Question

please explain each step

9) Find is, i2, i3, and is 3A v2 102 40 12 § 6A 40127 0 2 A

Answer 1

GA I 1 A www you to yol for Solving in, iz, i3, iy we have to use Super position Theorem Superposition Theorem if a number of voltage or Current Sources are acting Simultaneogly in a linear network, the resultant Current in any branch is the algebric sum of the Current that each Source acts alone would be produced in it, when replacing all other independent Sources by their inter -nal resistances. So let's take 6A Current Source first mas to Egon yol GA (1)

Now by Current division rule 1 10 - XGA 10+20 = 1/2 X 6A = QA 12 = 20 20+10 GA = 2 x 6A = UA here is, and it are zero because of open circuit - iz=0, iq=0 Now let's take 3A Current Source ЗА 13 Uols WWW-4 IOS YOSS from Current division rule *3A = IA 1, 10 *3 10+20 i= 20 X3A = 2 X 3A 2A 20 +30

90 - X3A = I x3A = 1.5A qoto yo - X3 A = 3 X 3A = 1.5A 40+40 Now let's take an Current Source alone 3 is 14 OJE ww LO vons son Das More D 2A 202 -By Corrent division rule 40 x 2A = 1X2A = 1A 40+40 - x. 2A = IA 40+40 i,, i = 0 because of open circuit

Using the principle of Superponition theorem. = 2A+ 1A + 0 = 3A 12 = 4A + 2A + 0 = 6A iz = of 1. 5A + (A = 2.5A 1y = 0ut 1.5A+ 1A = 2. 5A . 1,=3A, 12 = 6A1 iz=2. 5A, 14 = 2:5 A