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2.5kr 3. (12 points) (a) Find the current i, using source transformations. (8 points) (b) Verify your solution using either n

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2.SK чо Е ЧЕ 8.+mA 15kn 2oy :Gokn gokn 2 км - 18+ Ч - 3m A чE 2-ҫк — yok Зm A bol 8. А 15 к 9 DK 2 к чок (1) ок» (чо хео) 16

2.5K Isk 8.4 mA qok 3MA 24 km ?k IR : 3mA) (246) 12V 2.5lc 24k 41 tarto 8.4mA qok 2k 3ok a.sk 8.64 M A SISI 72v 39ok V-IR 72(

२.sk २.4mA इ30k 8.५MA ISK ३१० २.5c Isk २.५MA 8.4MS १ok 30 30||१० (४.५ - २.५) - 6MA -२२.5kr EMA 4A 21 २.5k 6MA ISK २२.564 (6mA

b) (1, -symA ) 11 чок ЧЕ 2.SK (1,-1- - 8:494-) -, -а,) 1 120V + 3 bobл + , 8:4mA gok ISK (1) 1, 2E ч kul х 12 ov- Чот, + bok(

Applying kul in о бок (1 - т, ) ЧЕ 12 чок із + 2 I2 bour, - bokІ, I, + Rok 13 + 2) I, &o1-ъч ki, — qok13\ - до 9,6% E боk II

solving 3 oroo38 215 б: ооч 3 6615 S — о. бо обSS .. To 2 (oооч 368 as + б• бъөЬs 25 — ч 4xto 3) Io -- - оо 3329 S 3.39 KA

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