Reilly (mar3978) – ohw16 – turner – (56725)
1
This
printout
should
have
15
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001 (part 1 of 2) 10.0 points
The switch
S
has been in position
b
for a
long period of time.
C
R
1
R
2
R
3
E
S
b
a
When the switch is moved to position “a”,
find the characteristic time constant.
1.
τ
=
radicalbig
R
1
R
2
C
2.
τ
=
R
1
+
R
2
2
C
3.
τ
= (
R
1
+
R
2
)
C
correct
4.
τ
=
1
R
2
C
5.
τ
=
1
R
1
C
6.
τ
=
1
(
R
1
+
R
2
)
C
7.
τ
=
R
2
C
8.
τ
=
2
(
R
1
+
R
2
)
C
9.
τ
=
R
1
C
10.
τ
=
1
√
R
1
R
2
C
Explanation:
In charging an
R C
circuit, the characteris
tic time constant is given by
τ
=
R C ,
where in this problem
R
is the equivalent
resistance, or
R
=
R
1
+
R
2
.
002 (part 2 of 2) 10.0 points
1
.
3
μ
F
0
.
7 MΩ
4 MΩ
2
.
5 MΩ
0
.
5 V
S
b
a
S
has been left at position “
a
” for a long
time. It is then switched from “
a
” to “
b
” at
t
= 0.
Determine the energy dissipated through
the resistor
R
2
alone from
t
= 0 to
t
=
∞
.
Correct answer: 0
.
1
μ
J.
Explanation:
Let :
E
= 0
.
5 V
,
R
1
= 0
.
7 MΩ
,
R
2
= 4 MΩ
,
R
3
= 2
.
5 MΩ
,
and
C
= 1
.
3
μ
F
.
The total energy dissipated
U
R
2
+
R
3
dissip
=
1
2
C
E
2
.
Since
R
2
and
R
3
are in series, the energy
dissipated by
R
2
is only a fraction
R
2
R
2
+
R
3
of
the total energy:
U
R
2
dissip
=
parenleftbigg
R
2
R
2
+
R
3
parenrightbigg parenleftbigg
1
2
C
E
2
parenrightbigg
We observe that power consumption consid
eration provides an independent check on the
fraction used.
Since the two resistors are in
series they share a common current,
I
.
The
corresponding power consumptions by
R
2
and
R
3
are respectively
P
2
=
I
2
R
2
and
P
3
=
I
2
R
3
.
This shows the correctness of the fraction, i.e.
P
2
/
(
P
2
+
P
3
) =
R
2
/
(
R
2
+
R
3
).
Alternate
solution:
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Reilly (mar3978) – ohw16 – turner – (56725)
2
More formally, noting that the initial current
is
I
0
=
E
R
2
+
R
3
, the total energy dissipated
by
R
2
is
U
R
2
=
integraldisplay
∞
0
I
(
t
)
2
R
2
dt
=
integraldisplay
∞
0
I
2
0
R
2
e
−
2
t/
[
C
(
R
2
+
R
3
)]
dt
=
parenleftbigg
E
R
2
+
R
3
parenrightbigg
2
R
2
bracketleftbigg
−
C
(
R
2
+
R
3
)
2
bracketrightbigg
×
e
−
2
t/
[
C
(
R
2
+
R
3
)]
vextendsingle
vextendsingle
vextendsingle
vextendsingle
∞
0
=
R
2
R
2
+
R
3
1
2
C
E
2
=
(4 MΩ)
(4 MΩ) + (2
.
5 MΩ)
1
2
(1
.
3
μ
F) (0
.
5 V)
2
=
0
.
1
μ
J
.
003 (part 1 of 2) 10.0 points
Assume:
The battery is ideal (it has no
internal resistance) and connecting wires have
no resistance.
Unlike most real bulbs, the
resistance of the bulb in the questions below
does not change as the current through it
changes.
A battery, a capacitor, a bulb, and a switch
are in the circuit as shown below. The switch
is initially open as shown in the diagram, and
the capacitor is uncharged.
C
E
R
S
Which of the following correctly describes
what happens to the bulb when the switch is
closed?
1.
The bulb is dim and remains dim.
2.
None of these is correct.
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 Spring '10
 TURNER,J
 Force, Magnetic Field, Electric charge, Reilly

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