28.1
ALTERNATING-CURRENT CIRCUITS
28
E
XERCISES
Section 28.1 Alternating Current
14.
Use of Equations 28.1 and 28.2 allows us to write
rms
2
2(230 V)
325 V,
p
=
=
=
V
V
and
2
f
ω
π
=
=
1
2
(50 Hz)
314 s
.
π
−
=
Then the voltage expressed in the form of Equation 28.3 is
V t
1
( )
(325 V)sin[(314 s
) ].
t
−
=
15.
I
NTERPRET
We’re given the rms voltage and asked to find the peak voltage.
D
EVELOP
The rms voltage
V
and the peak voltage
rms
p
V
are related by Equation 28.1,
rms
/
2.
p
V
V
=
E
VALUATE
Equation 28.1 gives
rms
2
2(6.3 V)
8.91 V.
p
V
V
=
=
=
A
SSESS
The rms (root-mean-square) voltage is obtained by squaring the voltage and taking its time average, and
then taking the square root. Therefore, it is smaller than the peak voltage by a factor of
2.
16.
(a)
2(208 V)
294 V
p
V
(Equation 28.1), and
(b)
(Equation 28.2).
=
=
3
1
2
(400 Hz)
2.51
10
s
ω
π
−
=
=
×
17.
I
NTERPRET
We’re given the AC current in terms of sinusoidal function, and asked to deduce the rms current and
the frequency of the current.
D
EVELOP
As shown in Equation 28.3, the AC current can be written as
p
sin(
)
I
I
I
t
ω
φ
=
+
where
p
I
is the peak current amplitude,
ω
is the angular frequency, and
I
φ
is the phase constant. Comparison of the
current with Equation 28.3 shows that its amplitude and angular frequency are
p
495 mA
I
=
and
1
9.43 (ms)
.
ω
−
=
E
VALUATE
(a)
Applying Equation 28.1 gives
rms
p
/
2
495 mA/
2
350 mA.
I
I
=
=
=
(b)
Similarly, using Equation 28.2 we have
1
/2
9.43 ms
/2
1.50 kHz.
f
ω
π
π
−
=
=
=
A
SSESS
The phase
I
φ
is zero in this problem. Note that since the rms (root-mean-square) current is obtained by
squaring the current and taking its time average, and then taking the square root, it is smaller than the peak current
by a factor of
2.
18.
The phase constant is a solution of Equation 28.3 for
t
0,
=
i.e.,
V
V
(0)
sin(
).
p
V
φ
=
Since
sin(
)
sin(
),
V
V
φ
φ
π
=
−
±
one
must also consider the slope of the sinusoidal signal function at
t
0.
=
In addition, the conventional range for
V
φ
usually runs from
to
or
180
−
°
180 ,
+
°
.
V
π
φ
π
−
≤
≤
Thus,
when
but
1
sin
[
(0)/
]
V
p
V
V
φ
−
=
0
(
/
)
0,
dV dt
≥
V
φ
=
1
sin
[
(0)/
]
p
V
V
π
−
−
±
when
according as
V
(Here, we are taking
|s
0
(
/
)
0
dV dt
≤
(0)
0.
>
<
in
[
(0)/
]|
/2
p
V
V
1
π
−
≤
or
90
as
common on most electronic calculators, since the sine function is one-to-one only in such a restricted range.) For
signal
(a)
in Figure 28.25, we guess that
,
°
(0)
/ 2
p
V
(since that curve next crosses zero about halfway
between
V
≈
/2
π
and
)
π
and the slope at zero is positive, so
1
sin
(1/
2)
/4 or 45 .
a
φ
π
−
=
=
°
(This signal is
sin(
)
sin(
/4),
p
a
p
V
t
V
t
ω
φ
ω
+
=
+
π
=
which leads a signal with zero phase constant by
For the other signals,
(b)
V
so
45 .)
°
(0)
0,
0
(
/
)
0,
dV dt
>
0;
b
φ
=
(c)
V
0
(0)
, (
/
)
0,
p
V
dV dt
=
=
so
1
1
sin
(1)
sin
(1)
/2
c
φ
π
π
−
−
=
= −
+
=
or
(d)
V
so
90 ;
°
(0)
0,
=
0
(
/
)
0,
dV dt
<
d
φ
π
= ±
or
180 ;
±
°
and
(e)
V
0
(0)
, (
/
)
0,
p
V
dV dt
= −
=
=
so
1
sin
( 1)
e
φ
−
=
−
1
sin
( 1)
/2
π
π
−
−
−
−
= −
or
90 .
−
°
Section 28.2 Circuit Elements in AC Circuits
19.
I
NTERPRET
In this problem we want to find the rms current in a capacitor connected to an AC power source.

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