815_Physics ProblemsTechnical Physics - Chapter 28*P28.68(a...

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