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howard (cah3459) – hw13 – turner – (56705)
1
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001 (part 1 of 3) 10.0 points
Consider the circuit
a
b
A
26 V
9V
I
1
3
.
6 Ω
2
.
5 Ω
I
3
3
.
8 Ω
I
2
3
.
7 Ω
±ind the current through the Amp meter,
I
3
.
Correct answer: 1
.
65506 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
= 26 V
,
E
2
= 9 V
,
r
1
=3
.
,
r
2
=2
.
,
r
3
.
,
r
4
.
,
and
r
5
= 6 Ω
.
We consider
R
1
=
r
1
+
r
2
.
6 Ω + 2
.
5 Ω = 6
.
1 Ω
,
and
R
2
=
r
4
+
r
5
.
7 Ω + 6 Ω = 9
.
.
±rom the junction rule,
I
1
=
I
2
+
I
3
.
Applying Kirchho²’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
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
.
7 Ω) (3
.
8 Ω)
+(6
.
1 Ω) (3
.
8 Ω + 9
.
7 Ω)
= 119
.
21 Ω
,
then
I
3
=
E
1
R
2
2
R
1
r
3
R
2
+
R
1
(
R
2
+
r
3
)
(5)
=
(26 V) (9
.
7 Ω)

(9 V) (6
.
1 Ω)
119
.
21 Ω
=
1
.
65506 A
.
002 (part 2 of 3) 10.0 points
±ind the current
I
1
.
Correct answer: 3
.
23127 A.
Explanation:
±rom Eq. 1 and
I
3
From Eq. 5, we have
I
1
=
E
1

I
3
r
3
R
1
=
26 V

(1
.
65506 A) (3
.
8 Ω)
6
.
=
3
.
23127 A
.
003 (part 3 of 3) 10.0 points
±ind the current
I
2
.
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View Full Document howard (cah3459) – hw13 – turner – (56705)
2
Correct answer: 1
.
57621 A.
Explanation:
From Kirchhof’s junction rule,
I
2
=
I
1

I
3
=3
.
23127 A

1
.
65506 A
=
1
.
57621 A
.
004 (part 1 of 4) 10.0 points
An energy plant produces an output potential
o± 17000 kV and serves a city 222 km away. A
highvoltage transmission line carries 1180 A
to the city. The efective resistance o± a trans
mission line [wire(s)] is 4
.
09 Ω
/
km times the
distance ±rom the plant to the city.
What is the potential provided to the city,
i.e.
, at the end o± the transmission line?
Correct answer: 15928
.
6 kV.
Explanation:
Let :
V
plant
= 17000 kV
,
±
= 222 km
,
I
= 1180 A
,
and
ρ
=4
.
09 Ω
/
km
.
The potential drop on the wire is
V
=
IR
=
I ρ ±
= (1180 A) (4
.
09 Ω
/
km) (222 km)
= 1071
.
42 kV
,
so the potential delivered to the city is
V
city
=
V
plant

V
wire
= 17000 kV

1071
.
42 kV
=
15928
.
6 kV
.
005 (part 2 of 4) 10.0 points
How much power is dissipated due to resistive
losses in the transmission line?
Correct answer: 1
.
26427
×
10
9
W.
Explanation:
P
=
I
2
R
=
I
2
ρ±
= (1180 A)
2
(4
.
09 Ω
/
km) (222 km)
=
1
.
26427
×
10
9
W
.
006 (part 3 of 4) 10.0 points
Assume the plant charges $ 0
.
117
/
kW
·
hr ±or
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: 1
.
4792
×
10
5
dollars
/
hr.
Explanation:
Let :
R
= $ 0
.
117
/
kW
·
hr
.
The cost o± energy is
Cost =
P
[(
R
) (time)]
= (1
.
26427
×
10
9
W)
×
(0
.
117
/
kW
·
hr) (1 hr)
±
1 kW
1000 W
²
=
1
.
4792
×
10
5
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 o± power; however, the output electric
potential o± the energy plant is 20% greater.
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This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.
 Spring '10
 Turner

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