Solutions 2.7Page 176
Problem 1
In the circuit of Fig. 2.7.7, suppose that
5
=
L
H,
Ω
=
25
R
, and that the source
E
of emf
is a battery supplying 100 V to the circuit.
Suppose also that the switch has been in
position 1 for a long time, so that steady current of 4A is flowing in the circuit.
At time
, the switch is thrown to position 2, so that
0
=
t
4
)
0
(
=
I
and
0
=
E
for
t
.
Find
.
0
≥
)
(
t
I
It is given that
describes the circuit.
Substituting given information
yields:
)
(
t
E
RI
I
L
=
+
′
0
25
5
=
+
′
I
I
Separating variables and integrating yields:
t
e
K
I
K
t
I
dt
I
dI
5
~
5
ln
5
−
=
+
−
=
−
=
∫
∫
The initial condition
is used to solve for the integration constant as follows:
4
)
0
(
=
I
∴
=
=
=
4
~
)
1
(
~
4
)
0
(
K
K
I
t
e
I
5
4
−
=
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Problem 3
Suppose that the battery in Problem 2 is replaced with an alternating current generator
that supplies a voltage of
volts.
With everything else the same, now
find
.
t
t
E
60
cos
100
)
(
=
)
(
t
I
The differential equation describing the circuit is
5
t
I
I
60
cos
100
25
=
+
′
.
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 Spring '11
 HAFTKA
 Alternating Current, Constant of integration, RC circuit, LC circuit

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