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Unformatted text preview: nd A2 by matching our symbolic solution to
dv ( t )
+
. We have already found v(0+) to
€ circuit values for v(0 ) and dt
+
t =0
be 0 V. Matching this to the symbolic solution for t = 0+, we have the
following:
€
A1 + A2 = 0V or A1 = –A2
For the value of the derivative in the circuit, we always try to first write
our variable, (v in this case), in terms of energy (or state) variables, iL
and/or vC. Here, this is a simple matter:
v ( t ) = vC ( t ) € Then we differentiate, and use the component equations involving d/dt for
L and/or C:
d
d
i (0+ ) −2.3 mA
v(t)
= vC ( t )
=C
=
+
+
dt
dt
C
5 µF
t =0
t =0 We equate this to the symbolic derivative: d
−2.3 mA
v(t)
= A1s1 + A2 s2 = A1 ( s1 − s2 ) = A1 2 α 2 − ω 2 =
o
dt
5 µF
t =0+ € or € A1 = −2.3 mA
2 α 2 − ω 2 ⋅ 5 µF
o −2.3 mA =
2 2
Mr/s ⋅ 5 µF
5 = −2.3 mV
≈ −257 µV
45 Plugging in values gives the solution for v(t > 0): € € −1+ v ( t ) = −257µV ⋅ e 2
Mr/s⋅t
5 −1− + 257µV ⋅ e 2
Mr/s⋅t
5...
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This document was uploaded on 03/17/2014 for the course ECE 2260 at University of Utah.
 Fall '08
 Cotter,N

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