Chapter33_34solns

Chapter33_34solns - Chapter 33 Solutions 1 Much of Europe...

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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 33-3. taking oi = 0. Solution Use of Equations 33-1 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 33-3 is Wt) = (325 V)x in[(314 s'l)t]. 3. An oscilloscope displays a sinusoidal signal whose peak-to-pealr voltage (see Fig. 33-1) is 28 V. What is the rms voltage? Solution is shown in Fig. 33-1, the peak-to—peak voltage is twice the peak voltage. so Equation 33-1 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 33-3 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 1.0-pF capacitor connected across the 120-V rms. (SO—Hz AC line? Solution Equation 33-5 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 1.8-kfl 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 50-mH inductor is connected across a ID-V rms AC generator. and an rrris current of 2.0 mA flows. What. is the generator frequency? Solution From Equation 33-7. 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 0.22-pF capacitor and a 1.7-mH inductor. Solution Equations 33-2 and 11 give f = I/Zin/L = rpm/(0.22 uF)(1-7 mH = 3.23 kHz. 29. You have a 2.0-rnH 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 33-11) 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 2000-1117 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 SUO-uF capacitor once you‘ve finished? 3 A 500 uF 100 H 112000 pF FIGURE 33-30 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 235-V line to the house. Solution (a) The turns ratio given in Equation 33-18 is Nm/Npri = Vm/Vpri = 235/4000 = 1/17. (b) If there are no transformer losses. Equation 33-19 gives Im = (Vpfi/Vmflpfl 2 (17)(2.3 A} = 39.1 A (rms). 59. A transformer steps up the 120-V 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 right-handed coordinate system, as shown. (The vector relationship is summarized in Equation 3420b.) Therefore, the given wave is headed in the —I-direction. 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 60-112 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 34-19 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 3J-r10fi V/m)? Solution Equation 34-21b 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 light-absorbing 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 34-24) is Pm; = S {c = {130 Vlr'/<:rn2)/(3x10a m/s) = 6 mPa. ...
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