ProbSolv_Chapter09

ProbSolv_Chapter09 - CHAPTER 9 - SINUSOIDS AND PHASORS List...

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Unformatted text preview: CHAPTER 9 - SINUSOIDS AND PHASORS List of topics for this chapter : Sinusoids Phasors Phasor Relationships for Circuit Elements Impedance and Admittance Impedance Combinations Applications SINUSOIDS Problem 9.1 Express the following as cosine functions. (a) 5 sin(27tt) (b) 4.3 sin(cot — 47°) (c) 2 sin(cot - 712/ 2) To convert the sine function to the cosine function, we will need a trigonometric identity. . -'F sin(x) = cos(x i 90°) (a) 5cos(27tt — 90°) (b) 4.3 cos((ot ~ 47° — 90°) = 4.3 cos(cot — 137°) (c) 7t/ 2 rad = 90° and 1t rad = 180° 25in((ot — 1t/2) = 2cos(03t — 71/2 — n/Z) = 2 cos(wt — Tli) Because sin(x i 90°) = i cos(x) or cos(x i180°) = — cos(x) this can also be written as - 2cos(wt) Problem 9.2 Find the magnitude, angular frequency, frequency, and phase angle of each of the following functions. (a) 5 sin(l 0t) (b) - 2.5 cos(27tt) (c) ‘ J5 cos(cot — 37°) 155 (a) Consider v(t) = Vm sin(cot + (1)). Also, note that to = 211 f . Vm=_5_ 03:10 f=10/27t, (1)=0° (b) Consider v(t) = Vm cos(wt + (1)) Vm=2.5 co=27r f=l, ¢=180° Note that (1) = 180° due to the negative sign in front of the function. (e) Consider v(t) = Vm cos(wt + (1)) vm=J§ (0:2 rem/21:, ¢=-37° Problem 9.3 [9.5] Given v1 = 20 sin(03t + 60°) and v2 = 60 cos(cot — 10°) , determine the phase angle between the two sinusoids and which one lags the other. ' v1 = 20 sin(cot + 60°) = 20 cos(cot + 60° — 90°) = 20 cos(wt — 30°) v2 = 60 cos(o)t ~ 10°) This indicates that the phase angle between the two signals is 20° and that vl lags v2 . . W PHASORS Problem 9.4 Convert the following into phasors. (a) 100 sin(cot) (b) 20 cos((ot) (c) 50 cos(cot — 80°) (d) 25 sin((ot + 45°) (a) 10040° assuming a reference of A sin((ot + (1)) (b) 2010° assuming a reference of Acos(03t + (1)) (c) 50.4 - 80° assuming a reference of A cos(c0t + (1)) (d) 25445° assuming a reference of A Sin(a)t + (1)) . 156 . Problem 9.5 [9.11] Let X = 8Z40° and Y = 104 — 30°. Evaluate the following quantities and express your results in polar form. (a) (X+ Y) X“ (b) (X - Y)* (c) (X+Y)/X (a) X+Y = 8440°+1oz—30° X+ Y = (6.128+j5.142)+ (8.66—j5) X+ Y =14.79+j0.142 =14.79zo.55° (X + Y) X‘ = (14.7940.55°)(8z - 40°) (x + Y) X“ = 118.34 - 39.45° (b) X—Y = 8440°—101-30° X— Y = (6.128+j5.142)——(8.66—j5) X— Y = -2532 + j10.142 =1o.45.4104° (X —— Y)* = 10.454 - 104° (c) From (a), X+ Y =14.79ZO.55° . X+Y _ 14.7940.55° _ x 8440° (x + Y)/X = 1.8494 - 39.45° Problem 9.6 If A sin(03t + 4)) is used as a common reference, what would be the phasors? (a) 100 sin(03t) (b) 20 cos((nt) (c) 50cos(wt - 80°) (d) 25 sin(cot + 45°) (a) 10040° (b) 20490° (c) 50410° (d) 25445° 157 W O PHASOR RELATIONSHIPS FOR CIRCUIT ELEMENTS Problem 9.7 Given the circuit of Figure 9.1, find the steady-state value of vc (t) when iS (t) = 5 sin(1000t) A. Figure 9.1 > Carefully DEFINE the problem. Each component is labeled completely. The problem is clear. > PRESENT everything you know about the problem. _In the time domain, . I VC=E ic dt If the circuit is transformed to its frequency domain equivalent, however, then 1 V = I Z - = ,—I ° ° JmC The final answer can then be converted to the time domain. > Establish a set of ALTERNATIVE solutions and determine the one that promises the greatest likelihood of success. The technique used to solve this problem is Ohm's law. The choice to be made is whether to use the time domain or the frequency domain. From what we know about the problem, converting the time-domain circuit into the frequency domain allows the use of algebra with complex numbers rather than calculus to analyze the circuit. Analysis of simple circuits can be done in the time-domain as a check of the answer. > ATTEMPT a problem solution. Transforming the circuit to the frequency domain, the current source is IS = 5 4 0° . The resistor converts to ZR = 1 k9, and the capacitor becomes =—=—————=—'1oo =1 — ° Z° jmc j(1000)(10'6) J 0 0004 90 . 158 Thus, . . Vc = (540°)(10001 - 90°) = 50004 - 90° or vc (t) = 5000sin(1000t — 90°) V > EVALUATE the solution and check for accuracy. Using KVL in the frequency domain, VR = IzR = (540°)(1000) = 500040° V Vc = 126 = 50004-90° VS = VR + V0 = 500040°+ 50004 - 90° VS = 5000 — j5000 = 5000\5 z - 45° or vs (t) = 5000J§ sin(1000t — 45°) V Using KVL in the time domain, 1 1 vc(t) = E Iic(t) dt = I676— I5sin(1000t) dt vc (t) = 106 103 (5)[- cos(1000t)] vc (t) = 50005in(1000t — 90°) V . vR (t) = Ri(t) = 50005in(1000t) V Vs : VR + Vc ‘ vs (t) = 5000 sin(1000t) + 5000sin(1000t — 90°) vs (t) = 500045 sin(1000t — 45°) V Our check for accuracy was successful. > Has the problem been solved SATISFACTORILY? If so, present the solution; if not, then return to “ALTERNATIVE solutions” and continue through the process again. This problem has been solved satisfactorily. vc (t) = 5000sin(1000t — 90°) V Problem 9.8 Given 0) = 100 , determine the frequency domain (8 = j0) ) values for the following elements. (a) R=1£2,10§2,1k£2,1MQ,100MQ (b) L=10H,5H,1H,5mH,40uH (c) C = 2mF, 333 uF, SuF,10pF 159 (a) R = 19,109,1kQ,lMQ,100MQ (b) coL : 10009, 5009, 1009, 500 m9, 4m!) where the units are ohms of inductive reactance 1 (c) ——=5s2,30r2,2kg2,1Go 00C ..__._.____.__ where the units are ohms of capacitive reactance Problem 9.9 Given R = 1009 , L = 1H , and C = 100 “F, calculate the values in the following table. Clearly, R = 100 which is not dependent upon the frequency. L = 1 and C = 10'4 implies that . X (0L (0 d X 1 104 = : an : — = _ L C ‘ 03C 03 and the table becomes IMPEDANCE AND ADMITTAN CE Problem 9.10 Assume that Z = R + jXL — jXC. For the values used in Problem 9.9, what would be the values of Z in rectangular coordinates? 160 Insert the values of R , XL , and XC into Z: R+j(XL —XC) and it is evident that Problem 9.11 [9.43] In the circuit of Figure 9.2, find VS if I0 = 240° A. Vs ‘ —j2 Q —j o o G L, 2 Q j4 o E E 1 (2 Figure 9.2 Since the left portion of the circuit is twice as large as the right portion, the equivalent circuit is shown below. Vs 2Z Z V1 =~10(1—j) =—2(1-j) V2 = 2V]: —4(1—j) 161 VS =V1+V2 =6(1——j) Vs = 8.48541350 V . Problem 9.12 . Using the values in Problem 9.10, what would be the values of Z in polar coordinates? — — — 1000 _ 10000 — OW o IMPEDANCE COMBINATIONS Problem 9.13 Given the circuit of Figure 9.3, find Zin for 0) = 1000 rad/s . 10 k9 Zin . ’ 20 H 0.1 uF Figure 9.3 The phasor domain equivalent includes ZR = 10 k9 ZL = j(oL = j(1000)(20) = j20 kQ Z ——1——————L—————-_J—~ 'IOkQ C "ij‘j(1000)(0.1x10-6)“10-4 ‘1 . 162 Zm = ZL H (ZR +Zc) = j20k || (10k—j10k) . ' Z _ (j20k)(10k—j10k)_ j20k(10—j10)_(200k)(1+j)' in j20k+10k—j10k _ 10+j10 ‘ (10)(1+j) zin = 20kg Problem 9.14 [9.47] Find Zeq in the circuit of Figure 9.4. Zeq Figure 9.4 All of the impedances are in parallel. Thus, z l—j 1+j2 j5 1+j3 = (0.5 + j0.5) + (0.2—- j0.4) + (—j0.2)+ (0.1— j0.3) = 0.8— j0.4 6‘1 ___l._ 6‘1 " 0.8—j0.4 zeq = 1+j0.SQ Z 163 W0 0 APPLICATIONS Problem 9.15 The circuit shown in Figure 9.5 is used to make a simple low-pass filter. An important part of choosing the appropriate value of C is to determine the highest frequency to be passed and then choose a value of C such that the output voltage is 1 /\/5 times the magnitude of the input at that frequency. What value of C makes this a low-pass filter for frequencies from 0 Hz to 1000 Hz ? 10 k9 + i ' + Vin“) C : ’ \ Vout(t) Figure 9.5 Transforming this circuit to the frequency domain yields 10k!) 1 0 , — ' coC Vout _ ‘ 1n lOk—j/OJC -j/coC _ V. iv” I "' 10k—j/coC f = 1000 Hz is the upper frequency limit, called the corner frequency. ’ _;j_/£°_C_ — —1— 10k - j/COC cDQ141000) 2 10k—j/27t(1000)C ” /108+Xc2 “J5 ‘ — j/2n(1000)C Xe 1 164 Problem 9.16 [9.61] Using the circuit of Figure 9.6, (a) Calculate the phase shift. (b) State whether the phase shift is leading or lagging (output with respect to input). (0) Determine the magnitude of the output when the input is 120 V. 209 409 309 Figure 9.6 (a) Consider the circuit as shown. 209 V2 409 V1 309 (j30)(30 + j60) 30 + j90 (j10)(43 + j21) 43 + j31 Z1=j30||(30+j60)= =3+j21 Z2 = leII (40+Z1)= =1.535+j8.896=9.028480.21° 165 (b) Let Vi =140°. V _ 22 V _(9.0Z8180.21°)(140°) 2‘z2+20 “ 21.535+j8.896 V2 = 0.3875457.77° _ 21 V _ 3+j21V "z1+40 2—43+j21 2 v1 =0.17184113.61° V1 47.85426.03° '60 '2 J V] J 2 = = V=~2 'v V° 30+j60 1+j2 ' 5( “)1 Vo = (0.8944426.56°)(0.17184113.6°) V0 = 0.15364140.2° Therefore, the phase shift is 140.2° The phase shift is leading. If Vi = 120 V, then Vo = (120)(0.15364140.2°)=18.431140.2°V and the magnitude is 18.43 V. 166 _ (21.213481.87°)(0.3875457.77°) ...
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ProbSolv_Chapter09 - CHAPTER 9 - SINUSOIDS AND PHASORS List...

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