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Problem 4.60 < 2 of 13 M Review Constants Consider the circuit shown in (Figure 1). Suppose that R = 7 k2 Part A Find the cur

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sod given that R=7ke yoke A uk2 2-slah 20 8.4 MHD 160k2 mas ER & 2k-2 • Source transformation technique: Free I os jou Vc TENow circuit - 2.5kn 4K2 (246.2 28.uma egoke R$ dio 3m A 262 24ke Step 2: Again Apply Source transformation technique; $2146.2Now circuit becomes- 2.5k_2 3okela 2@ 2.4 mA Douma * goke R&cod equivalent of two parallel current sources - 12.4 mA duma $18so that circuit 22 •5k 2.5л 25. 135V ioli Roissv Revio KUL in АРНЯ above crreuit 20 – 135 -135 25xto - кL = 0 135 + (25k te)By current division wile- io 90 qot9.5 I 99.5 x 4.21875 ma 90 99.5xio 90 = 4.664ma Il circuit- Now Apply RCL at node A in abozo 120 – 40K-2x I bokex (I-3.730 mA) 120-yoKeyI boke XI + boks 43.736ht=0 120 + 224.16 (40+60)k2 XI 344.16 = look 2 x 344.16

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