ALTERNATINGCURRENT CIRCUITS
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
ϖ
π
=
=
1
2
(50 Hz)
314 s
.
π

=
Then the voltage expressed in the form of Equation 28.3 is
1
( )
(325 V)sin[(314 s
) ].
V t
t

=
15.
I
NTERPRET
We’re given the rms voltage and asked to find the peak voltage.
D
EVELOP
The rms voltage
rms
V
and the peak voltage
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 (rootmeansquare) 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)
3
1
2
(400 Hz)
2.51
10
s
ϖ
π

=
=
×
(Equation 28.2).
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
ϖ
φ
=
+
where
p
I
is the peak current amplitude,
ϖ
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)
.
ϖ

=
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
ϖ
π
π

=
=
=
A
SSESS
The phase
I
φ
is zero in this problem. Note that since the rms (rootmeansquare) 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
0,
t
=
i.e.,
(0)
sin(
).
p
V
V
V
φ
=
Since
sin(
)
sin(
),
V
V
φ
φ
π
=

±
one
must also consider the slope of the sinusoidal signal function at
0.
t
=
In addition, the conventional range for
V
φ
usually runs from
180

°
to
180 ,
+
°
or
.
V
π
φ
π

≤
≤
Thus,
1
sin
[
(0)/
]
V
p
V
V
φ

=
when
0
(
/
)
0,
dV dt
≥
but
V
φ
=
1
sin
[
(0)/
]
p
V
V
π


±
when
0
(
/
)
0
dV dt
≤
according as
(0)
0.
V
<
(Here, we are taking
1
sin
[
(0)/
]
/2
p
V
V
π

≤
or
90 ,
°
as
common on most electronic calculators, since the sine function is onetoone only in such a restricted range.) For
signal
(a)
in Figure 28.25, we guess that
(0)
/ 2
p
V
V
≈
(since that curve next crosses zero about halfway between
/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
ϖ
φ
ϖ
π
+
=
+
which leads a signal with zero phase constant by
45 .)
°
For the other signals,
(b)
(0)
0,
V
=
0
(
/
)
0,
dV dt
so
0;
b
φ
=
(c)
0
(0)
, (
/
)
0,
p
V
V
dV dt
=
=
so
1
1
sin
(1)
sin
(1)
/2
c
φ
π
π


=
= 
+
=
or
90 ;
°
(d)
(0)
0,
V
=
0
(
/
)
0,
dV dt
<
so
d
φ
π
= ±
or
180 ;
±
°
and
(e)
0
(0)
, (
/
)
0,
p
V
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.