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