P26_067.PDF
(
63 KB
)
Pobierz
Chapter 26 - 26.67
67. (a) The equivalent capacitance is
C
eq
=
C
1
C
2
/
(
C
1
+
C
2
)
.
Thus the charge
q
on each capacitor is
q
=
C
eq
V
=
C
1
C
2
V
C
1
+
C
2
=
(2
.
0
µ
F)(8
.
0
µ
F)(300 V)
2
.
0
µ
F+8
.
0
µ
F
=4
.
8
×
10
−
4
C
.
The potential differences are:
V
1
=
q/C
1
=4
.
8
×
10
−
4
C
/
2
.
0
µ
F = 240 V,
V
2
=
V
−
V
1
= 300 V
−
240 V = 60 V
.
(b) Now we have
q
1
/C
1
=
q
2
/C
2
=
V
(
V
being the new potential difference across each capacitor)
and
q
1
+
q
2
=2
q
.Wesolvefor
q
1
,
q
2
and
V
:
q
1
=
C
1
+
C
2
=
2(2
.
0
µ
F)(4
.
8
10
−
4
C)
2
.
0
µ
F+8
.
0
µ
F
×
=1
.
9
×
10
−
4
C
,
q
2
=2
q
−
q
1
=7
.
7
×
10
−
4
C
,
V
=
C
1
=
1
.
92
10
−
4
C
2
.
0
µ
F
×
=96V
.
(c) In this circumstance, the capacitors will simply discharge themselves, leaving
q
1
=
q
2
=0and
V
1
=
V
2
=0.
2
C
1
q
q
1
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Inne foldery tego chomika:
chap01
chap02
chap03
chap04
chap05
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