)t(i RR +=+

==

For the inductor, )0(it)0(it551

d551

d)(vL1)t(i LL

t

0LL+=+===

For the capacitor, 5t)25t5(51)t25t5.2(dt

d51

dt)t(dv

C)t(i 2CC +=+=+==

Hence,)t(i)t(i)t(i)t(i CLR === , when )0(i5)0(i L==

Our check for accuracy was successful.

Has the problem been solved SATISFACTORILY? If so, present the solution; if not,then return to ALTERNATIVE solutions and continue through the process again.This problem has been solved satisfactorily.

=)t(v )0(vV]55t35t5.2[ C2 ++++++++++++

Problem 6.17 Given the circuit in Figure 6.1, find )t(v for V)t510()t(vC += .

Figure 6.1

A21)5(

101)t510(dt

d101

dt)t(dv

C)t(i Cc =

=+==

At21

110

t510R

)t(v)t(i C10

+=

+==

At21

23

t21

121)t(iL

+=

++=

V5.221)5(t

21

23

dtd)5(dt

)t(diL)t(vL =

=

+==

V)t1030()20(t21

23

R)t(iR)t(i)t(v L2020 +=

+===

)t510()5.2()t1030()t(v)t(v)t(v)t(v CL20 ++++=++= =)t(v V)5.42t15( ++++

Problem 6.18 [6.67] Design an analog computer to simulate

)t2sin(10vdtdv

2dtvd

o

oo2

=++

where 2)0(vo = and 0)0(vo = .

o

oo2

vdtdv

2)t2sin(10dtvd

=

Thus, by combining integrators with a summer, we obtain the appropriate analog computeras shown below.

v(t)

20 5 H

1/10 F+ 10

+

vc(t)

Problem 6.19 Calculate )t(v and )t(vL for the circuit shown in Figure 6.1 andvC(0) = 10 Volts (with the plus side of vC at the top of the capacitor).

Figure 6.1

=)t(v V0 =)t(vL V)tcos(10

+

v(t)

10 H

1/10 Fsin(t)

+ vL(t)

+

C

R

+

C

R

t = 0

2 V

+

d2vo/dt2d2vo/dt2

dvo/dtvo

+

R

R

dvo/dt

R/2

+

R

R

+

R

R

R/10

sin(2t)+

sin(2t)

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PrefaceChapter 1 Basic ConceptsChapter 2 Basic LawsChapter 3 Methods of AnalysisChapter 4 Circuit TheoremsChapter 5 Operational AmplifierChapter 6 Capacitors and InductorsCapacitorsSeries and Parallel CapacitorsInductorsSeries and Parallel InductorsApplications

Chapter 7 First-Order CircuitsChapter 8 Second-Order CircuitsChapter 9 Sinusoids and PhasorsChapter 10 Sinusoidal Steady-State AnalysisChapter 11 AC Power AnalysisChapter 12 Three-Phase CircuitsChapter 13 Magnetically Coupled CircuitsChapter 14 Frequency ResponseChapter 15 Laplace TransformChapter 16 Fourier SeriesChapter 17 Fourier TransformChapter 18 Two-Port Networks

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