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Chapter 5
For the top juncture, all the currents ﬂow out, and the Kirchhof’s current law gives
−
I
1
−
I
2
−
I
3
=0
⇒
I
1
+
I
2
+
I
3
=0
.
There±ore, the system, describing the current in
RLC
,is
q
C
=
R
2
I
2
,
R
2
I
2
=
R
1
I
3
+
LI
0
3
,
I
1
+
I
2
+
I
3
=0
.
With given data,
R
1
=
R
2
=1Ω
,
L
=1H
,and
C
= 1 F, and the relation
I
1
=
dq/dt
,th
is
system becomes
q
=
I
2
,
I
2
=
I
3
+
I
0
3
,
q
0
+
I
2
+
I
3
=0
.
Replacing in the last two equations
I
2
by
q
,weget
I
0
3
+
I
3
−
q
=0
,
q
0
+
q
+
I
3
=0
.
We eliminate
q
by substituting
q
=
I
0
3
+
I
3
into the second equation and obtain
I
0
3
+2
I
0
3
+2
I
3
=0
.
The characteristic equation,
r
2
+2
r
+ 2 = 0, has roots
r
=
−
1
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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