Et e a t i1 3 1 r1 e t eb t i3 a 3 2 r3 de 3 3 i3

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Unformatted text preview: (t ) − e A (t ) i1 = (3-1) R1 e (t ) − eB (t ) i3 = A (3-2) R3 de (3-3) i3 = C B dt e (t ) i2 = A (3-4) R2 i1 (t ) − i2 (t ) − i3 (t ) = 0 (3-5) R1 i1 eA i3 eB R3 i2 Switch C R2 e(t) Check: the number of equations should equal the number of unknowns. Here, we have 5 equations with 5 unknowns, i1, i2, i3 eA and eB. e(t ) − e A (t ) e A (t ) e A (t ) − e B (t ) Eliminate i1, i2 and i3 using Equation (3-5), − − =0 that leads to R1 R2 R3 1 R3 R3 R 1 1 e(t ) e B (t ) + or + 1 e A (t ) = 3 e(t ) + e B (t ) + e A (t ) = + + R1 R1 R3 R1 R2 R1 R2 R3 Substitute the given values: 4e A (t ) = 2e(t ) + e B (t ) or e A (t ) = Equating Equations (3-2) and (3-3), i3 = 1 1 e(t ) + e B (t ) 2 4 e A (t ) − e B (t ) de =C B R3 dt (3-6) that leads to e A (t ) de e (t ) =C B + B R3 dt R3 de B + e B (t ) = e A (t ) (3-7) dt We now have two equations, Equations (3-6) and (3-7), and two unknowns eA and eB. Eliminating eA de B 3 1 + e B (t ) = e(t ) dt 4 2 e B (t = 0) = 0 or CR3 3t − 1 1 12 Solve for eB using LT: E B ( s ) = . From th...
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This note was uploaded on 10/15/2012 for the course ME 3015 taught by Professor Ueda during the Spring '08 term at Georgia Tech.

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