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Problem 4 ( 20 points): The applied voltage in the following transformer circuit is 10 cos (100t + 30). Find voltage across 1

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Phabor Domain can written Equivalent circuit 1000 500 Vome/30 (N) -jXcT 10/30 jXL 5:1 10 cos (100t +30) 10 /30 2 phaser Domaifinding the the rinins Equivalent of source side TODO Zth slo 30 -j1000 Vth B Vm= open circuit voltage (OOO + 10/30 -31000 Valo, 6/02 = ab Loito2 51-60-(-45) 0 0 0,- Vn 5L-15 blor To determine Zth. Deactivate voltage source short circuit. by 1000 4500 - 3500 Zth 5/15 Vih O.1 5.1 Apply KVL to loop i 51-15 +(500-j500)I, + Vi = 0 500-j 500 I, +V, = 54-15 Apply HVL loop 2 -Vsubstitute VI Iz = 5 I in 5 v2 = (500+jo. 1) Iz Vi (500 +jo.1) 5 II Vi = (500+ jo 1) 25 I, Vi = (12500 +2.5) 11 substitute Vwe know I=5I, Ij = 1,924–12.8 -12.81 MA I 2 500 الوهی + jool V Y = I2 (jool) 1292_12.81 x 0.1190 V2 0.192 (77.19 v aloi blo2

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