Higher Engineering Mathematics

COMPOUND ANGLES 187

B

p i v + 0 −

π

ω 2πω t (seconds)

p

v

i

Figure 18.10

Rearranging givesp=^12 VmIm(2 sinωtcosωt).

Thuspower,p=^12 VmImsin 2ωt.

The waveforms of v, i and p are shown in
Fig. 18.10. Over a complete cycle the average power
Pis zero. When the voltage across a capacitor is
increasing, energy is transferred from the circuit to
the electric field, but this energy is returned when the
voltage is decreasing.
Summarizing,the average powerPin a purely
capacitive a.c. circuit is zero.

(d)R–LorR–Ca.c. circuits

Let a voltage v=Vmsinωt be applied to a cir-
cuit containing resistance and inductance or resis-
tance and capacitance. Let the resulting current be
i=Imsin(ωt+φ), where phase angleφwill be posi-
tive for anR–C circuit and negative for anR–L
circuit. The corresponding instantaneous power,p,
is given by:

p=vi=(Vmsinωt)Imsin(ωt+φ)

i.e. p=VmImsinωtsin(ωt+φ)

Products of sine functions may be changed into
differences of cosine functions as shown in Sec-
tion 18.4,
i.e. sinAsinB=−^12 [cos(A+B)−cos(A−B)].

Substitutingωt=Aand (ωt+φ)=Bgives:

power, p=VmIm{−^12 [cos(ωt+ωt+φ)

−cos(ωt−(ωt+φ))]}

i.e. p=^12 VmIm[cos(−φ)−cos(2ωt+φ)]

However, cos(−φ)=cosφ

Thusp=^12 VmIm[cosφ−cos(2ωt+φ)]

The instantaneous powerpthus consists of

(i) a sinusoidal term,−^12 VmImcos (2ωt+φ) which

has a mean value over a cycle of zero, and

(ii) a constant term,^12 VmImcosφ(sinceφis constant

for a particular circuit).

Thus the average value of power,P=^12 VmImcosφ.

SinceVm=

√

2 VandIm=

√

2 I, average power,

P=^12 (

√

2 V)(

√

2 I) cosφ

i.e. P=VIcosφ

The waveforms ofv,iandp, are shown in Fig. 18.11

for anR–Lcircuit. The waveform of power is seen to

pulsate at twice the supply frequency. The areas of