E117CW19 - "AP 9.10 v1l = 24053.13“ = 144 + j192V v2...

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Unformatted text preview: "AP 9.10 v1l = 24053.13“ = 144 + j192V v2 = {QM—90° = ——j96V ij = j(4000)(15 x 10-3) = jfiOQ 1 __. 6x106 ' ij _ 3(4000)(25) Perform source transfonnatiens: E 144 + j 192 = ————-—-——-——— = 3.2 — '2.4A j60 j60 3 V2 _96 . 3.2-j2.* Combine the parallel impedances: 1 1 1 1 3'5 1 y = __..._ __ _ _ = _ = _ 3660 + 30 + +-—j60 + 20 3'60 12 1 Z = — == 129 Y .. v. = 12(3.2 + 32.4) = 33.4 + j28.3V = 43 /_36.87° V v... = 43 cos(4000t + 36.87") v , The node voltage equation is v V v v — 1001—90m ~10 _ ———— — -———————————-— = 0 + 5 + —j(20/9) + 3'5 + 20 Therefore e = 31.62 eee(50,000t -— 71.57°) V AP 9.13 Let 15, lb, and IC be the three clockwise mesh currents going from left to right. Summing the voltages around meshes a and b gives 33.8 = (1 + j2)’Ia + (3 — j5)(IEL — Ib) and I 0 = (3 — j5)(Ib — In) + 2(Ib —— 1C). But . V: = —2'5(Ia —— It), therefore L = -0.75[—j5(1fl — Ib)]t Solving for I = In = 29 + 3'2 = 29.07@.95° A. P 9.59 (12 — j12)Ia —— 1215. — 5(——j8) = 0 ~12Ia + (‘12 + j4)Ig + 2‘20 — 504) = 0 Solving, Ig == 4 — 3'2 = 4.47! — 26*57°A ...
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This note was uploaded on 08/12/2010 for the course ENGR 117 taught by Professor Nickarnold during the Fall '08 term at Santa Barbara City.

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E117CW19 - "AP 9.10 v1l = 24053.13“ = 144 + j192V v2...

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