Solve: (a)
The 20.0 pF, 30.0 pF and 40.0 pF capacitors are in series and their equivalent capacitance
is given by
After a long time the voltage across the 10.0 pF capacitor is 50.0 V. The charge on this capacitor is
The voltage across
is 50.0 V and
This is the charge on each of the capacitors in series, so
(b)
Note that
as it should.
(c)
(d)
The 10.0 pF capacitor and
are in parallel so their equivalent is
Reflect:
The charges on the capacitors start at zero and increase to their final values. The current starts at its maxi
mum value and then decays to zero.
19.70.
Set Up:
The time constant for the circuit is
While charging,
and
In the discharging circuit both the charge and current decrease to
of their initial values in time
Solve:
(a)
gives
and
(b)
(c)
19.71.
Set Up:
Equations (19.17) give
i
and
q
as functions of time during charging. During discharging, both
i
and
q
decay exponentially, as in Eq. (19.19) for
i.
t
5 t 5
188
m
s
i
5
I
0
e
2
t
/
RC
5
I
0
e
2
1
86.1
m
s
2
/
1
188
m
s
2
5
0.633
I
0
t
5 2
RC
1
ln
3
1
2
1
/
e
42
5 2
1
188
m
s
21
2
0.458
2
5 1
86.1
m
s
e
2
t
/
RC
5
1
2
1
e
1
e
5
1
2
e
2
t
/
RC
.
q
5
Q
final
e
q
5
Q
final
1
1
2
e
2
t
/
RC
2
.
t 5
RC
5
1
125
V
21
1.50
m
F
2
5
188
m
s
t
5 t
.
1
/
e
i
5
I
0
e
2
t
/
RC
.
q
5
Q
final
1
1
2
e
2
t
/
RC
2
t 5
RC
.
t 5
RC
5
1
20.0
V
21
19.2 pF
2
5
384 ps
C
eq
5
10.0 pF
1
9.23 pF
5
19.2 pF.
C
s
I
5
E
R
5
50.0 V
20.0
V
5
2.50 A
V
20
1
V
30
1
V
40
5
50.0 V,
V
40
5
q
40
C
40
5
461 pC
40.0 pF
5
11.5 V.
V
30
5
q
30
C
30
5
461 pC
30.0 pF
5
15.4 V.
V
20
5
q
20
C
20
5
461 pC
20.0 pF
5
23.1 V.
V
10
5
50.0 V.
q
20
5
q
30
5
q
40
5
461 pC.
1
50.0 V
2
5
461 pC.
q
5
5
C
5
V
5
5
1
9.23 pF
2
C
s
q
10
5
C
10
V
10
5
1
10.0 pF
21
50.0 V
2
5
500 pC.
C
s
5
9.23
m
F.
1
C
s
5
1
20.0 pF
1
1
30.0 pF
1
1
40.0 pF
.
C
s
i
5
I
0
e
2
t
/
RC
.
t 5
RC
,
q
5
C
/
V
.
i
5
0.05
I
0
.
e
2
t
/
t
5
0.05
t
t
5
3.0.
2
t
t
5
ln
1
0.05
2
.
e
2
t
/
t
5
0.05.
0.95
5
1
2
e
2
t
/
RC
.
q
5
0.95
Q
final
i
5
I
0
e
2
t
/
RC
.
q
5
Q
final
1
1
2
e
2
t
/
RC
2
t 5
RC
,
I
0
5
E
R
5
12.0 V
75.0
V
5
0.160 A
E
5
I
0
R
.
q
5
0,
I
0
Current, Resistance, and DirectCurrent Circuits
1917
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Solve: (a)
The graphs are the same as in Figure 19.33a and b in the textbook and are given in Figure 19.71a.
(b)
The graphs of
i
and
q
versus
t
for a discharging capacitor are sketched in Figure 19.71b.
Figure 19.71
19.72.
Set Up:
There is a single current path so the current is the same at all points in the circuit. Assume the
current is counterclockwise.
Solve: (a)
Apply the loop rule, traveling around the circuit in the direction of the current:
Our calculated
I
is positive so
I
is counterclockwise, as we assumed.
(b)
19.73.
Set Up:
The power for a resistor
R
is
The power for an emf is
Solve: (a)
The loop rule for a counterclockwise direction gives
(b)
The net power output of the battery is
(c)
In the emf of this battery electrical energy is being consumed at the rate
and in the internal resistance of the
battery electrical energy is being consumed at the rate
(d)
Chemical energy is being converted to electrical energy at a rate of
in the 12.0 V
battery and electrical energy is being converted to chemical energy at a rate of
in the 4.0 V
battery. The net rate of conversion of chemical energy to electrical energy is
The overall
rate of dissipation of electrical energy (conversion to heat) is
The two rates
are equal, as required by conservation of energy.
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 Spring '09
 RODRIGUEZ
 Physics, Charge, Current, Resistance, Resistor, Electrical resistance, Series and parallel circuits, DirectCurrent Circuits

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