Compound angles 173
In Problems 7 and 8, solve forθin the range 0◦≤
θ≤ 180 ◦.
- cos6θ+cos2θ= 0
[22. 5 ◦, 45 ◦, 67. 5 ◦, 112. 5 ◦, 135 ◦, 157. 5 ◦] - sin3θ−sinθ= 0
[0◦, 45 ◦, 135 ◦, 180 ◦]
In Problems 9 and 10, solve in the range
0 ◦to 360◦. - cos2x=2sinx [21. 47 ◦or 158. 53 ◦]
- sin4t+sin2t= 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=Vmsinωt be applied to a cir-
cuit comprising resistance only. The resulting current
isi=Imsinωt, and the corresponding instantaneous
power,p, is given by:
p=vi=(Vmsinωt)(Imsinωt)
i.e. p=VmImsin^2 ωt
From double angle formulae of Section 17.3,
cos2A= 1 −2sin^2 A,from which,
sin^2 A=^12 ( 1 −cos2A)thus
sin^2 ωt=^12 ( 1 −cos2ωt)
Then power p=VmIm
[
1
2 (l−cos2ωt)
]
i.e. p=^12 VmIm(1−cos2ωt)
The waveforms ofv,iandpare shown in Fig. 17.8. The
waveform of power repeats itself afterπ/ωseconds and
hence the power has a frequency twice that of voltage
andcurrent.Thepowerisalwayspositive,havingamax-
imum value ofVmIm. The average or mean value of the
power is^12 VmIm.
The rms value of voltageV= 0. 707 Vm,i.e.V=
Vm
√
2
,
from which,Vm=
√
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=
Im
√
2
, from
which,Im=
√
2 I. Hence the average power,P,devel-
oped in a purely resistive a.c. circuit is given by
P=^12 VmIm=^12 (
√
2 V)(
√
2 I)=VIwatts.
Also, powerP=I^2 RorV^2 /Ras for a d.c. circuit,
sinceV=IR.
Summarizing, the average power P in a purely
resistive a.c. circuit given by
P=VI=I^2 R=
V^2
R
whereVandIare rms values.
(b) Purely inductive a.c. circuits
Let a voltagev=Vmsinωtbe applied to a circuit con-
tainingpure inductance (theoretical case). The resulting
current isi=Imsin
(
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=(Vmsinωt)Imsin
(
ωt−
π
2
)
i.e. p=VmImsinωtsin
(
ωt−
π
2
)
However,
sin
(
ωt−
π
2
)
=−cosωtthus
p=−VmImsinωtcosωt.
Rearranging gives:
p=−^12 VmIm(2sinωtcosωt).
However, from double-angle formulae,
2sinωtcosωt=sin2ωt.
Thus power,p=− 21 VmImsin2ωt.