Higher Engineering Mathematics, Sixth Edition

Compound angles 173

In Problems 7 and 8, solve forθin the range 0◦≤

θ≤ 180 ◦.

7. cos6θ+cos2θ= 0
   [22. 5 ◦, 45 ◦, 67. 5 ◦, 112. 5 ◦, 135 ◦, 157. 5 ◦]


8. sin3θ−sinθ= 0
   [0◦, 45 ◦, 135 ◦, 180 ◦]
   In Problems 9 and 10, solve in the range
   0 ◦to 360◦.


9. cos2x=2sinx [21. 47 ◦or 158. 53 ◦]


10. 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.