hw7 - 22.23 828 CHAPTER 22. SINUSOIDAL AC CIRCUIT ANALYSIS...

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Unformatted text preview: 22.23 828 CHAPTER 22. SINUSOIDAL AC CIRCUIT ANALYSIS 22.3 a) The magnitudes ar (D b) To put .21 and 22 in polar form, we need their magnitudes and their angles. We found their magnitudes in part a), now to find their angles. 7? , 7r Since 21 is in the second quadrant of the complex plane, its angle is not between —E rad and 5 red. In fact, its angle is between E red and 7: rad. The arctangent function will not give us its angle, but instead will give us an angle 7r rad ‘ less than it should be, so We’ll find its angle by Im 21 Re 21 . . 3 _ polar angle of 21 = arctan + 1r rad : arctan -«_—2 ~l— 7r rad = 2.16 rad. 71' 7;" Since 22 is in the fourth quadrant, its angle is between ——E rad and -2— rad, 80 We may 1136 the arctangent function directly to find its angle. Em 22 R8 22 —4 polar angle of 22 2 arctan = arctan ~3— = —0.92’? rad. So, using these angles and the magnitudes found in part a), the polar forms are 2:1 : 3.61 d (2.15 rad) and .32 m 5 A (—0.927 rad). c) Using the magnitudes and angles found in parts a) and b), the exponential forms are 21 = 3 615m“ rad) and £90927 rad)- 22:56 d) Here they are, sitting in the complex plane. .45.... “gnu - hum; I...” ..l . “wuaenrhwm a) The impedance of a resistor is independent of frequency and is equal to the resistance, 20 I 33 = 100 Ed). b) The impedance of the capacitor is I _“ 1 _ —i __ -i _ - p 3 MO _ nit—yo ‘ awe “ 2mm ><103 Hz (00100 x 1043 F) “ 3(10'9 X 10 m : —z‘(15.9 in). c) The impedance of the inductor is z; = iwL 2 2'2me 2 man X103 Hz}{0.500 H) 2113.14 x 103 n) 2 113.14 km). d) The impedance is related to the capacitive reactance by \- . 20 = "gyro :a 2:6 2 2'20 equities x 103 m} = 15.9 ><103 Q e) The impedance is related to the inductive reactance by 23L : tag; 24> 25L = —z'ZL = 4959.14 x 103 9)} z 3.14 x 103 o. 22.26. 839 i l l 22.26 W’e’ll write the equivalent impedance as a function of the unknown capacitance, set the result equal I to 150 Q, and solve for the capacitance. ' 3:" Since the circuit elements are in series, their equivalent impedance is the sum of the individual impedances, I. so - I 1509=ZR+ZL+ZC l : R l g ' —.-._. I T w T 211.10 a : R + L H — I “U 0.90 — R J— 3 col} — l I ‘—' I we . # 1 1 1 1 — -~«—- : Q 2 —~— : 2 = . *6 : .1 . “L wt? 0 :fi" C Lw2 L(27w)2 (o 250 H ){meoe Hz);2 28 1 X 10 F 28 “F 22.27 was in series with a capacitor. Therefore the total impedance will el combination of resistor and inductor, be the sum of the impedance of the parall Z Z mi—L-fi and the impedance of the capacitor, 30. Thus, 33 + 21, ZRZL Z = Z . ZR +315 + 0 The individual impedances are 1 2' 2:53, Z:'L:‘2 L, ed Z:—_—~=- . R L aw i m; n C tavC 27rI/C‘ Substitute these expressions into the expression for Z, and separate the real and imaginary parts. 3 _ Ri27wL m __2'_ R + tEZmJL 27'er Ri27rI/L R — i2Tl'I/L 2' Z R + i271‘UL (R — em) " awe iR227rz/L + Egan/L)? 7; R2 + (271—1432 _ i2qu Ram/LP a , 2922va l 2 R2 +(271‘VLJE T4122 + (21ij2 “ 2771/0)- OW Substitute the known values R = 1.00 X 103 Q, L N =0.25GH,andv=1.DOxlD3I-Iz. Z H {1.00 x 103 Q)[27r(l.00 x 103 Hz)(0.250 HM2 (1.00 x 103 m2 +£271'(1.00 x 103 Hz)(0.250 H);2 +1, (100 x 103 9)22w(1.00 x 103 Hz)(0.250 H) _ 41 ) (1.00 x 103 m2 + {27r(1.(}0 x 103 Hz)(0.250 Hm 2?r(1.00 x 103 HZ)C’ ' After doing alI the arithmetic, this simpiifies to 1 2 ' 4' — ~—-—-—~— - Z 712 Q 3( as 9 2741.00 X 103 Hm?) 22.35 a) The impedance of the capacitor is ' —rr — —1 — -— med—fi——6-— : —1i(26.5 = (25.5 it) A rad). ‘30 ‘ iwC' * wC _ (3T7rad/s)(100 x 10— F) b) The capacitive reactance {to is related to the capacitive impedance by 20 : HM’C, so X0 : —Zc. = Mafia.” 0“) : 26.5 G. —2 —z c) First we’ll find the current phasor I for the circuit. Then we’il look at its magnitude to find the peak current. The independent source voitage phasor is Vsource : A rad/mt}. Choose the clockwise direction for the current in the circuit, apply the KVL in that directlon, and solve for the current phasor I. ' IWVSOUI'CE nV+IZG=OV==> _ 30 EW (25.5 c) .4 rad) rad + (377rad/S)t)- 7:“ :(0755 A) z (2 30, the peak value of the current is Ipeak : 0.7.55 d) The real current fit) is the real part of the current phasor that we just found, so rm 2 (0.755 A) cos rad + (377 rad/5):). When t z 0 s, the argument of the cosine is {l A. For values of it slightly above 0 s the cosine is 11 reaches "1 at the time when its argument reaches 7r r 2 egative. It will become increasincrl é: rad, so the current at that time is (0.755 A) cos (Tr) : c y negative untii it ad. This occurs when 77 5 red + (377 rad/s)t = 7r rad e—J—r‘ t: 4.1? X 10"3 s x 4.17 me. At this time, [(t} : (0.755 A) cosh rad) : —?.55 A. nterclockwise direction, rather than clockwise Tired + 2 {377 rad/s)t : 27¢ rad 2:; t: 1.25 x 10—2 s : 12.5 ms. At this time, [(1‘) : (0.?55 A) cos(27r rad) = 7.55 A. 22.37 . . - .F ‘ , a) The angular frequency or" the source is the coefiiment of t 111 the argument or the cosme, so a; R 377 rad/s _ _ : 60.0 Hz. 2n“ 271‘ red w 2 377 rad/’5 => :2 2 b) The impedance of an inductor is 21; x int x 21377 red/s}{0.i50 H) : i(55.6 o). e) The independent voltage source phasor is Va) = (170 A rad/s )t} = V)€if377 rad/5):: cl) Choose the direction of the current phasor as shown below. The current enters the positive polarity terminai of the inductor. I—+ e) Apply the KVL CIOCkwise around the loop. V IZZ: V _2'V Britt/2 raflam V)ei(377 rad/s): 2156-69) * 55.5 o ‘ W : A)ei[—7r/2 rad +(377 rad/Ski : z rad + rad/S)j) f) The potential difference phasor across the inductor is VL I IZL = A)eif—Tr/2 rad +(3'?’r' rad/s)tj(i55‘6 2 A)ez'[—7r/2 rad-H377“ rad/s}t}ei(7r/2 rad)(55l6 :(1707V)ei(377md/==J*= (170 V) g [erred/Sm. g) The reai current through the inductor is the real part of the current phasor. [(15) = (3.00 A ) cos rad + (377 rad/s Jr) . h) The real potential difference across the inductor is the real part of the potential difference phasor acrose it. VL(t) = (170 V) cos{(377 rad/s)t]. i) The phase difference between the potential difference across the inductor and the current through it is ,3 z phase difference 2 (377 rad/s)t — (—12: red + (377 rad/s )t) : g rad. i) The peak current through the inductor is the coeficient of the cosine term in part g). s 'Ipea = 3.00 A. k) The rrns current through the inductor is the peak value divided by \f2— U The peak potential difference across the inductor is the coefficient of the cosine term in part h). V; peak 2 170 V. ml The rms potential difference across the inductor is the peak value divided by NOV Vers I ‘72."— n) The power factor associated with the 1' potential difference across the inductor an power factor = cosfl = cos rad) = G. o) The average power absorbed by the inductor is I Vrmsfrms C055 : Vl‘msjrmso : O 22.39 a) The impedance of the inductor is 23 :iwL=i(377rad/s)(0.05{}H)271199,) :(199) 4 rad). b) The inductive reactance XL is related to the inductive impedance by 2;, = iXL, SO ' 9 XL=§£=Z(1, ):199. ’t 2 C) First we’ll find the current phaeor I for the circuit. Then we’ll look at its magnitude to find the peak current. The independent voltage source phasor is VSUUI‘CE 3L (EODV) 4 [(377rad/sjt} : —~—_.___._,__ (199) 4 rad) “Vsource ‘l‘ IZL : O V d I = 2 (1.1 A) 4 rad +(377rad/s)t]. So, the peak value of the current is rpm = 1.1 A. d) The reai current fit) is the real part of the current phasor that we just found, so In) : (1.1 A)cos rad + {377 rad/sfi). When 75 z 0 s, the argument of the cosine is _“ (W!) —«2-— rad, so the current at that time is (1.1 A ) cos 0 A, For vaiues of t slightly above 0 s, the coeine will become increasingiy positive, unti time when its argument reaches 0 rad. This 0 1 it reaches 1 at the ccurs when mewmwfl~. *7!" 2 red +(377rad/s)t:0rad 2:» t=4.17x10*3s 24.171113. At this time, Kt): (1.1A)cos(0 rad) = 1.1 A. 22.47 a) _The resonant angular frequency of the circuit must be ab] e to be varied from 500 kHz to 1600 kHz. The resonant angular frequency is given by Equation 22.90 on page 1029 of the text. (.00 = "1— 2> O = ‘1— : l o M £02}; Oar/PL In order to tune in the 500 kHz frequency, the capacitance must be : 3 :-l__:*——~—~—-1—»~__.__~21.56x10—10F =156pF. weL (27ry)2L [277(500 x 103 Hz)]2(550 X 10“5 H) In order to tune in the 1600 kHz frequency, the capacitance must be Ce—L— 1 ~%1 ~152x10—“Fe152 s ‘ * * [27r(16(}0><1031-12)}2(6z’3{)><10“6 H)h ' ‘ ' p ' SO we need a variable capacitor whose capacitance can be varie 5} To find Q, use Equation 22.100 on page 1030 of the text. Ci from 15.2 pF to 156 pF. u we 274500 x103 Hz) Aw “ 27r(2.0 x 103 Hz) _ ’11 solve Equation 22.101 on page 1030 of the text for R. L L stcm rs Rfl/eee in order to tune in to 500 kHz, ue for C in the last equation, along = 2.5 X 102. To find the resistance needed, we From part a), we know that 1.56 X 10‘10 F. So use this val L: 650 X 10—6 H. the capacitor must be adjusted to C = with the known values Q r— 25 x 102 and I L / 650 x 10-5 H R _ 092 H (1.56 x 10-10 F)(2.5 x 102)2 “ 8'2 ‘Q‘ c) Repeat the anaiysis of part b}, but with 500 kHz repiaced by 1500 kHz. To find Q, use Equation 22.100 on page 1030 of the text. Q m we _ 277(1600 x 1031a) —~———._ z .0 x 102. Am Mao x 103 Hz) 8 To find the resistance needed, we’ll soive Equation 22.101 on page 1030 of the text for R. r. u /L Q2V0R2 fiR‘ one From part a), we know that in order to tune in to 1.52 x 10—11 F. So use this value for C in the 1 L=550 X 10‘6H. 1600 kHz, the capacitor must be adjusted to C’ = ast equation, along with the known vaiues Q 2 8.0 X 102 and v —e R: L 2 {WM 829, Go? to .52 x 10-11 F)(8.U ><102)2 2 ...
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hw7 - 22.23 828 CHAPTER 22. SINUSOIDAL AC CIRCUIT ANALYSIS...

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