Munoz (gm7794) – HW04 – TSOI – (92515)
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001
(part 1 of 2) 10.0 points
Consider the circuit
100 V
2
μ
F
4
μ
F
3
μ
F
5
μ
F
a
b
c
What is the equivalent capacitance for this
network?
1.
C
equivalent
=
10
7
μ
F
2.
C
equivalent
=
7
3
μ
F
3.
C
equivalent
= 14
μ
F
4.
C
equivalent
= 7
μ
F
correct
5.
C
equivalent
=
3
2
μ
F
Explanation:
E
B
C
1
C
2
C
3
C
4
a
b
c
Let :
C
1
= 2
μ
F
,
C
2
= 4
μ
F
,
C
3
= 3
μ
F
,
C
4
= 5
μ
F
,
and
E
B
= 100 V
.
The equivalent capacitance of capacitors
C
1
and
C
2
(parallel) is
C
12
=
C
1
+
C
2
= 6
μ
F
.
C
12
and
C
3
are in series, so
1
C
123
=
1
C
12
+
1
C
3
=
C
3
+
C
12
C
12
C
3
C
123
=
C
12
C
3
C
3
+
C
12
=
(6
μ
F) (3
μ
F)
6
μ
F + 3
μ
F
= 2
μ
F
.
C
123
and
C
4
are parallel, so
C
=
C
4
+
C
123
=
7
μ
F
.
002
(part 2 of 2) 10.0 points
What is the charge stored in the 5
μ
F lower
right capacitor?
1.
Q
1
= 1
,
100
μ
C
2.
Q
1
= 1
,
800
μ
C
3.
Q
1
= 500
μ
C
correct
4.
Q
1
= 360
μ
C
5.
Q
1
= 710
μ
C
Explanation:
Let :
C
4
= 5
μ
F
and
E
B
= 100 V
.
The charge stored in a capacitor is given by
Q
=
C V
, so,
Q
4
=
C
4
V
= (5
μ
F) (100 V)
=
500
μ
C
.
003
10.0 points
Consider the capacitor circuit
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Munoz (gm7794) – HW04 – TSOI – (92515)
2
E
B
6
μ
F
2
μ
F
8
μ
F
3
μ
F
b
a
c
Calculate the equivalent capacitance
C
ab
between points
a
and
b
.
Correct answer: 4
.
63158
μ
F.
Explanation:
Let :
C
1
= 6
μ
F
,
C
2
= 8
μ
F
,
C
3
= 2
μ
F
,
and
C
4
= 3
μ
F
.
E
B
C
1
C
3
C
2
C
4
b
a
c
For capacitors in series,
1
C
series
=
1
C
i
V
series
=
V
i
,
and the individual charges are the same.
For parallel capacitors,
C
parallel
=
C
i
Q
parallel
=
Q
i
,
and the individual voltages are the same.
Let
c
be the point between the two parallel
sections.
Let
c
be the point between the two parallel
sections. The capacitance between
a
an
c
is
C
l
=
C
1
+
C
3
= 6
μ
F + 2
μ
F
= 8
μ
F
.
The capacitance between
b
and
c
is
C
r
=
C
2
+
C
4
= 8
μ
F + 3
μ
F
= 11
μ
F
.
Thus the capacitance between
a
and
b
is
1
C
ab
=
1
C
l
+
1
C
r
=
C
r
+
C
l
C
l
C
r
C
ab
=
C
l
C
r
C
l
+
C
r
=
(8
μ
F) (11
μ
F)
8
μ
F + 11
μ
F
=
4
.
63158
μ
F
.
004
(part 1 of 2) 10.0 points
Consider two conductors 1 and 2 made of the
same ohmic material;
i.e.
,
ρ
1
=
ρ
2
.
Denote
the length by
, the cross sectional area by
A
.
The same voltage
V
is applied across the
ends of both conductors. The field
E
is inside
of the conductor.
V
1
E
1
I
1
1
r
1
V
2
E
2
I
2
2
r
2
If
A
2
= 2
A
1
,
2
= 2
1
and
V
2
=
V
1
, find
the ratio
E
2
E
1
of the electric fields.
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 Summer '08
 morrison
 Physics, Resistor, SEPTA Regional Rail, Correct Answer, Series and parallel circuits

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