homework 06 – HOFFMA, DAVID – Due: Feb 25 2008, 4:00 am
1
Question 1, chap 28, sect 4.
part 1 of 2
10 points
A 23 V battery has an internal resistance
r
.
1 A
23 V
r
10 Ω
90 Ω
internal
resistance
What is the value of
r
?
Correct answer: 14 Ω (tolerance
±
1 %).
Explanation:
I
1
I
2
I
3
E
r
R
2
R
3
internal
resistance
Let :
E
= 23 V
,
R
2
= 10 Ω
,
R
3
= 90 Ω
,
and
I
1
= 1 A
.
Since
R
2
and
R
3
are connected parallel,
their equivalent resistance
R
23
is
1
R
23
=
1
R
2
+
1
R
3
=
R
3
+
R
2
R
2
R
3
R
23
=
R
2
R
3
R
2
+
R
3
=
(10 Ω) (90 Ω)
10 Ω + 90 Ω
= 9 Ω
.
Using Ohm’s law, we have
E
=
I
1
r
+
I
1
R
23
r
=
E −
I
1
R
23
I
1
=
23 V
−
(1 A) (9 Ω)
1 A
= 23 Ω
−
9 Ω
=
14 Ω
.
Question 2, chap 28, sect 4.
part 2 of 2
10 points
Determine the magnitude of the current
through the 90 Ω resistor on the righthand
side of the circuit.
Correct answer: 0
.
1 A (tolerance
±
1 %).
Explanation:
The potential drop across the 90 Ω resistor
on the righthand side of the circuit is
E
3
=
E −
I
1
r
= 23 V
−
(1 A) (14 Ω)
= 23 V
−
14 V
= 9 V
,
so the current through the resistor is
I
3
=
E
3
r
3
=
9 V
90 Ω
=
1
10
A
= 0
.
1 A
,
and
I
2
=
I
1
−
I
3
= 1 A
−
1
10
A
=
9
10
A
=
0
.
9 A
.
Question 3, chap 28, sect 4.
part 1 of 3
10 points
9
.
9 V
1
.
7 V
3
.
1 V
I
1
0
.
1 Ω
3
.
1 Ω
I
2
6
.
5 Ω
I
3
9
.
7 Ω
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homework 06 – HOFFMA, DAVID – Due: Feb 25 2008, 4:00 am
2
Find the current
I
1
in the 0
.
1 Ω resistor
at the bottom of the circuit between the two
power supplies.
Correct answer:
1
.
37392
A (tolerance
±
1
%).
Explanation:
E
1
E
2
E
3
I
1
R
A
R
B
I
2
R
C
I
3
R
D
At a junction (Conservation of Charge)
I
1
+
I
2
−
I
3
= 0
.
(1)
Kirchhoff’s law on the large outside loop gives
(
R
A
+
R
B
)
I
1
+
R
D
I
3
=
E
1
+
E
2
.
(2)
Kirchhoff’s law on the righthand small loop
gives
R
C
I
2
+
R
D
I
3
=
E
3
.
(3)
Let :
R
A
= 0
.
1 Ω
,
R
B
= 3
.
1 Ω
,
R
C
= 6
.
5 Ω
,
R
D
= 9
.
7 Ω
,
E
1
= 9
.
9 V
,
E
2
= 1
.
7 V
,
and
E
3
= 3
.
1 V
.
Using determinants,
I
1
=
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
0
1
−
1
E
1
+
E
2
0
R
D
E
3
R
C
R
D
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
1
1
−
1
R
A
+
R
B
0
R
D
0
R
C
R
D
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
Expanding along the first row, the numera
tor is
D
1
=
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
0
1
−
1
E
1
+
E
2
0
R
D
E
3
R
C
R
D
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
= 0
−
1
vextendsingle
vextendsingle
vextendsingle
vextendsingle
E
1
+
E
2
R
D
E
3
R
D
vextendsingle
vextendsingle
vextendsingle
vextendsingle
+ (
−
1)
vextendsingle
vextendsingle
vextendsingle
vextendsingle
E
1
+
E
2
0
E
3
R
C
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
−
[(
E
1
+
E
2
)
R
D
− E
3
R
D
]
−
[
R
C
(
E
1
+
E
2
)
−
0]
=
R
D
(
E
3
− E
1
− E
2
)
−
R
C
(
E
1
+
E
2
)
= (9
.
7 Ω) (3
.
1 V
−
9
.
9 V
−
1
.
7 V)
−
(6
.
5 Ω) (9
.
9 V + 1
.
7 V)
=
−
157
.
85 V Ω
.
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 Spring '08
 Turner
 Resistance, Work, Resistor, SEPTA Regional Rail, Electrical resistance, DAVID

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