Question

(a) Write the relationship between the instantaneous time voltage i(t) = 1, cos(wt+p) and its phasor transform I. Write phasor transforms of the following voltages and currents. All answers should be in the form of AZB where A is a positive number A 20 and B is given in degrees over the range of -180° <B S 180°. Make sure that you include the units. Show all work to receive partial credit. (6) v(t) = VŽ cos(7t + 1/4 - 75") (Volts) (c) i(t) = -40 sin(2xft +0.75x) + 20 cos(2 wft) (milliAmps) Write the instantaneous time functions corresponding to the following phasors. Write all answers in terms of a single cosine. Make sure that you include the units. Show all work to receive partial credit. (d) 1 = -1 -1 + 2ejm/2 (mA) (e) D = (70 - 80 - 302-90° + 2017) (V)

Answer 1

page-2 (t) = – 40 (caso 757) Sin (20ft) - 40 Sinarsa Colraft + 20 cosaft dict-[-401-0.707) BOS(2144-90)-40x:707 +2060320ft] sing Sino cosco-go) coscaft -cost cas Sin e(t) =(28.284L-90 - 28.707L0 + 2040 Se Tz Let coswta cosant as reference og -Loo i (t) = 728.284 12 8.284 2720 - 5.857 i latsz dang-kaliman? JC-201²+ (5.857 (180 + fan 1/20 = 20.839 / 180'+7367 = 20. 5.857 839 (253.67 220.839/360+ $53.672 202-106.32 mA

" I = 2 cos(ut + 1350 mA v=vm collect +0=12 Vans Cos (cut to ina page.3 (d) I = -1 -j + 2 2-1-j+2 ( costs tj sine) mA 2-1-j+2(0+; 1) =-1-j tj 2= (1+j1) = 51-1024 (132 (180- tant, ma = 1.414 (135 mA=Irms 20 mA I= Im Cas (cent to) where anal dos 8(1.414) = 2 270 (e) T = 70-280-301-90 + 20eja 70-j80 - 30 [j1) +20 Cost tisina) 70-j80 tj 30 + 20 (-1) v o = 50-350 = 550² +502 +502 [- tema u = 70.71 <-45 Very Lb. 270 V V V = 100 cos Cent -45°) v