This preview shows page 1. Sign up to view the full content.
83. When the switch is open, we have a series
LRC
circuit involving just the one capacitor near the upper
right corner. Eq. 3365 leads to
ω
d
L
−
1
ω
d
C
R
=tan
φ
o
=tan(
−
20
◦
)=
−
tan 20
◦
.
Now, when the switch is in position 1, the equivalent capacitance in the circuit is 2
C
.Inth
i
sc
a
s
e
,w
e
have
ω
d
L
−
1
2
ω
d
C
R
=tan
φ
1
= tan 10
.
0
◦
.
Finally, with the switch in position 2, the circuit is simply an
LC
circuit with current amplitude
I
2
=
E
m
Z
LC
=
E
m
r
³
ω
d
L
−
1
ω
d
C
´
2
=
E
m
1
ω
d
C
−
ω
d
L
whe
rew
This is the end of the preview. Sign up
to
access the rest of the document.
Unformatted text preview: description of the ±rst situation, when the switch was open). We solve for L , R and C from the three equations above: R = −E m I 2 tan φ o = 120 V (2 . 00 A) tan 20 . ◦ = 165 Ω C = I 2 2 ω d E m ³ 1 − tan φ 1 tan φ o ´ = 2 . 00 A 2(2 π )(60 . 0 Hz)(120 V) ( 1 + tan 10 . ◦ tan 20 . ◦ ) = 1 . 49 × 10 − 5 F L = E m ω d I 2 µ 1 − 2 tan φ 1 tan φ o ¶ = 120 V 2 π (60 . 0 Hz)(2 . 00 A) µ 1 + 2 tan 10 . ◦ tan 20 . ◦ ¶ . 313 H...
View
Full
Document
This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Capacitance

Click to edit the document details