malsam (wgm329) – hw13 – turner – (56705)
1
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001 (part 1 of 3) 10.0 points
Consider the circuit
a
b
A
11 V
10 V
I
1
4
.
7 Ω
4 Ω
I
3
5 Ω
I
2
4
.
1 Ω
5
.
1 Ω
Find the current through the Amp meter,
I
3
.
Correct answer: 0
.
083756 A.
Explanation:
a
b
A
E
1
E
2
I
1
r
1
r
2
I
3
r
3
I
2
r
4
r
5
Let :
E
1
= 11 V
,
E
2
= 10 V
,
r
1
= 4
.
7 Ω
,
r
2
= 4 Ω
,
r
3
= 5 Ω
,
r
4
= 4
.
1 Ω
,
and
r
5
= 5
.
1 Ω
.
We consider
R
1
=
r
1
+
r
2
= 4
.
7 Ω + 4 Ω = 8
.
7 Ω
,
and
R
2
=
r
4
+
r
5
= 4
.
1 Ω + 5
.
1 Ω = 9
.
2 Ω
.
From the junction rule,
I
1
=
I
2
+
I
3
.
Applying Kirchhoff’s loop rule, we obtain
two equations.
E
1
−
I
1
R
1
−
I
3
r
3
= 0
E
1
=
I
1
R
1
+
I
3
r
3
(1)
E
2
−
I
3
r
3
+
I
2
R
2
= 0
E
2
=
I
2
R
2
−
I
3
r
3
= (
I
1
−
I
3
)
R
2
−
I
3
r
3
=
I
1
R
2
−
I
3
(
R
2
+
r
3
)
,
(2)
Multiplying Eq. (1) by
R
2
and Eq. (2) by
−
R
1
and adding,
E
1
R
2
=
I
1
R
1
R
2
+
I
3
r
3
R
2
(3)
−E
2
R
1
=
−
I
1
R
1
R
2
+
I
3
R
1
(
R
2
+
r
3
)
(4)
E
1
R
2
− E
2
R
1
=
I
3
[
r
3
R
2
+
R
1
(
R
2
+
r
3
)]
.
Since
r
3
R
2
+
R
1
(
R
2
+
r
3
) = (9
.
2 Ω) (5 Ω)
+(8
.
7 Ω) (5 Ω + 9
.
2 Ω)
= 169
.
54 Ω
,
then
I
3
=
E
1
R
2
− E
2
R
1
r
3
R
2
+
R
1
(
R
2
+
r
3
)
(5)
=
(11 V) (9
.
2 Ω)
−
(10 V) (8
.
7 Ω)
169
.
54 Ω
=
0
.
083756 A
.
002 (part 2 of 3) 10.0 points
Find the current
I
1
.
Correct answer: 1
.
21623 A.
Explanation:
From Eq. 1 and
I
3
from Eq. 5, we have
I
1
=
E
1
−
I
3
r
3
R
1
=
11 V
−
(0
.
083756 A) (5 Ω)
8
.
7 Ω
=
1
.
21623 A
.
003 (part 3 of 3) 10.0 points
Find the current
I
2
.
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malsam (wgm329) – hw13 – turner – (56705)
2
Correct answer: 1
.
13248 A.
Explanation:
From Kirchhoff’s junction rule,
I
2
=
I
1
−
I
3
= 1
.
21623 A
−
0
.
083756 A
=
1
.
13248 A
.
004 (part 1 of 4) 10.0 points
An energy plant produces an output potential
of 1600 kV and serves a city 88 km away. A
highvoltage transmission line carries 1520 A
to the city. The effective resistance of a trans
mission line [wire(s)] is 1
.
74 Ω
/
km times the
distance from the plant to the city.
What is the potential provided to the city,
i.e.
, at the end of the transmission line?
Correct answer: 1367
.
26 kV.
Explanation:
Let :
V
plant
= 1600 kV
,
ℓ
= 88 km
,
I
= 1520 A
,
and
ρ
= 1
.
74 Ω
/
km
.
The potential drop on the wire is
V
=
I R
=
I ρ ℓ
= (1520 A) (1
.
74 Ω
/
km) (88 km)
= 232
.
742 kV
,
so the potential delivered to the city is
V
city
=
V
plant
−
V
wire
= 1600 kV
−
232
.
742 kV
=
1367
.
26 kV
.
005 (part 2 of 4) 10.0 points
How much power is dissipated due to resistive
losses in the transmission line?
Correct answer: 3
.
53768
×
10
8
W.
Explanation:
P
=
I
2
R
=
I
2
ρ ℓ
= (1520 A)
2
(1
.
74 Ω
/
km) (88 km)
=
3
.
53768
×
10
8
W
.
006 (part 3 of 4) 10.0 points
Assume the plant charges $ 0
.
087
/
kW
·
hr for
electric energy.
At this rate, how much does it cost to trans
mit energy to the city (by the transmission
line heating the atmosphere) each hour?
Correct answer: 30777
.
9 dollars
/
hr.
Explanation:
Let :
R
= $ 0
.
087
/
kW
·
hr
.
The cost of energy is
Cost =
P
[(
R
) (time)]
= (3
.
53768
×
10
8
W)
×
(0
.
087
/
kW
·
hr) (1 hr)
parenleftbigg
1 kW
1000 W
parenrightbigg
=
30777
.
9 dollars
/
hr
.
007 (part 4 of 4) 10.0 points
Consider the money lost by the transmission
line heating the atmosphere each hour.
As
sume the energy plant produces the same
amount of power; however, the output electric
potential of the energy plant is 20% greater.
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 Spring '08
 Turner
 Resistor, SEPTA Regional Rail, Correct Answer, Jaguar Racing, University City, JenkintownWyncote

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