Platt, David – Homework 12 – Due: Oct 5 2005, 4:00 am – Inst: Ken Shih
1
This
printout
should
have
8
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
Consider the following circuit containing
identical bulbs
A
,
B
,
C
,
D
, and
E
respec
tively. Denote the potentials across bulbs as
V
A
,
V
B
,
V
C
,
V
D
, and
V
E
.
E
A
B
C
D
E
Rank the potentials across the bulbs.
1.
V
A
=
V
B
=
V
C
=
V
D
=
V
E
2.
V
A
=
V
B
> V
A
=
V
D
=
V
E
3.
V
A
=
V
D
=
V
E
> V
B
> V
C
4.
V
C
> V
B
> V
A
> V
D
> V
E
5.
V
A
=
V
B
=
V
C
> V
D
> V
E
6.
V
A
=
V
B
=
V
C
> V
D
=
V
E
correct
7.
V
E
=
V
D
> V
A
> V
B
=
V
C
8.
V
A
=
V
D
=
V
E
> V
B
=
V
C
9.
V
B
=
V
C
> V
A
=
V
D
=
V
E
10.
V
A
=
V
C
> V
B
> V
D
=
V
E
Explanation:
The potential across bulbs
A
,
B
and
C
is
V
. The potential across bulbs
D
and
E
is
V
2
.
Thus
i
A
=
i
B
=
i
C
=
V
R
i
D
=
i
E
=
V
2
R
.
002
(part 1 of 2) 10 points
Four resistors are connected as shown in
the figure.
96 V
S
1
c
d
a
b
19 Ω
32 Ω
64 Ω
79 Ω
Find the resistance between points
a
and
b
.
Correct answer: 38
.
7151 Ω.
Explanation:
E
B
S
1
c
d
a
b
R
1
R
2
R
3
R
4
Let :
R
1
= 19 Ω
,
R
2
= 32 Ω
,
R
3
= 64 Ω
,
R
4
= 79 Ω
,
and
E
B
= 96 V
.
Ohm’s law is
V
=
I R .
A good rule of thumb is to eliminate junc
tions connected by zero resistance.
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Platt, David – Homework 12 – Due: Oct 5 2005, 4:00 am – Inst: Ken Shih
2
E
B
c
a
b
R
1
R
2
R
3
R
4
The parallel connection of
R
1
and
R
2
gives
the equivalent resistance
1
R
12
=
1
R
1
+
1
R
2
=
R
2
+
R
1
R
1
R
2
R
12
=
R
1
R
2
R
1
+
R
2
=
(19 Ω) (32 Ω)
19 Ω + 32 Ω
= 11
.
9216 Ω
.
E
c
a
b
R
12
R
3
R
4
The series connection of
R
12
and
R
3
gives
the equivalent resistance
R
123
=
R
12
+
R
3
= 11
.
9216 Ω + 64 Ω
= 75
.
9216 Ω
.
E
B
a
b
R
123
R
4
The parallel connection of
R
123
and
R
4
gives the equivalent resistance
1
R
ab
=
1
R
123
+
1
R
4
=
R
4
+
R
123
R
123
R
4
R
ab
=
R
123
R
4
R
123
+
R
4
=
(75
.
9216 Ω) (79 Ω)
75
.
9216 Ω + 79 Ω
= 38
.
7151 Ω
.
or combining the above steps, the equivalent
resistance is
R
ab
=
R
1
R
2
R
1
+
R
2
+
R
3
¶
R
4
R
1
R
2
R
1
+
R
2
+
R
3
+
R
4
=
•
(19 Ω) (32 Ω)
19 Ω + 32 Ω
+ 64 Ω
‚
(79 Ω)
(19 Ω) (32 Ω)
19 Ω + 32 Ω
+ 64 Ω + 79 Ω
=
38
.
7151 Ω
.
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 Fall '08
 Turner
 Physics, Work, Resistor, SEPTA Regional Rail, Electrical resistance, Series and parallel circuits, R1 R2 R3, Ken Shih

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