P31_060.PDF

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Chapter 31 - 31.60
60. (a) The energy delivered by the battery is the integral of Eq. 28-14 (where we use Eq. 31-43 for the
current):
t
t
2
R
1
e − Rt/L dt = E
2
R
t + L
R
e − Rt/L
1
P battery dt =
−
−
0
0
2 . 00 s + (5 . 50 H) e − (6 . 70 Ω)(2 . 00 s) / 5 . 50 H
1
=
(10 . 0V) 2
6 . 70 Ω
−
6 . 70 Ω
=18 . 7J .
(b) The energy stored in the magnetic field is given by Eq. 31-51:
2 Li 2 ( t )= 1
2 L R
2
U B =
(1
−
e − Rt/L ) 2
2 (5 . 50 H) 10 . 0V
2 1
e − (6 . 70 Ω)(2 . 00 s) / 5 . 50 H 2
=
1
−
6 . 70 Ω
=5 . 10 J .
(c) The difference of the previous two results gives the amount “lost” in the resistor: 18 . 7J
−
5 . 10 J =
13 . 6J .
E
1
643282727.001.png

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