sanne (as42476) – Homework 06 – Yao – (59110)
1
This
printout
should
have
26
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
(part 1 of 2) 10.0 points
Consider
the
combination
of
resistors
shown in the figure.
4
.
5 Ω
1 Ω
4
.
9 Ω
9 Ω
7
.
8 Ω
2
.
2 Ω
3
.
7 Ω
a
b
What is the resistance between point
a
and
point
b
?
Correct answer: 7
.
92496 Ω.
Explanation:
Let’s redraw the figure
i
R
1
i
R
2
i
R
5
i
R
4
i
R
3
i
R
6
i
R
7
a
b
Let :
R
1
= 4
.
5 Ω
,
R
2
= 1 Ω
,
R
3
= 4
.
9 Ω
,
R
4
= 9 Ω
,
R
5
= 7
.
8 Ω
,
R
6
= 2
.
2 Ω
,
and
R
7
= 3
.
7 Ω
.
Basic Concepts:
•
Equivalent resistance.
•
Ohm’s Law.
There are two rules for adding up resis
tances. If the resistances are in series, then
R
series
=
R
1
+
R
2
+
R
3
+
· · ·
+
R
n
.
If the resistances are parallel, then
1
R
parallel
=
1
R
1
+
1
R
2
+
1
R
3
+
· · ·
+
1
R
n
.
Solution:
The key to a complex arrange
ments of resistors like this is to split the prob
lem up into smaller parts where either all the
resistors are in series, or all of them are in
parallel. It is easier to visualize the problem
if you redraw the circuit each time you add
them.
i
R
1
i
R
2
i
R
5
i
R
4
i
R
367
a
b
Step 1:
The three resistors on the right are
all in series, so
R
367
=
R
3
+
R
6
+
R
7
= (4
.
9 Ω) + (2
.
2 Ω) + (3
.
7 Ω)
= 10
.
8 Ω
.
i
R
1
i
R
2
i
R
3675
i
R
4
a
b
Step 2:
R
5
and
R
367
are connected paral
lel, so
R
3675
=
parenleftbigg
1
R
5
+
1
R
367
parenrightbigg
−
1
=
R
5
R
367
R
5
+
R
367
=
(7
.
8 Ω) (10
.
8 Ω)
18
.
6 Ω
= 4
.
52903 Ω
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
sanne (as42476) – Homework 06 – Yao – (59110)
2
i
R
1
i
R
36752
i
R
4
a
b
Step 3:
R
2
and
R
3675
are in series, so
R
23675
=
R
2
+
R
3675
= (1 Ω) + (4
.
52903 Ω)
= 5
.
52903 Ω
.
Step 4:
R
23675
and
R
4
are parallel, so
R
236754
=
parenleftbigg
1
R
4
+
1
R
23675
parenrightbigg
−
1
=
R
4
R
23675
R
4
+
R
23675
=
(9 Ω) (5
.
52903 Ω)
14
.
529 Ω
= 3
.
42496 Ω
.
i
R
1
i
R
367524
a
b
Step 5:
Finally,
R
1
and
R
236754
are in se
ries, so the equivalent resistance of the circuit
is
R
eq
=
R
1
+
R
236754
= (4
.
5 Ω) + (3
.
42496 Ω)
=
7
.
92496 Ω
.
002
(part 2 of 2) 10.0 points
If the current in the righthand 2
.
2
Ω resis
tor is 6 A
,
what is the potential difference
between points
a
and
b
?
Correct answer: 183
.
046 V.
Explanation:
Let :
R
6
= 6 A
.
Current, resistance, and voltage drop are
related by
V
=
I R .
Since
all the resis
tances and the current through one resistor
are known, all the voltages and currents can
be calculated.
Then the same current through
R
6
must go
through
R
3
and
R
7
since they are in series.
This is the same as saying there is 6 A going
through
R
367
(see previous part).
V
8
=
I
8
R
367
= (6 A) (10
.
8 Ω) = 64
.
8 V
.
Now since
R
367
and
R
5
share the same end
points (they are connected in parallel), the
potential difference across each must be the
same.
So,
V
5
= 64
.
8 V
,
and the current
through
R
5
is
I
5
=
V
5
R
5
=
64
.
8 V
7
.
8 Ω
= 8
.
30769 A
.
A total of 14
.
3077 A goes through
R
2
and is
split, with 8
.
30769 A going through
R
5
and
6 A going through
R
367
.
From Step 3 in the previous part, we know
that the
R
2
,
R
5
,
R
367
combination is equiva
lent to
R
23675
. This means that the potential
difference across
R
23675
is
V
10
=
R
23675
I
10
= (14
.
3077 A) (5
.
52903 Ω)
= 79
.
1077 V
,
which is also the voltage drop across
R
4
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Physics, Work, Resistor, SEPTA Regional Rail, Correct Answer, Jaguar Racing, Electrical resistance

Click to edit the document details