Baixe sadiku 5ª edição circuitos eletricos e extras - chapt09pp 120121 e outras Notas de estudo em PDF para Engenharia Elétrica, somente na Docsity! Monday, June 27.2011 CHAPTER 9 P.P.9.1 amplitude = 30 phase = –75 angular frequency () = 4 = 12.57 rad/s period (T) =  2 = 0.5 s frequency (f) = T 1 = 2 Hz P.P.9.2 )9055tcos(4)55tsin(4-i1  )145tcos(4i1  , 377 rad/s Compare this with )65tcos(5i2  indicates that the phase angle between and is 1i 2i 145 + 65 = 210 Thus, i1 leads i2 by 210 P.P.9.3 (a) (5 + j2)(-1 + j4) = -5 + j20 – j2 – 8 = -13 + j18 560 = 2.5 + j4.33 (5 + j2)(-1 + j4) – 560 = -15.5 + j13.67 [ (5 + j2)(-1 + j4) – 560 ]* = -15.5 – j13.67 = 20.67221.41 (b) 340 = 2.298 + j1.928 10 + j5 + 340 = 12.298 + j6.928 = 14.11529.39 –3 + j4 = 5126.87       48.97-823.2 87.1265 39.29115.14 4j3- 4035j10 2.823-97.48 = –0.3675 – j2.8 1030 = 8.66 + j5    5j3010 4j3- 4035j10 8.293 + j7.2 P.P.9.4 (a) v = 7 cos(2t + 40) The phasor form is V = 740 V (b) Since –sin(A) = cos(A + 90), i = –4 sin(10t + 10) = 4 cos(10t+10 + 90) i = 4 cos(10t + 100) The phasor form is I = 4100 A P.P.9.5 (a) Since –1 = 1±180 (we can use either sign) V = –2540 = 25(40–180) = 25–140 The sinusoid is v(t) = 25 cos(t – 140) V or 25 cos(ωt+220˚) V (b) I = j (12 – j5) = 5 + j12 = 1367.38 The sinusoid is i(t) = 13 cos(t + 67.38) A P.P.9.6 Let v(t) = –10sin(ωt–30˚) + 20cos(ωt+45˚) = 10cos(ωt–30˚+90˚) + 20cos(ωt+45˚) Taking the phasor of each term V = 1060 + 2045 V = 5 + j8.66 + 14.142 + j14.142 V = 19.142 + j22.8 = 29.7749.98˚ Converting V to the time domain v(t) = 29.77 cos(t + 49.98) V P.P.9.7 Given that )30t5cos(50dtv10v5 dt dv 2   we take the phasor of each term to get 2j V +5 V + j 10 V = 50-30,  = 5 V [j10 + 5 – j(10/5)] = V (5 + j8) = 50-30 V =      58434.9 30-50 8j5 30-50 V = 5.3-88 Converting V to the time domain v(t) = 5.3 cos(5t – 88)V P.P.9.8 For the capacitor, V = I / (jC), where V = 1030,  = 100 I = jC V = (j100)(50x10-6)(1030) I = 50120 mA i(t) = 50 cos(100t + 120) mA P.P.9.9 Vs = 2030,  = 10 ACMG=45V ACPHASE P.P.9.13 To show that the circuit in Fig. (a) meets the requirement, consider the equivalent circuit in Fig. (b). Z = -j10 || (10 – j10) = j2010 j10)(10j10-   = j21 j10)(10j-   = 2 – j6  10  -j10  + Vo  10  + Vi  (a) -j10  V1 10  Z = 2j6 +  Vi = 60 V + V1  (b) =0 V1 = )1( 3 60 )60( 6210 62 j j j    Vo = 10j10 10j-  V1 = )1( 3 60 1 - j j j              = 02j- Vo = 20–90˚ This implies that the RC circuit provides a 90 lagging phase shift. The output voltage is = 20 V P.P.9.14 the 1-mH inductor is jL = = j31.42 )101)(105)(2(j 3-3  the 2-mH inductor is jL = = j62.83 )102)(105)(2(j 3-3  Consider the circuit shown below. + Vo  Z = 10 || (50 + j62.83) = 83.62j60 )83.62j50)(10(   Z = 9.205 + j0.833 = 9.2435.17 V1 = Z / (Z + j31.42) Vi = )10( 253.32205.9 17.5243.9 j  = [(9.2435.17˚)/(33.5474.07˚)]10 = 2.756–68.9˚ Vo = 83.62j50 50  V1 =   49.51297.80 )9.68-756.2(50 = 1.7161–120.39 Therefore, magnitude = 1.7161 V phase = 120.39 phase shift is lagging j31.42  10  + Vi  j62.83  50  V1 P.P.9.15 Zx = (Z3 / Z1) Z2 Z3 = 12 k Z1 = 4.8 k Z2 = 10 + jL = = 10 + j9.425 )1025.0()106)(2(j10 6-6  Zx = k8.4 k12 (10 + j9.425) = 25+ j23.5625  Rx = 25, Xx = 23.5625 = Lx H625.0 )106(2 5625.23 f2 X L 6 x x     i.e. a 25- resistor in series with a 0.625-H inductor.