81.

(a) We find L from X

L

= ωL = 2πf L:

f =

X

L

2πL

=

1.30

× 10

3

Ω

2π(45.0

× 10

−3

H)

= 4.60

× 10

3

Hz .

(b) The capacitance is found from X

C

= (ωC)

−1

= (2πf C)

−1

:

C =

1

2πf X

C

=

1

2π(4.60

× 10

3

Hz)(1.30

× 10

3

Ω)

= 2.66

× 10

−8

F .

(c) Noting that X

L

∝ f and X

C

∝ f

−1

, we conclude that when f is doubled, X

L

doubles and X

C

reduces by half. Thus, X

L

= 2(1.30

×10

3

Ω) = 2.60

×10

3

Ω and X

C

= 1.30

×10

3

Ω/2 = 6.50

×10

2

Ω.