2260
F 07
HOMEWORK #1 prob 3 solution
E
X
:
+
–
+
–
2.3 V
t = 0
1 k
Ω
1 μH
100 m
Ω
v
(
t
)
5 μF
After being closed for a long time, the switch opens at
t
= 0.
Find
v
(
t
) for
t
> 0.
S
OL
'
N
:
We may perform the following initial steps in any order:
1)
Find characteristic roots for the parallel RLC circuit (for
t
> 0)
2)
Find the final value of
v
(
t
) as
t
>
∞
, which is the
A
3
(constant)
term in the solution.
3)
Find the initial values of energy variables:
i
L
(0
+
) =
i
L
(0
–
) and
v
C
(0
+
) =
v
C
(0
–
)
We will perform them in the order listed.
First, we find the characteristic
roots.
s
1,2
=
−α
±
α
2
− ω
o
2
For a parallel RLC circuit, the value of
α
is onehalf the inverse RC time
constant:
α
=
1
2
RC
=
1
2
⋅
100m
Ω⋅
5
μ
F
=
1 M/s
For both parallel and series RLC circuits, the resonant frequency,
ω
o
, is
the inverse of the square root of the product of L and C:
ω
o
2
=
1
LC
=
1
1
μ
H
⋅
5
μ
F
=
0.2 M
2
r/s
(
)
2
Substituting into the equation for the roots yields the following:
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s
1,2
=
−
1 Mr/s
±
1
−
0.2
Mr/s
=
−
1
±
2
5
Mr/s
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 Fall '08
 Cotter,N
 RLC, RC circuit, Voltage source, parallel rlc circuit

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