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Physics 21
Fall, 2007
Solution, Hour Exam #2
The graders for the problems were:
1 McGeehan, 2 Belony, 3 Sweeney, 4 Lyu, 5 Kanofsky
For questions about the grading, see the grader by Nov. 20.
Problem 1.
For the following circuit,
R
=5
.
0Ω
,
L
=
40
.
0mH
,and
C
= 300 nF, and the frequency is 1000 Hz.
R
L
C
V
rms
5
I
rms
530.5
I
rms
251.3
I
rms
5
I
rms
279.2
I
rms
V
rms
φ
(a)
(b)
(a) (5 pts.) Draw a phasor diagram for the voltages in this
circuit. What is the rms value of the current if
V
rms
=
6V?
The phasor diagram is shown in panel (a). The quanti
ties are determined from
V
R
=
RI
rms
=5
I
rms
V
L
=
ωLI
rms
=2
π
×
1000
×
0
.
040 = 251
.
3
I
rms
V
C
=
I
rms
ωC
=
I
rms
2
π
×
1000
×
300
×
10
−
9
= 530
.
5
I
rms
Then add
V
L
+
V
C
[panel (b)] and solve for
V
rms
=
p
5
2
+ 279
.
2
2
I
rms
= 279
.
2
I
rms
I
rms
=
V
rms
279
.
2
=21
.
5mA
(b) (5 pts.)
What is the phase relationship between the
current and the applied ac voltage?
In other words,
does the current lag or lead the applied voltage?
by
how much?
tan
φ
=
−
279
.
2
5
=
−
55
.
8
⇒
φ
=
−
88
.
97
◦
The current leads the voltage by almost 90
◦
.
(c) (5 pts.) What is the power delivered to the circuit by
the power supply?
P
av
=
V
rms
I
rms
cos
φ
=6(
.
0215)(
.
0179) = 2
.
3
×
10
−
3
W
(d) (5 pts.) At what frequency would the rms current be
the maximum for the given
V
rms
? (It is not necessary
to derive this result; just show the expression that you
use to calculate it.)
ω
=
1
√
LC
=
1
p
.
040(300
×
10
−
9
)
=9
.
13
×
10
3
rad
/
s
Problem 2.
Consider a straight segment of wire of length
a
carrying a current
I
. (The rest of the wire need not be
considered and is not shown.)
P
=(
x,
0,0)
(0,

a
/2,0)
(0
,+a
/2,0)
x
y
I
d
l
r

r'
r'
r
(a) (5 pts.)
Use the BiotSavart law to write an expres
sion for the magnetic ±eld
d
B
at the point
P
on the
x
axis due to the current
I
in the in±nitesimal segment of
wire
d
l
located at (0
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 Spring '08
 Hickman
 Physics

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