Exercises 5.6
voltage source. We apply Kirchhof’s voltage law,
E
L
+
E
R
+
E
C
=
E
(
t
), to Loop 1 and
Loop 2 to get two equations connecting currents in the network. (Similarly to Example 2 and
Problem 11, there is no need to apply Kirchhof’s voltage law to Loop 3 because the resulting
equation is just a linear combination oF those For other two loops.)
Loop 1:
E
L
+
E
R
=0
⇒
0
.
5
dI
1
dt
+1
·
I
2
⇒
dI
1
dt
+2
I
2
.
(5.37)
Loop 2:
E
L
+
E
C
=cos3
t
⇒
0
.
5
dI
1
dt
+
q
3
0
.
5
t
⇒
dI
1
dt
+4
q
3
=2cos3
t.
(5.38)
Additionally, at joint points, by Kirchhof’s current law,
−
I
1
+
I
2
+
I
3
⇒−
I
1
+
I
2
+
dq
3
dt
.
(5.39)
Putting (5.37)–(5.39) together yields the Following system:
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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 Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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