Chapter 28
155
*P28.68
(a)
With the switch closed, current exists in a simple series
circuit as shown. The capacitors carry no current. For
R
2
we have
P
=
I R
2
2
I
R
=
=
⋅
=
P
2
2 40
18 5
.
.
V A
7 000 V A
mA .
The potential difference across
R
1
and
C
1
is
∆
V
IR
=
=
×
=
−
1
2
1 85
10
4 000
74 1
.
.
A
V
A
V
e
j
b
g
.
The charge on
C
1
FIG. P28.68(a)
Q
C
V
=
=
×
=
−
1
6
3 00
10
74 1
222
∆
.
.
C V
V
C
e
j
a
f
µ
.
The potential difference across
R
2
and
C
2
is
∆
Ω
V
IR
=
=
×
=
−
2
2
1 85
10
7 000
130
.
A
V
e
j
b
g
.
The charge on
C
2
Q
C
V
=
=
×
=
−
2
6
6 00
10
130
778
∆
.
C V
V
C
e
j
a
f
µ
.
The battery emf is
IR
I R
R
eq
=
+
=
×
+
=
−
1
2
2
1 85
10
7 000
204
b
g
b
g
.
A
4 000
V
A
V
.
(b)
In equilibrium after the switch has been opened, no current
exists. The potential difference across each resistor is zero. The
full 204 V appears across both capacitors. The new charge
C
2
Q
C
V
=
=
×
=
−
2
6
6 00
10
204
1 222
∆
.
C V
V
C
e
j
a
f
µ
for a change of
1 222
778
444
C
C
C
µ
µ
µ
−
=
.
FIG. P28.68(b)
*P28.69
The battery current is
150
45
14
4
213
+
+
+
=
a
f
mA
mA
.
(a)
The resistor with highest resistance is that
carrying 4 mA. Doubling its resistance will
reduce the current it carries to 2 mA. Then
the total current is
FIG. P28.69
150
45
14
2
211
+
+
+
=
a
f
mA
mA
, nearly the same as before. The ratio is
211
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- Fall '11
- Staff
- Physics, Current, Resistor, Potential difference, Electrical resistance, Series and parallel circuits
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