Rö 8 /_2o11 Yo arftfpol*<x &l .lt31.51' A High-Pass Filter.One application of L-R-C seriescircuits is to high-pass or low-pass filters, which filter outeither the low- or high-frequencycomponents of a signal. A high-pass filter is shown in Fig.P3 1.51, where the output volt-

31.65 " Si..i The currenr in a

certain circuit varies with time asshown in Fig. P31.65. Find theaverage curent and the nns cur-rcnt in terms of .16.

age is taken across the L-Ä combination. (The L_R combinationrepresents an inductive coil that also has resistance due to the largelength of wire in the coil.) Derive an expression for V"u,/(, theratio of the output and source voltage amplitudes, as a function ofthe angular tiequency a.r of the source. Show that when a.r is small,this ratio is proportional to a.r and thus is small, and show that theratio approaches unity in the limit of large frequency.

3t.5Sl -100-f) resistor and a 6.00-pF .upu.ito. are connectedin parallel to an ac generator that supplies an rms voltage of220yat an angular frequency of 360 rad/s. Use the results of problem3 L56. Note that since there is no inductor in the circuit, the If uLterm is not present in the expression for Z. Find (a) the currentamplitude in the resistor; (b) the current amplitude in the capacitor;(c) the phase angle of the source current with respect to the sourcevoltage; (d) the amplitude of the current through the generator.

!e] _D_oes the source current lag or lead the source voltage?

Figure P31.51

Figure F31.6f;

i

quantity. Consider the L-R-C series circuir shown in Fig. p31.75. Thevalues ol the circuit elements, the source voltage amplitude, antl thesource angular frequency are as shown. Use the phasor diagram tech_niques presented in Section 3l.l to solve tbr (a) rhe current amplitudeand (b) the phase angle @ of the current with respect to rhe sourcevoltage. (Note that this an_qle is the negative of the phase angledefined in Fig.3l.l3.) Now analyze the same circuit using ihecomplex-number approach. (c) Determine the complex impedanceof the circuit, 2.p". Tuke the absolute vulue to obtiin Z. the actual

irnpetlunce ol the circuit. rtl) Take the voltage umplitude rrf rhesource, /.0*, to be real, and find the complex cuffent ilmplitude /cp".Find the actual currenr amplitude by taking the absolute value of/.n,.(e) Find the phase angle ry' of the current with respect to the sourcevoltage by using the real and imaginary parts of 1.n*, as explainedabove. (f) Find the complex representations of the voltages across theresistance. the inducrance, and the capacirance. (g) Adding theanswers tbund in part (f ), verify that the sum of these complex num_bers is real and equal to 200 V the voltage ofthe source.

32.43 A small helium-neon laser ernits red u;-iribt. ligt,, *i,t upower of 4.60 mW in a beam that has a diameter of 2.50 mm.(a) What are the amplitudes of the electric and magnetic fields ofthe light? (b) What are the average energy densities associatedwith the elecrric field and with the magneric field? (c) What is thetotal energy contained in a 1.00-m length of the beam?

32.48 , i e circular loop of r.r'ire has radius 7.50 cm], sinu-soidal electromagnetic plane wave traveling in air passes throughthe loop, with the direction of the magnetic field of the wave per_pendicular to the plane of the loop. The intensity of the wave at thelocation of the loop is 0.0195 W/m2, and the wavelengrh of thewave is 6.90 m. What is the maximum emf induced in the loop?

Io

3f .75 .'' Complex Numbers in a Circuit. The volrage acrossa circuit element in an ac circuit is not necessarily in phase withthe current through that circuit rj_..-_ h._d -relement. Therefore the voltage rlgure l'J'l ' '

iamplitudes across the circuitelements in a branch in an ac

circuit do not add algebraically.A method that is commonlyemployed to simplify the analy-sis of an ac circuit driven by a

sinusoidal source is to representthe impedance Z as a complex number. The resistance R is taken tobe the real part ofthe impedance, and the reactance X : Xt - Xcis taken to be the imaginary part. Thus, for a branch containing a

resistor, inductor, and capacitor in series, the complex impedanceis Z"r* : R + iX, where 12 : - l. If the voltage amplitude acrossthe branch ir %p*, we define a complex current amplitude byIrp, : Vp*lZrp^. The actual cunent amplitude is the absolutevalue ofthe complex current amplitude; that is, / : (1.0**1.0^)l/2.The phase angle @ of the current with respect to the source voltageis given by tan@ = Im(1"0*)/Re(1.0*). The voltage amplitudesVp-"p^, V2-gpy, and V6-.0^ across the resistance, inductance, andcapacitance, respectively, are found by multiplying Irr^by R, iX1,and -r'Xc, respectively. From the complex representation lbr thevoltage amplitudes, the voltage across a branch is just the algebraicsum of the voltages across each circuit element: %p, : VR-.p, *Vtrp* * V6'-.0^. The actual value of any current amplitude or volt-age amplitude is the absolute value of the corresponding complex

L : 0.500 H

-rUtrffL-Au: ,onu I.:\Y. = lfiroraa/rf 125/rF

I ...:lR:4m{]

fi,ö g % uupQTra,Qttprl\-

Äo )tu\rqlr

-

:t

?l'il .. IErl.,l3 Y 5

[*

VK:Vr=v-:Vort

V:3%u+V1

Du uI

Vut ,\.,

Vt cv

RIlx^II=l.j-llt = *'rJ-r

ixir= l-*l r =#- r1/mr"-r

\/TVruutLcL

,w "0;&,t lL'-^ &Lfr- s L,-; u c"/) å

n\rxr! d" RcuCÅ1,Y)l

0w w ", rl"r{, [*" h''ttn ; r]e/?t "

%.1+ Nr<--e-'t\/

Vs

lciI

gz' 1 N^L'R n (w L**,d

V:

V 'r, i3

Vl,rz-

P;rY' + F

U

fr=40-oJLC * 6,o*Fcrl * t {0 raäf )

U : 8"ZoV hms\

A,LL= \cqc^rt /

å';" il= U'ile

)rL= 3llV ' Col L>e o {" t)

o.)

b)

c)

:/

A/tR

/})"g

: TDZ'?JL-

LMKtr^,

l+@* v,Q,u+u A

<...--. i'- ,lrLN-

\7- ZA.-a )'L- ,be= R-4;r->pc^

X"Xc=k

a fru"E

ACLc=bwC

Z-

uJCR =D q= 4A,8o

Al-

Je l,o, A;LLÅl

-) Vällwrr S yt*,*;*;t{t a )ffi,il'^4',

t;)fq

I*t= "(1p

?- q%'r åJlo7'7 r7Jo

= *-'tryt'c+t

3tmE"!, /j '+6

1)

2)

MerLilrilrcd tr" 0 ['l,L ryn't"*u{ri '

l*,:&;A"^ ue o( ",/tt vvlsauLu*1rn/'1 *'u '

fio h u h pT'Å',':i;trUi^;"'];Å"' f P)

lrnnn I

.Tto=E

Ut')*

9 t,*t

ll : looV

lgl :lO00t-l

L=QfooHl,t;7ff oo-fL

c=R=

XLXc

7= 6;aL

\-cuc

=

7Kl =-

Xrr >X.r --

= Toö-{L

c/ =_ 7 6,7/

rueÅ ()6.36,g0 &

qV :- c1\o A

'-----

l60vL>oVSLbv

UL

uc-

Uq-U;=U.--

C'nV = ffi =)

;ly;,,r,^rr^ b;f ) f*A'Lh; hf ork{,^,,^*tu-

ryqn'vi,t";ra,

a)7 =. Rti@L-å)=-

L?K

Aors

lro4ooi

T -- lz*"1t-) **^q = Q:a\

t - c9,92-

uur"

utcf K

Utaf r

z=lz"v.l :6,t n -3 o oi (-p-)

= t-aT*{LC )" cr-t ovq'w)

U-z*.

A) T" =_' tF( 7 0,ZL+Azrt; (A)

D

:2 ? +6,1ot'Fxft#',tr)T q = lzf t16i tv)*tY* I

= LV.*L,L = * lLotl6t; CtJ )*-[--^ :! = \1f-2rli CV)- -.fn tl C

U, +U. -t-c+^ -"f *

kco V:

9l,qg

a) r=4ar(-:hT-

'/, Go ^ /1/

2 ,5o htn.-?

4,60, lo 'W : 1S+ V[/7,,",tr,f l,Lf ,lo'u^)'t J:---J:a^Å/t't A'Åo^ AAt ',

Ee f*d eo (kx-*4jB '- lJ*l* cq k,r-ut+)z

{o = 8'f J'rc ''C/t"

frl ot^

.4/-./*

q

-L4-FJ-b /r4q<

B^u* =

\IE\/ eoc

Tl-l'w< A-t-

C

l----

-6: 2'80'/0 T

c) [/*u : Ur**()r*)'t'rr%, -/l

=f,n/0J-?=

?L1r DIl- =VB ry,

-(Ä ? %; =. ,Fuo^=-. = €oE-*rtL7^orJ* 1rr, Y":,,-Å 1

u -L=+az*

?- '!n'[fr-WaWrFu, 'l

(_

vih6", lr"'nTJJ_A

a'ttf U qn l- au'{4 C :

3(,qt ,Fu= l,zr6e 'tr-'T^/n

/ " =1,focvn

) = C,qo yn

Gir,ru ", f

= fr,

J /')" '

l;t = W"Co>Tt r-wL)

T=*.':W=W (r=r4_2.Bro*=fry:-W,/4toT

<-rt---

F : A'I p,a,c ''-

Lh,W,/,1r,6,1a n

< 6/,*

Mf 'B^o*

Q,or;qT

{-

*,

{ar

^/