EECE 210 Lecture, Spring 2008

Homework 10 P10.9 I, =40°mA jwL = 710,000 2; F007 —j10,000 2 +f 7j10,000,,. 5 I= 4 = mA Fp (40d) = 80,00" P= 5 ltol*(5000) = 5 (0-08)2(5000) =16W Q= 5 Ht*(--10,000) = —32 VAR S=P4+jQ=16—j32VA |S] = 35.78 VA P 10.13 Z; = —j10,000]|20,000 = 4000 — 780002 Z, = 2000 — 720002 % _ 4000-j8000_, Z, 2000-72000. ” Z Vo = AL V, =10°V i V, = (3 — jl)(1) = 3 — jl = 3.16/— 18.43° V 2 = 1m 100) 5 10-3 =5mw P=5R 7 21000 -310kQ P 10.21 340 = 7, qt, tT, Ty ° 2400/0 : 962 yy v(rms) L, L, 1440 £3 e + + | 240017 = 60,000 + 740,000 Ii =25 4516.67; 9°. Ty = 25— j16.67 A(rms) 240013 = 20,000 — 710,000 Ij = 8.33 — 4,167; ”. Ip = 8.33 + j4.167 A(rms) 2400 /0° ak. _ 2400/02 ag 7 1867+ 0A, k= 796 I = =0-j25A I, =h+h+h+l=50— 375A V, = 2400 + (74)(50 — j37.5) = 2550 + 7200 = 2557.83/4.48° V(rms) P1025 fa] I= Oo 9.4 — 51.8 = 3/— 36.87° A(rms) 124 + 593 P = (8)?(4) = 36W 1 b] Y% = ———— = 5.33 — j4mS (b] Y= 359-4990 2388 4m - Xp=—L— =-2500 6 OO 4x 1073 fel Z, = —— 1 —~ = 18759 b"5.33x103° 465 (0° 3 {d] I= 15 +73 7 2.43/—0.9° A P = (2.43)(4) = 23.58 W [el % = 2% (100) = 65.5% Thus the power loss after the capacitor is added is 65.6% of the power loss before the capacitor is added. P 10.38 [a] First find the Thévenin equivalent: jwL = 530002 Zyn = 6000||12,000 + 73000 = 4000 + 730002 it 12,000 =" (180) = 120° V Vom = 0004 12,000 SDS eae =) —~=-jl —7, = ~ 710009 I AWN LY \—_@ AW 40002 330002 20009 wv -310000- - 120 l= ——"__ = 18 — j6mA 6000 + j2000 ~ 18 76™ P= 5l1i2(2000) = 360 mW [b] SetC,=0.1pF so —j/wC =—j20002 Set R, as close as possible to R, = V 4000? + 1000? = 4123.12 Ro = 40002 120 = — 4.77 51.85 mA lel T= 5060-4 j1000 ~ M477 — 31.85 m 1 Pe 5 HI? (4000) = 443.1 mW Yes; 443.1 mW > 360 mW 120 I= — =15mA {d] 3000 5 Pe 5(0.015)?(4000) = 450 mW [e] R, = 4000 2; C, = 66.67 nF [f] Yes; 450mW > 443.1 mW 53000 — j2000 = 71000 2