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21.56. Set Up:
Problem 21.55 shows that
initially is zero and rises to 25.0 V, that
starts at 25.0 V and
decreases to zero, and that
i
starts at zero and rises to 1.67 A.
so is proportional to
i. A
reads the current
i.
Solve:
The graphs are sketched in Figure 21.56.
Figure 21.56
21.57. Set Up:
Current decay in an
R

L
circuit is described by Eq. (21.28).
Solve:
and
and
Reflect:
As
R
is decreased the time constant increases and it takes longer for the current to decay. For this circuit the
time constant is
After one time constant the current has decayed to about 37% of its original value.
It is reasonable for it to take a little over two time constants for the current to decay to 10% of its original value.
21.58. Set Up:
The current as a function of time is given by
The energy stored in the inductor
is
U
reaches
its maximum value when
i
is
times its maximum value.
Solve: (a)
The maximum current is
gives
and
(b)
and
21.59. Set Up:
The resonant angular frequency of an
L

C
circuit is
Solve:
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 Spring '07
 Shoberg

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