251
HOMEWORK
5:
SOLUTIONS
Q26.19: ANSWER: think of relations to the known SI units. For example, energy stored in a capacitor is U=CV
2
/2
[J], while the dissipated power in a resistor is P=I
2
R [J/s]. Divide U/P= CV
2
/( 2I
2
R). From the Ohm’s law:
V=RI, therefore U/P= CR/2 [J/(J/s)=s], so RC has units of seconds.
26.41.
I
DENTIFY
:
The capacitors, which are in parallel, will discharge exponentially through the resistors.
S
ET
U
P
:
Since
V
is proportional to
Q
,
V
must obey the same exponential equation as
Q
,
V
=
V
0
e
–
t
/
RC
. The current is
I = (V
0
/
R) e
–
t
/
RC
.
E
XECUTE
:
(a)
Solve for time when the potential across each capacitor is 10.0 V:
t
=
−
RC
ln(
V
/
V
0
) = –
(80.0 Ω)(35.0
μ
F) ln(10/45) = 4210
μ
s = 4.21 ms
(b)
I = (V
0
/
R) e
–
t
/
RC
. Using the above values, with
V
0
= 45.0 V, gives
I
= 0.125 A.
E
VALUATE
:
Since the current and the potential both obey the same exponential equation, they are both reduced
by the same factor (0.222) in 4.21 ms.
26.48.
I
DENTIFY
:
When the capacitor is fully charged the voltage
V
across the capacitor equals the battery emf and
Q
CV
=
. For a charging capacitor,
(
)
/
1
t
RC
q
Q
e
−
=
−
.
S
ET
U
P
:
ln
x
e
x
=
E
XECUTE
:
(a)
6
4
(5.90
10
F)(28.0 V)
1.65
10
C.
Q
CV
−
−
=
=
×
(b)
/
(1
)
t
RC
q
Q
e
−
=
−
, so
/
1
t
RC
q
e
Q
−
=
−
and
.
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 Fall '05
 Ogilvie
 Energy, Power, Work, Magnetic Force, Magnetic Field, Circular Area, electric field direction, equivalent branches

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