terry (ect328) – oldhomework 30 – Turner – (59130)
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001
10.0 points
A transformer has input voltage and current
of 16 V and 5 A respectively, and an output
current of 0
.
7 A.
If there are 1397 turns turns on the sec
ondary side of the transformer, how many
turns are on the primary side?
Correct answer: 195
.
58 turns.
Explanation:
Let :
n
s
= 1397 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
= (1397 turns)
0
.
7 A
5 A
=
195
.
58 turns
.
002
10.0 points
An ideal transformer shown in the figure
below has a primary with 20 turns and sec
ondary with 8 turns.
The load resistor is 72 Ω.
The source voltage is 98
.
7 V
rms
.
98
.
7 V
rms
R
S
20 turns
8 turns
72 Ω
If a voltmeter across the load measures
18
.
3 V
rms
, what is the source resistance
R
S
?
Correct answer: 520
.
82 Ω.
Explanation:
Let :
R
S
= Source Resistor
,
R
L
= 72 Ω
,
N
1
= 20 turns
,
N
2
= 8 turns
,
N
1
N
2
=
5
2
,
and
E
= 98
.
7 V
rms
.
The rms voltage across the transformer pri
mary is
V
1
=
N
1
N
2
V
2
,
(1)
using
V
1
from Eq. 1, the source voltage is
V
S
=
V
R
S
+
V
1
=
I
1
R
S
+
N
1
N
2
V
2
.
(2)
The secondary current is
I
2
=
V
2
R
L
.
(3)
The primary current, in terms of the sec
ondary current, is given by
I
1
=
I
2
N
2
N
1
.
(4)
Substituting the expression for
I
2
from Eq.
3 into the expression for
I
1
from Eq. 4, we
obtain
I
1
=
V
2
R
L
N
2
N
1
.
(5)
Substituting the expression
I
1
from Eq. 5 into
the the expression for
V
S
from Eq. 2, we obtain
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 Fall '09
 Turner
 Alternating Current, Trigraph, Inductor, Electrical resistance

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