Patel, Kinal – Homework 17 – Due: Apr 4 2008, 7:00 pm – Inst: Weathers
1
This
printout
should
have
10
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
A transformer has input voltage and current
of 12 V and 5 A respectively, and an output
current of 0
.
7 A.
If there are 1365 turns turns on the sec
ondary side of the transformer, how many
turns are on the primary side?
Correct answer: 191
.
1 turns.
Explanation:
Let :
n
s
= 1365 turns
,
I
p
= 5 A
,
and
I
s
= 0
.
7 A
.
Energy is conserved, so
P
p
=
P
s
I
p
V
p
=
I
s
V
s
V
p
V
s
=
I
s
I
s
For the transformer
V
∝
n
n
p
n
s
=
V
p
V
s
=
I
s
I
p
n
=
n
s
I
s
I
p
= (1365 turns)
0
.
7 A
5 A
=
191
.
1 turns
.
keywords:
002
(part 1 of 2) 10 points
A stepup transformer is connected to a gen
erator that is delivering 133 V and 110 A.
The ratio of the turns on the secondary to the
turns on the primary is 1040 to 5.
What voltage is across the secondary?
Correct answer: 27
.
664 kV.
Explanation:
Let :
n
s
n
p
=
1040
5
= 1040
and
V
p
= 133 V
.
V
∝
n
so
V
s
V
p
=
n
s
n
p
The stepped up voltage in secondary is
V
s
=
V
p
·
n
s
n
p
= (133 V) (1040)
·
1 kV
1000 V
=
27
.
664 kV
.
003
(part 2 of 2) 10 points
What current flows in the secondary?
Correct answer: 528
.
846 mA.
Explanation:
Let :
I
p
= 110 A
.
Energy is conserved, so
P
p
=
P
s
I
s
V
s
=
I
p
V
p
I
s
=
I
p
V
p
V
s
=
(110 A) (133 V)
27664 V
·
10
3
mA
1 A
=
528
.
846 mA
.
keywords:
004
(part 1 of 1) 10 points
Consider an audio amplifier with an output
impedance of 663
Ω
and a speaker that has an
input impedance of 36
Ω
. A circuit replicating
this is shown in the figure below.
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 Spring '10
 Turner
 Alternating Current, Speed of light, Correct Answer, Input impedance, patel, Impedance matching

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