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86. (a) We note that we obtain the maximum value in Eq. 3128 when we set
t
f
d
==
=
=
π
2
ω
1
4
1
460
0 00417
()
.s
or 4.17 ms. The result is
ε
mm
sin(
sin (
)
.
π/2) =
90
36 0
°
=
V.
(b) At
t
= 4.17 ms, the current is
sin (
)
sin (90
( 24.3 ))
(0.164A) cos(24.3 )
0.1495A
0.150 A.
d
iI
t
I
φ
=−
=
°
−
−
°
=
°
=≈
using Eq. 3129 and the results of the Sample Problem. Ohm’s law directly gives
(0.1495A)(200 )
29.9V.
R
vi
R
Ω
=
(c) The capacitor voltage phasor is 90° less than that of the current. Thus, at
t
= 4.17 ms,
we obtain
sin(90
( 24.3 ) 90 )
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Current

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