Inductance 227

*Q32.14 (a) The instant after the switch is closed, the situation is

as shown in the circuit diagram of Figure (a).The requested quantities are:

11=01=£01=£°1L 'c R'R R

I ~VL =£0' ~Vc =0, ~VR =£0 I

Q=O

~ Vc = 0

h=O+

(b) After the switch has been closed a long time, thesteady-state conditions shown in Figure (b) willexist. The currents and voltages are:

I 1L = 0, 1C = 0, 1R = 0 I

I ~VL =0, ~Vc =£0' ~VR =0 I

Figure (a)

h = 0 + _ ~ VL = 0

~VR= 0

+ 1 _

Co

Figure (b)

FIG. Q32.14

) ,

*Q32.15 (i) Answer (a). The mutual inductance of two loops in free space-that is, ignoring the use ofcores-is a maximum if the loops are coaxial. In this way, the maximum flux of the primary loopwill pass through the secondary loop, generating the largest possible emf given the changingmagnetic field due to the first.

(ii) Answer (c). The mutual inductance is a minimum if the magnetic field of the first coil lies inthe plane of the second coil, producing no flux through the area the second coil encloses.

Q32.16 When the capacitor is fully discharged, the current in the circuit is a maximum. The inductance ofthe coil is making the current continue to flow. At this time the magnetic field of the coil containsall the energy that was originally stored in the charged capacitor. The current has just finisheddischarging the capacitor and is proceeding to charge it up again with the opposite polarity.

Q32.17 If R> {¥' 'then the oscillator is overdamped-it will not oscillate. If R < {¥' then theoscillator is underdamped and can go through several cycles of oscillation before the radiatedsignal falls below background noise.

Q32.18 An object cannot exert a net force on itself. An object cannot create momentum out of nothing.A coil can induce an emf in itself. When it does so, the actual forces acting on charges in

different parts of the loop add as vectors to zero. The term electromotive force does not refer toa force, but to a voltage.

Section 32.1 Self-Induction and Inductance

P32.1£=-L M =(-2.00 H{ 0-0.500 A)( ~) =1100 V I~t 0.0100s IH·A .--.

Treating the telephone cord as a solenoid, we have:

L_l1oN2A (4nxlO-7 T.m/A)(70.0)2n(6.50XIO-3 mr _I I- f! 0.600 m - 1.3611H

228 Chapter 32

*P32.3 £=-L ~~ =-L :t (frnax sin cat) =-LcaIrnax cos cot =-(10.0 X 1O-3)(120n)(5.00)coscat

£= -( 6.00n)cos(120nt) = 1-(18.8 V)cos(377t) I

@ From I£I=L(~} we have

Ncl>From L = __ B , we haveI

£ 24.0 X 10-3 V = 2.40 X 10-3 HL=(Mj/),.t)= 1O.0Ajs

LI _ (2.40 X 10-3 H)( 4.00 A) -119.2 .uT. m2 Icl>B=li- 500

P32.5 .uoN2A .uo(420? (3.00 X 10--4) --4L= __ = =4.16xlO H£ 0.160

£=-L dI ~ dI = -£ = -175 X 10-6 V -1-0.421 A/s Idt dt L 4.16xlO--4 H .----.

FIG. P32.8

I = £0 e-kt = dqkL dt

\IQI= ~~ \

B= .uonI = .uo(~}0.040 0 A)= 1188.uT I

cl>B=BA=13.33XlO-s T.m2\

co_ co -kt --L dIc-coe - -dt

dI - £0 -kt d---e tL

If we require I ~ 0 as t ~ 00, the solution is

Cd) I B and cl> B are proportional to current; L is independent of current.

I£I=L dI =(90.0X10-3)~(t2-6t) Vdt dt

Ca) At t = 1.00 s, £=1360 mV I

Cb) At t=4.00 s, £=1180 mV I

Cc) £=(90.0xlO-3)(2t-6)=0

when I t = 3.00 s I

P32.9

P32.6

Ncl>

Cc) L=--t-=I 0.375 mH I

j~ Ca)i Ij I'--. ../ Cb)

Inductance 229

Section 32.2 RL Circuits

P32.10 Taking l' = L, / = re-t/r:R I

/R + L d/ = 0 will be true ifdt

d/ = /ie-t/r 0.!. )dt l'

/iR e-t/r + L( /ie-t/r)( -~ ) = 0

Because l' = !:..., we have agreement with 0 = O.R

t=-1'ln(1-0.900)

t = -(0.200 s)In (0.100) = I 0.461 s I

P32.11 (a)At time t,

whereAfter a long time,At /(t) = 0.500/maxsoIsolating the constantson the right,

and solving for t,or(b)

Similarly, to reach 90% of / max'

andThus, Ll' = R = 0.200 s

/ _ e( 1-e~)max- eR =-R

(0.500)e = e(1_e-t/0.200S)R R

0.500 = 1- e-t/0.200 S

In (e -t/0.200s) = In (0.500)

_ t0.200 s = -0.693

t = I 0.139 s I

0.900 = 1- e-t/r

leA)

0.5

oo 0.2 0.4

FIG. P32.11

t (s)--r0.6

The current increases from 0 to asymptotically

approach 500 mA. In case (a) the current jumps upessentially instantaneously. In case (b) it increaseswith a longer time constant, and in case (c) the increaseis still slower.

500 mA

o

I

o

FIG. P32.12

FIG. P32.13

'V,o T 27

FIG. P32.15

4.00128.00Q

(1)

II;~-1I3-12~

(2) J..10.0 ~4.00 1.00~ g H(3) Q\

SFIG. P32.18

E

1max =Rand

dl R-=-1 e-t/rdt L max

dl =~e-t/r =(6.67 A/s)e-L50/(0500) =(6.67 Ajs)e-3.OO =1 0.332 Aj~dt L

_ . dl RoE 100 V _I It-O. -=-1 . e =-=--- 6.67 Ajsdt L max L 15.0 H L....-.---

~ VL = t:- ~VR = 36.0 V -16.0 V = 20.0 V

~ VR = 16.0 V = I 0.800 I~VL 20.0 V

~ VR = IR = (4.50 A)(8.00 Q) = 36.0 V

~VL =t:-~VR =~

~VR =IR=(8.00 Q)(2.00 A)=16.0 V

and

Therefore,

L 15.0 H = 0.500 s:'r = R = 30.0 Q

(b) t = 1.50 s:

(a)

(b)

1 = Imax (1- e-t/r) = (~}1- e--Q250/2.00)= I 0.176 A I

1 = ~ = 6.00 V = 11.50 A Imax R 4.00 Q ~.-_.

(d) 0.800 = 1_e-t/200ms -7t=-(2.00 ms)ln(0.200) = I 3.22 ms I

L = 'r R = (5.00 xlO-4 s)(60.0 V/A) = 130.0 mH I

(c)

(b)

continued on next page

( ) dl3+10.0 V - 4.0011 - 8.00/3 - 1.00 - = 0dt

+10.0 V - 4.00/1 - 4.00/2 = 0

P32.18 Name the cunents as shown. By Kirchhoff's laws:

1(=/2+/3

P32.17

P32.16 After a long time, 12.0 V = (0.200 A)R. Thus, R = 60.0 Q. Now, 7 = ~ givesR

P32.15 (a)

P32.14 1=~(l_e-t/r)= 120 (1_e-1.80/700) = 3.02 AR 9.00

~ VR = IR = (3.02) (9.00) = 27.2 V

~VL = t:- ~VR = 120-27.2= \92.8 V I

230 Chapter 32

Q (a),

! '

-

cR

FIG. P32.42

10.0 g 0.100 Hr.C-lhJ

V T1lF

Ifo = 2nJLC

I I I IC----- - 608pF- (2nfo)2 L - [2n(6.30X106 Hz)T (1.05 x 10-6 H) ~-

I I I If = - - 135 Hz

2nJLC 2n~(0.0820 H)(17.0xlO-6 F) --

Q = Qmaxcoswt = (180 pC)cos(847 xO.OOI 00) = 1119 pC I

1= ~7=-wQmax sinwt =-(847)(180)sin(0.847) =1_-_l_14_mA_1

I I I If=--=--======= 503Hz

2nJLC 2n~(O,100 H)(1.00XlO-6 F) ~-

Q=C£=(1.00xlO-6 F)(12.0 V)=112.0 pC I

u = Q2 _ ([105 X 1O-6]cos[(1.899 x 104 rad/s)(2.00XlO-3 s)Jtc 2C- 2(840xlO-12) 16·03JI

W=_I_= I -1.899x104 rad/s

JLC ~(3.30 H)(840XIO-12 F)

Q = Qmaxcos Wt, 1= dQ = -wQmaxsin wtdt

(c)

(b)

en the switch has been closed for a long time, battery, resistor,

and coil carry constant current Imax=!i.When the switch is opened,R

current in battery and resistor drops to zero, but the coil carries thissame current for a moment as oscillations begin in theLC loop.We interpret the problem to mean that the voltage amplitude of these

'11' . AV' I C( AV? I L/2 FIG. P32.39OSCI atlOns IS Ll ,Ill"2 Ll = "2 max'

_ C( L1v)2 _ C (L1V)2 R2 _ (0.500 X10-6 F) (150 V)2 (250 Q)2 _I IThen, L - 2 2 2 - 0.281 H .Imax £ (50.0 V) --

(b)

(d) At all times

The resonance frequency is

(c)

Thus,

(a)

236 Chapter 32

P32.42 (a)

P32.41 This radio is a radiotelephone on a ship, according to frequency assignments made byinternational treaties, laws, and decisions of the National Telecommunications and InformationAdministration.

P32.43

continued on next page

n1)32.40 (a)U

I I

(b)

(~

U _ 1 LI2 _ 1 Lro2Q2 . 2( ) _ Q~ax sin2 (rot)-- -- sm rot -----L 2 2 max 2C

U ==(105 X 10--6ct sin2 [(1.899 x 104 radjs)( 2.00 X10-3 s) ] ==[0.529 J IL 2(840 X10-12F) -.

(c) U,o,al==UC+UL==!6.56J I

Inductance 237

The RLC Circuit

( 7.60 )'2( 2.20 X 10-3) ==1.58 X 104 radjs

Therefore,id ==~; ==\2.51 kHz I

(b) Rc ==f§ ==169.9 Q I

P32.45 (a)

(b)

(c)

__ 1_ == 1 ==14.47 kradjs Iroo- J[C ~(0.500)(0.100XI0--6)

ro == 1_1 - ( R Y ==\4.36 kradjs Id 1 LC 2L

/).ro ==\2.53% lower Iroo

P32.46 Choose to call positive current clockwise in Figure 32.15. It drains charge from the capacitor

d· I' dQ A 1 k" d h .. h .accor mg to ==--. c oc WIse tnp aroun t e cIrcmt t en gIvesdt

Q dI+--IR-L-== 0

C dt

+Q + dQ R+L~ dQ ==0, identical with Equation 32.28.C dt dt dt

0.500 ==e-Rt/2L

so

and

Imax cc e- Rt/2L

~==-ln(0.500)2L

so

((~.. \~

2L \ eL )t==-Rln(0.707)== 0.347 R I (half as long)