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 ω=

Figure 6.1

(a) =×=+

== 7-7-cc 102

dt

)6t2(d10

dt

dvC)t(i A2.0 µµµµ

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

))tcos(2(d10

dt

dvC)t(i 7-7-c

c ωω=ω

==

=)t(ic At)sin(0.2- µµµµωωωωωωωω

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

(a)

<<<

<=

t5

5t0

0t

A0

A1

A0

)t(i (b)

<<<

<=

t2

2t0

0t

A0

At

A0

)t(i 2

+− 0.10 µµµµF

ic(t)v(t)

Figure 6.1

(a) ∫= dt)t(iC

1)t(v cc

For 0t < , V0)t(vc =

For 5t0 << , V10t10

tdt1

10

1)t(v 5

5-

t

05-c ×=== ∫For t5 < , V105)t(v 5

c ×=

=)t(vc

<<<<××××<<<<<<<<××××

<<<<

t5V105

5t0V10t

0tV0

5

5

These voltages are quite large. This is due to the large currents and smallcapacitances. Normally, the currents would be quite small, in the µA range.

(b) ∫= dt)t(iC

1)t(v cc

For 0t < , V0)t(vc =

For 2t0 << , V103

tdtt

10

1)t(v 5

3t

0

25-c ×== ∫

For t2 < , V103

8)t(v 5

c ×=

=)t(vc

<<<<××××

<<<<<<<<××××

<<<<

t2V10667.2

2t0V103t

0tV0

5

53

Problem 6.3 [6.11] A voltage of V)t4cos(60 π appears across the terminals of a3-mF capacitor. Calculate the current through the capacitor and the energy stored in it from

0t = to s125.0t = .

i(t) 10 µµµµF

+

vc(t)

−−−−

dt

))t4cos(60(d)103(

dt

dvCi 3-

π×==

=ππ×= t)]-sin(4)[4)(60)(103(i -3 At)sin(40.72- ππππππππ

=πππ== t))]sin(4-0.72[)]t4cos(60[vip Wt)sin(821.6- ππ

∫∫ ππ==81

0

t

0dtt)sin(8-21.6dtpw

=ππ

π= 81

0)t8cos(8

6.21w J5.4-

Problem 6.4 Find )t(vc , as shown in Figure 6.1, given that

Figure 6.1

(a) =)t(vc

<<<<<<<<<<<<−−−−++++<<<<<<<<−−−−

<<<<<<<<

<<<<

t6kV9

6t5kV)27t12-t(

5t2kV)2t2(

2t0kV2t

0tV0

2

2

(b) =)t(vc

<<<<<<<<<<<<++++<<<<<<<<++++

<<<<

t2V0

2t0kV8)-4t(

0t2-kV8)4t(

2-tV0

(a)

i(t)

2 A

2 5 6 t

(b)

i(t)

4 A

t

2

–2–4

i(t) 1 mF

+

vc(t)

−−−−

SERIES AND PARALLEL CAPACITORS

Problem 6.5 Given the circuit in Figure 6.1, V0)0(v1 =− , V100)0(v2 =− , calculatethe voltages after the switch closes.

Figure 6.1

Carefully DEFINE the problem.Each component is labeled completely. The problem is clear.

PRESENT everything you know about the problem.Since the capacitors are in parallel, the charge, CVq = , must remain the same.

Also, when the switch is closed, the voltages across the capacitors are the same.

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 can be used to solve this problem.

ATTEMPT a problem solution.

For 0t < , J5)100)(10(2

1VC

2

1w 232

22 === − (there is no initial charge on C1)

For 0t < , C1.0)100)(10(VCq 322 === −

For 0t > , 2CCC 21 =+= mF

V50102

1.0V 3 =

×= −

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

1CV

2

1w 232 =×== −

+

v2

−−−−

+

v1

−−−−

1 mF 1 mF

t = 0

20 µµµµF

C1 10 µµµµF 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 µµµµ

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

Figure 6.1

10 µµµµF10 µµµµF 10 µµµµF

20 µµµµF

There is an infinite number of solutions.To find one solution, let F10C2 µ= , 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 µµµµ and =2C F10 µµµµ produce F30Ceq µ=

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 +ω=

(a) =+

==dt

)6t5(d5

dt

)t(diL)t(vL V25

(b) dt

))30tsin(3(d5

dt

)t(diL)t(vL

°+ω==

=)t(vL V)30tcos(15 °°°°++++ωωωωωωωω

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(diL)t(v L

L =

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

)2(d3)t(vL V0

For 0t3- << , ==dt

)3t2-(d3)t(vL V2-

For 2t0 << , ==dt

)t(d3)t(vL V3

For t2 < , ==dt

)2(d3)t(vL V0

(b) For -6t < , ==dt

)0(d3)t(vL V0

For -5t6- << , =+

=dt

)6t(d3)t(vL V3

For 0t5- << , ==dt

)1(d3)t(vL V0

For 1t0 << , =+

=dt

)1t(d3)t(vL V3

For 4t1 << , ==dt

)2(d3)t(vL V0

(a)

i(t)

2 A

t2–3

(b)

i(t)

2 A

1–5

1 A

4 5–6 t

i(t) 3 H

+

vL(t)

−−−−

For 5t4 << , =+

=dt

)10t2-(d3)t(vL V6-

For t5 < , ==dt

)0(d3)t(vL V0

Problem 6.11 [6.35] The voltage across a 2-H inductor is V)e1(20 t2-− . If theinitial current through the inductor is 0.3 A, find the current and the energy stored in the inductorat s1t = .

3.0dt)e1)(20(2

1)0(idtv

L

1i

t

0

2t-t

0+−=+= ∫∫

A)7.4e5t10(3.0e2

1t)10(i 2t-t

02t- −+=+

+=

At s1t = ,

=−+= 7.4e510i -2 A977.5

== 2iL2

1w J72.35

Problem 6.12 For the circuit shown in Figure 6.1, calculate )t(iL given that

Figure 6.1

(a)

v(t)

20 V

2 5 6 t

2

(b)

v(t)

40 V

t–2

–40 V

+− 4 H

iL(t)

v(t)

(a) =)t(iL

<<<<<<<<<<<<−−−−++++<<<<<<<<−−−−<<<<<<<<

<<<<

t6A5.22

6t5A)5.67t30-2.5t(

5t2A)5t5(

2t0At)45(

0tA0

2

2

(b) =)t(iL

<<<<<<<<<<<<++++<<<<<<<<++++

<<<<

t2A0

2t0A)20t10-(

0t2-A)20t10(

-2tA0

SERIES AND PARALLEL INDUCTORS

Problem 6.13 Given the collection of inductors shown in Figure 6.1, find the value of theequivalent inductance.

Figure 6.1

5 in series with 5 1055 =+=

10 in parallel with 10 51010

)10)(10(=

+=

5 in series with 10 15105 =+=

Thus, =eqL H15

Problem 6.14 Given the collection of inductors shown in Figure 6.1, find the values of 1L

and 2L , when the equivalent inductance is 20 H.

10 H

5 H

10 H

5 H

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 ΩΩΩΩ 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 ++=

∫ ττ++= d)(iC

1

dt

)t(diL)t(Ri)t(v

)0(vd)5(5)1)(5()5t)(10()t(v C

t

0+τ+τ+++= ∫

)0(vt52

t)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

50t10

10

)t(v)t(i R

R +=+

==

For the inductor, )0(it)0(it55

1d5

5

1d)(v

L

1)t(i LL

t

0LL +=+⋅=τ=ττ= ∫∫

For the capacitor, 5t)25t5(5

1)t25t5.2(

dt

d

5

1

dt

)t(dvC)t(i 2C

C +=+=+⋅==

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

A2

1)5(

10

1)t510(

dt

d

10

1

dt

)t(dvC)t(i C

c =

=+==

At2

11

10

t510

R

)t(v)t(i C

10

+=+

==Ω

At2

1

2

3t

2

11

2

1)t(iL

+=

++=

V5.22

1)5(t

2

1

2

3

dt

d)5(

dt

)t(diL)t(vL =

=

+==

V)t1030()20(t2

1

2

3R)t(iR)t(i)t(v L2020 +=

+=== ΩΩ

)t510()5.2()t1030()t(v)t(v)t(v)t(v CL20 ++++=++= Ω

=)t(v V)5.42t15( ++++

Problem 6.18 [6.67] Design an analog computer to simulate

)t2sin(10vdt

dv2

dt

vdo

oo2

=++

where 2)0(vo = and 0)0(vo =′ .

ooo

2

vdt

dv2)t2sin(10

dt

vd−−=

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

v(t)

20 ΩΩΩΩ 5 H

1/10 F+− 10 ΩΩΩΩ

+

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/dt2

d2vo/dt2

–dvo/dt

vo

−+

R

R

dvo/dt

R/2

−+

R

R

−+

R

R

R/10

–sin(2t)+−

sin(2t)