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Example 10.3-2 (See Example 10.3-2 in the textbook for the solution to a similar problem.) Consider these phasors: V, = 17 (1
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Solution:- Vi = 17 <106 (sts) 17 (as (306) + j sn (sa6) ] Sin (A06) -) Vi =-) VI jarssz] ·0• 2.756 + 17x 0.9612 (-9.6358)Ind. Quadamant Vz = -424 j15 (roltd (o-p8i)-pi- 44.598 15 -42 tan () = 19.6s3- -2) 42 as v, is in 2nd- Quaosont ; s0:- (e0-

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