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Unformatted text preview: Chapter 33 Solutions 1. Much of Europe uses AC power at 230 V rms and
50 Hz. Express this AC voltage in the form of
Equation 333. taking oi = 0. Solution Use of Equations 331 and 2 allows us to write Vp = (i V”... = «5(230 V} = 325 v, and w = 2:” = M50 Hz) = 314 5". Then the voltage expressed in the form of Equation 333 is Wt) = (325 V)x in[(314 s'l)t]. 3. An oscilloscope displays a sinusoidal signal whose
peaktopealr voltage (see Fig. 331) is 28 V. What
is the rms voltage? Solution is shown in Fig. 331, the peakto—peak voltage is
twice the peak voltage. so Equation 331 gives it... = V,/\/§ = VW/zﬁ = 28 woo/'2 = 9.90 V. An AC current is given by I = 495 sin(9.43t), with
l in milliamperes and t in milliseconds. Find
(a) the rms current and (b) the frequency in Hz. Solution Comparison of the current with Equation 333 shows
that its amplitude and angular frequency are I =
495 mA and w = 9.43 (marl. Application of Equa
tions 33] and 2 give (a) Inns = 495 mA/x/i = 350 mA. and (h) f 2 9.43/2rr(rns)=1.50 kHz. 13. What is the rms current in a 1.0pF capacitor
connected across the 120V rms. (SO—Hz AC line? Solution Equation 335 can be used with the rms current and
loltage. since both are 1/\/2_ times their peak values. 13115er = wCVms = (2W x 60 Hz)(1pF)(120 V) =
15.211121. 1?. A capacitor and a 1.8kﬂ resistor pass the same
current when each is separately connected across a
(SO—Hz power line. What is the capacitance? Solution The currents (nus or peak} are the same if X0 = R =
l/wC. so C = i/wR = [27r(60 Hz)(1.8 kiln—1 =
1.47 JuF. 19. A 50mH inductor is connected across a IDV rms AC generator. and an rrris current of 2.0 mA ﬂows. What. is the generator frequency? Solution From Equation 337. f = tau/2n = VP/erIPL. Since the
ratio of the peak values of voltage and current is the
same as that of the rms values, 3" = 10 V/21r(2 mA}x
(50 mH) = 15.9 kHz. 27. Find the resonant frequency of an LC circuit consisting of a 0.22pF capacitor and a 1.7mH
inductor. Solution
Equations 332 and 11 give f = I/Zin/L = rpm/(0.22 uF)(17 mH = 3.23 kHz. 29. You have a 2.0rnH inductor and wish to make an
LC circuit whose resonant frequency spans the AM
radio band [550 kHz to 1600 kHz). What range of
capacitance should your variable capacitor cover? Solution The resonant frequency of an LC circuit is
(Equation 3311) to = Um. so the capacitance
should cover a range from C = l/wQL = lierr x 550 idiz)2(2 mil) 2 41.9 pF down to C :
”(2: x 1.6 MHz)2(2 mH) = 4.95 pF. 39. The 20001117 capacitor in Fig. 3330 is initially
charged to 200 V. (a) Describe how you would
manipulate switches A and B to transfer all the
energy from the 2000,uF capacitor to the 500,(1F
capacitor. Include the times you would throw the
switches. (b) What will be the voltage across the
SUOuF capacitor once you‘ve ﬁnished? 3 A
500 uF 100 H 112000 pF FIGURE 3330 Problem 39. Solution (a) The energy initially stored in the first capacitor is
%(2 rnF)(200 V)2 = 40 .1. First close switch B for one
quarter of a period of the LC circuit containing the
2000 pF capacitor, or t5 2 %T3 = %(27r/w3) =
{en/1‘0— = %1’T‘/(10{} H)[2 mF) = 702 ms. This
transfers 40 J to the inductor. Then open switch B
and close switch A for one quarter of a period of the
LC circuit containing the 500 ,uF capacitor. or M =%n‘/(100 H){0.5 mF) = §t3 = 351 ms. This
transfers 40 J to the second capacitor from the l
inductor. Finally. open switch A. (b) When the second
capacitor has 40 J of stored energy, its voltage is #2010 .1) nos mF) = 400 v. Chapter 33 Solutions continued 58. A rurai power line carries 2.3 A rms at 4000 V.
A stepdown transformer reduces this to 235 V rms
to supply a house. Find (a) the turns ratio of the
transformer and (b) the current in the 235V line
to the house. Solution (a) The turns ratio given in Equation 3318 is
Nm/Npri = Vm/Vpri = 235/4000 = 1/17. (b) If there
are no transformer losses. Equation 3319 gives Im = (Vpﬁ/Vmﬂpﬂ 2 (17)(2.3 A} = 39.1 A (rms). 59. A transformer steps up the 120V rms AC power
line voltage to 23 kV rms for a TV picture tube. If
the rms current in the primary is 1.0 A, and the
transformer is 95% efﬁcient. what is the secondary
current? Solution Only 95% of the power in the primaryr is transformed
to power in the secondary. so Equation 33~19 can be
modiﬁed to yield (0.95)(120 V}(l A) = (23 kVHm. or
1m 2 4.96 mA (rms). Chapter 34 Solutions ’f. At a particular point the instantaneous electric field
of an electromagnetic wave points in the +1;
direction, while the magnetic ﬁeld points in the —z
direction. In what direction is the wave
propagating? Solution For electromagnetic waves in vacuum, the directions of
the electric and magnetic ﬁelds. and of wave
propagation, form a righthanded coordinate system,
as shown. (The vector relationship is summarized in Equation 3420b.) Therefore, the given wave is headed
in the —Idirection. 8. The ﬁelds of an electromagnetic wave are
E 2 E1, sinsz + wt)j and B = Br sin(kz + wt)'i.
Give a unit vector in the direction of propagation. Solution In an electromagnetic wave, when E is parallel toj
andBtoi. SisparalleltoEXBorin: —fr. {See
previous solution.) 21. A 60112 power line emits electromagnetic
radiation. What is the wavelength? Solution The wavelength in a vacuum [or air) is A = CH =
(3x108 In/s)/(60 Hz) = 5><10Ei In, almost as large as
the radius of the Earth. 31. A polarizer blocks 75% of a polarized light beam.
What is the angle between the beam‘s polarization
and the polarizer‘s axis? Solution
Equation 3419 gives 0 = cos“: ﬁS/Sg =
cos" 1— 15% = cos—1 ﬂ = 60°. 39. What would be the average intensity of a laser
beam so strong that its electric ﬁeld produced
dielectric breakdown of air (which requires
Ep 2 3Jr10ﬁ V/m)? Solution Equation 3421b for the average intensity of
electromagnetic waves gives 5' : E3/2noc =
{3x10'5 V/m)2(81rx10'7 H/rn)‘1(3><10a In/s)‘l =
11.9 GW/mz. 55. What is the radiation pressure exerted on a
lightabsorbing surface by a laser beam whose 2
intensity is 180 W / cm ? Solution The radiation pressure generated bv a totally
absorbed electromagnetic wave of given average
intensity (from Equation 3424) is Pm; = S {c =
{130 Vlr'/<:rn2)/(3x10a m/s) = 6 mPa. ...
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 Summer '08
 schuller

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