83. When the switch is open, we have a series LRC circuit involving just the one capacitor near the upper
right corner. Eq. 33-65 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 2C. In this case, we
have
ω
d
L
−
1
2ω
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
ω
d
L
−
1
ω
d
C
2
=
E
m
1
ω
d
C
− ω
d
L
where we use the fact that
1
ω
d
C
> ω
d
L in simplifying the square root (this fact is evident from the
description of the first 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
=
120V
(2.00 A) tan 20.0
◦
= 165 Ω
C
=
I
2
2ω
d
E
m
1
−
tan φ
1
tan φ
o
=
2.00 A
2(2π)(60.0Hz)(120V)
1 +
tan 10.0
◦
tan 20.0
◦
= 1.49 × 10
−5
F
L
=
E
m
ω
d
I
2
1
− 2
tan φ
1
tan φ
o
=
120V
2π(60.0Hz)(2.00 A)
1 + 2
tan 10.0
◦
tan 20.0
◦
0.313 H