Question

2. For the ideal transformer in the figure below, find the amplitudes and phase angles of V1, V2, 11, and 12. 10 k12 0.5 H 10

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Answer #1

Given, Loka 0.5H 10:1 -> I to + Indir Vs V 5000 MG where Vs = 1o Cos (100t) we As know from the det convention of transform t300r Loke 0.5H mm -> 1, 500 mA we have Woloo as source frequency (raclisee) XL = jwLa j 100 x 0.5 = jsor Xc = -jx1000 WC 10010 LO iki 10300.121 20:278 IG 0.9708 Z-0.278 MA - isy I, (10000 + 350) > - ... = 10 LO - 0.9708 Z-0.278 ( 10 + joios) = 10data we already have Now, from the previous Vi -lo 2 V2 = - 0:1 = -0.0292 L-0-279 0.0292_180-0-279 ♡ 0.0292 2179.72] also, sphяле Magnitude 0.292 V > -0.279 0.0292 V 179.721 ১ (त 0.9708 MA -0.278 Iz 9.708 m 179.722

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