60. (a) The energy delivered by the battery is the integral of Eq. 2814 (where we use Eq. 3143 for the
current):
Z
t
0
P
battery
dt
=
Z
t
0
E
2
R
³
1
−
e
−
Rt/L
´
dt
=
E
2
R
·
t
+
L
R
³
e
−
Rt/L
−
1
´
¸
=
(10
.
0V)
2
6
.
70 Ω
·
2
.
00 s +
(5
.
50 H)
(
e
−
(6
.
70 Ω)(2
.
00 s)
/
5
.
50 H
−
1
)
6
.
70 Ω
¸
=1
8
.
7J
.
(b) The energy stored in the magnetic Feld is given by Eq. 3151:
U
B
=
1
2
Li
2
(
t
)=
1
2
L
µ
E
R
¶
2
(1
−
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Current, Energy

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