Compound angles 173
In Problems 7 and 8, solve for θ in the range 0 ◦ ≤
θ ≤ 180 ◦ .
7. cos 6θ + cos 2θ = 0
[22.5 ◦ , 45 ◦ , 67.5 ◦ , 112.5 ◦ , 135 ◦ , 157.5 ◦ ]
8. sin 3θ − sin θ = 0
[0 ◦ , 45 ◦ , 135 ◦ , 180 ◦ ]
In Problems 9 and 10, solve in the range
0
◦ to 360
◦ .
9. cos 2x = 2 sin x
[21.47 ◦ or 158.53 ◦ ]
10. sin 4t + sin 2t = 0
[0 ◦ , 60 ◦ , 90 ◦ , 120 ◦ , 180 ◦ , 240 ◦ ,
270 ◦ , 300 ◦ , 360 ◦ ]
17.6 Power waveforms in a.c. circuits
(a) Purely resistive a.c. circuits
Let a voltage v = V m sin ωt be applied to a circuit comprising resistance only. The resulting current
is i = I m sin ωt , and the corresponding instantaneous
power, p, is given by:
p = vi = (V m sin ωt )(I m sin ωt )
i.e. p = V m I m sin
2
ωt
From double angle formulae of Section 17.3,
cos 2 A = 1 − 2 sin
2 A, from which,
sin
2 A =
1
2 (1 − cos 2 A) thus
sin
2
ωt =
1
2 (1 − cos 2ωt )
Then power p = V m I m
1
2 (l − cos 2ωt )
i.e.
p =
1
2 V m I m (1 − cos 2ω t)
The waveforms of v, i and p are shown in Fig. 17.8. The
waveform of power repeats itself after π/ω seconds and
hence the power has a frequency twice that of voltage
and current. The power is always positive, having a maximum value of V m I m . The average or mean value of the
power is
1
2 V m I m .
The rms value of voltage V = 0.707V m , i.e. V =
V m
√
2
,
from which, V m =
√
2 V .
2
Maximum
power
Average
power
t (seconds)
v
i
p
0
p
i
v
1
2
Figure 17.8
Similarly, the rms value of current, I =
I m
√
2
, from
which, I m =
√
2 I . Hence the average power, P, developed in a purely resistive a.c. circuit is given by
P =
1
2 V m I m =
1
2 (
√
2V )(
√
2I ) = V I watts.
Also, power P = I 2 R or V 2 /R as for a d.c. circuit,
since V = I R.
Summarizing, the average power P in a purely
resistive a.c. circuit given by
P = V I = I
2 R =
V 2
R
where V and I are rms values.
(b) Purely inductive a.c. circuits
Let a voltage v = V m sin ωt be applied to a circuit containing pure inductance (theoretical case). The resulting
current is i = I m sin
ωt −
π
2
since current lags voltage
by
π
2
radians or 90 ◦ in a purely inductive circuit, and
the corresponding instantaneous power, p, is given by:
p = vi = (V m sin ωt )I m sin
ωt −
π
2
i.e. p = V m I m sin ωt sin
ωt −
π
2
However,
sin
ωt −
π
2
= −cos ωt thus
p = −V m I m sin ωt cos ωt.
Rearranging gives:
p = −
1
2 V m I m (2 sin ωt cosωt ).
However, from double-angle formulae,
2 sinωt cos ωt = sin 2ωt.
Thus power, p =−
1
2 V m I m sin 2ω t.
In Problems 7 and 8, solve for θ in the range 0 ◦ ≤
θ ≤ 180 ◦ .
7. cos 6θ + cos 2θ = 0
[22.5 ◦ , 45 ◦ , 67.5 ◦ , 112.5 ◦ , 135 ◦ , 157.5 ◦ ]
8. sin 3θ − sin θ = 0
[0 ◦ , 45 ◦ , 135 ◦ , 180 ◦ ]
In Problems 9 and 10, solve in the range
0
◦ to 360
◦ .
9. cos 2x = 2 sin x
[21.47 ◦ or 158.53 ◦ ]
10. sin 4t + sin 2t = 0
[0 ◦ , 60 ◦ , 90 ◦ , 120 ◦ , 180 ◦ , 240 ◦ ,
270 ◦ , 300 ◦ , 360 ◦ ]
17.6 Power waveforms in a.c. circuits
(a) Purely resistive a.c. circuits
Let a voltage v = V m sin ωt be applied to a circuit comprising resistance only. The resulting current
is i = I m sin ωt , and the corresponding instantaneous
power, p, is given by:
p = vi = (V m sin ωt )(I m sin ωt )
i.e. p = V m I m sin
2
ωt
From double angle formulae of Section 17.3,
cos 2 A = 1 − 2 sin
2 A, from which,
sin
2 A =
1
2 (1 − cos 2 A) thus
sin
2
ωt =
1
2 (1 − cos 2ωt )
Then power p = V m I m
1
2 (l − cos 2ωt )
i.e.
p =
1
2 V m I m (1 − cos 2ω t)
The waveforms of v, i and p are shown in Fig. 17.8. The
waveform of power repeats itself after π/ω seconds and
hence the power has a frequency twice that of voltage
and current. The power is always positive, having a maximum value of V m I m . The average or mean value of the
power is
1
2 V m I m .
The rms value of voltage V = 0.707V m , i.e. V =
V m
√
2
,
from which, V m =
√
2 V .
2
Maximum
power
Average
power
t (seconds)
v
i
p
0
p
i
v
1
2
Figure 17.8
Similarly, the rms value of current, I =
I m
√
2
, from
which, I m =
√
2 I . Hence the average power, P, developed in a purely resistive a.c. circuit is given by
P =
1
2 V m I m =
1
2 (
√
2V )(
√
2I ) = V I watts.
Also, power P = I 2 R or V 2 /R as for a d.c. circuit,
since V = I R.
Summarizing, the average power P in a purely
resistive a.c. circuit given by
P = V I = I
2 R =
V 2
R
where V and I are rms values.
(b) Purely inductive a.c. circuits
Let a voltage v = V m sin ωt be applied to a circuit containing pure inductance (theoretical case). The resulting
current is i = I m sin
ωt −
π
2
since current lags voltage
by
π
2
radians or 90 ◦ in a purely inductive circuit, and
the corresponding instantaneous power, p, is given by:
p = vi = (V m sin ωt )I m sin
ωt −
π
2
i.e. p = V m I m sin ωt sin
ωt −
π
2
However,
sin
ωt −
π
2
= −cos ωt thus
p = −V m I m sin ωt cos ωt.
Rearranging gives:
p = −
1
2 V m I m (2 sin ωt cosωt ).
However, from double-angle formulae,
2 sinωt cos ωt = sin 2ωt.
Thus power, p =−
1
2 V m I m sin 2ω t.