Chapter 31
page 1
CHAPTER 31 – AC Circuits
1.
(
a
) The reactance of the capacitor is
X
C1
= 1/2p
f
1
C
= 1/2p(60 Hz)(7.2
×
10
–6
F) =
3.7
×
10
2
Ω
.
(
b
) For the new frequency we have
X
C2
= 1/2p
f
2
C
= 1/2p(1.0
×
10
6
Hz)(7.2
×
10
–6
F) =
2.2
×
10
–2
Ω
.
2.
We find the frequency from
X
L
=
ω
L
= 2p
fL
;
660
Ω
= 2p
f
(22.0
×
10
–3
H), which gives
f
= 4.77
×
10
3
Hz =
4.77 kHz
.
3.
We find the frequency from
X
C
= 1/2p
fC
;
6.70
×
10
3
Ω
= 1/2p
f
(2.40
×
10
–6
F), which gives
f
=
9.90 Hz
.
4.
The impedance is
Z
=
X
C
= 1/2p
fC
.
3.0
0
0
1000
f
(Hz)
X
C
(k
Ω
)
1.5
500
5.
The impedance is
Z
=
X
L
=
ω
L
= 2p
fL
.
f
(Hz)
10,000
5000
0
0
150
300
X
L
(
Ω
)

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