nguyen (jmn727) – oldmidterm 04 – Turner – (59070)
1
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printout
should
have
18
questions.
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before answering.
001
10.0 points
In a series
RLC
AC circuit, the resistance is
10 Ω, the inductance is 13 mH, and the capac
itance is 14
μ
F.
The maximum potential is
70 V, and the angular frequency is 100 rad
/
s.
Calculate the maximum current in the cir
cuit.
Correct answer: 0
.
098169 A.
Explanation:
Let :
R
= 10 Ω
,
L
= 13 mH = 0
.
013 H
,
C
= 14
μ
F = 1
.
4
×
10
−
5
F
,
V
max
= 70 V
,
and
ω
= 100 rad
/
s
.
The capacitive reactance is
X
C
=
1
ω C
=
1
(100 rad
/
s) (1
.
4
×
10
−
5
F)
= 714
.
286 Ω
and the inductive reactance is
X
L
=
ω L
= (100 rad
/
s) (0
.
013 H)
= 1
.
3 Ω
,
so the maximum current is
I
max
=
V
max
Z
=
V
max
radicalbig
R
2
+ (
X
L
−
X
C
)
2
=
70 V
radicalbig
(10 Ω)
2
+ (1
.
3 Ω
−
714
.
286 Ω)
2
=
0
.
098169 A
.
002
10.0 points
A certain capacitor in a circuit has a capaci
tive reactance of 36
.
3 Ω when the frequency is
110 Hz.
What capacitive reactance does the capac
itor have at a frequency of 5810 Hz?
Correct answer: 0
.
687263 Ω.
Explanation:
Let :
X
C
= 36
.
3 Ω
,
f
low
= 110 Hz
,
and
f
high
= 5810 Hz
.
The capacitive reactance is
X
C
=
C
2
π f
,
so
X
C
(
high
)
X
C
(
low
)
=
2
π f
low
C
2
π f
high
C
=
f
low
f
high
.
Thus
X
C
(
high
)
=
X
C
(
low
)
f
low
f
high
= (36
.
3 Ω)
parenleftbigg
110 Hz
5810 Hz
parenrightbigg
=
0
.
687263 Ω
.
003
10.0 points
A 71 mH inductor is connected to a outlet
where the
rms
voltage is 95
.
8 V and the fre
quency is 15 Hz.
Determine the energy stored in the inductor
at
t
= 7
.
6 ms, assuming that this energy is
zero at
t
= 0.
Correct answer: 6
.
27355 J.
Explanation:
Let :
t
= 7
.
6 ms = 0
.
0076 s
,
f
= 15 Hz
,
L
= 71 mH = 0
.
071 H
,
and
V
rms
= 95
.
8 V
.
The inductive reactance is
X
L
=
ω L
= 2
π f L
and the maximum current is
I
max
=
√
2
V
rms
X
L
=
√
2
V
rms
2
π f L
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nguyen (jmn727) – oldmidterm 04 – Turner – (59070)
2
The current at time
t
is
I
=
I
max
sin(
ω t
)
=
√
2
V
rms
2
π f L
sin(2
π f t
)
=
√
2 (95
.
8 V)
2
π
(15 Hz) (0
.
071 H)
×
sin[(2
π
(15 Hz) (0
.
0076 s)]
=
13
.
2936 A
,
so the potential energy stored in the inductor
is
U
=
1
2
L I
2
=
1
2
(0
.
071 H) (13
.
2936 A)
2
=
6
.
27355 J
.
004
10.0 points
A ideal transformer shown in the figure
below having a primary with 45 turns and
secondary with 36 turns.
The load resistor is 74 Ω.
The source voltage is 75 V
rms
.
75 V
rms
45 turns
36 turns
74 Ω
What is the
rms
electric potential across
the 74 Ω load resistor?
Correct answer: 60 V
rms
.
Explanation:
Let :
N
1
= 45 turns
,
N
2
= 36 turns
,
and
V
1
= 75 V
rms
.
The rms voltage across the transformer’s
secondary is
V
2
=
N
2
N
1
V
1
=
36 turns
45 turns
(75 V
rms
)
=
60 V
rms
,
which is the same as the electric potential
across the load resistor.
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 Spring '08
 Turner
 Resistance, Energy, Light

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