This preview shows page 1. Sign up to view the full content.
42. (a) We note that we obtain the maximum value in Eq. 3328 when we set
t
=
π
2
ω
d
=
1
4
f
=
1
4(60)
=0
.
00417 s
or 4
.
17 ms. The result is
E
m
sin(
π/
2) =
E
m
sin(90
◦
)=3
6
.
0 V. We note, for reference in the
subsequent parts, that at
t
=4
.
17 ms, the current is
i
=
I
sin(
ω
d
t
−
φ
)=
I
sin(90
◦
−
(
−
29
.
4
◦
)) = (0
.
196 A) cos(29
.
4
◦
)=0
.
171 A
using Eq. 3329 and the results of the Sample Problem.
(b) At
t
=4
.
17 ms, Ohm’s law directly gives
v
R
=
iR
=(
I
cos(29
.
4
◦
))
R
(0
.
171 A)(160 Ω) = 27
.
3V
.
(c) The capacitor voltage phasor is 90
◦
less than that of the current. Thus, at
t
=4
.
17 ms, we obtain
v
C
=
I
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '08
 SPRUNGER
 Physics, Current

Click to edit the document details