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Unformatted text preview: (t ) − e A (t )
i1 =
(31)
R1
e (t ) − eB (t )
i3 = A
(32)
R3
de
(33)
i3 = C B
dt
e (t )
i2 = A
(34)
R2
i1 (t ) − i2 (t ) − i3 (t ) = 0
(35) 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 (35),
−
−
=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 (32) and (33), i3 = 1
1
e(t ) + e B (t )
2
4 e A (t ) − e B (t )
de
=C B
R3
dt (36) that leads to e A (t )
de
e (t )
=C B + B
R3
dt
R3 de B
+ e B (t ) = e A (t )
(37)
dt
We now have two equations, Equations (36) and (37), 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.
 Spring '08
 UEDA
 Mechanical Engineering

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