Higher Engineering Mathematics

186 GEOMETRY AND TRIGONOMETRY

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(2 sinωtcosωt).

However, from double-angle formulae,

2 sinωtcosωt=sin 2ωt.

Thus power,p=−^12 VmImsin 2ωt.

The waveforms ofv,iandpare shown in Fig. 18.9.
The frequency of power is twice that of voltage and
current. For the power curve shown in Fig. 18.9, the
area above the horizontal axis is equal to the area

p i v + 0 −

π

ω

2π

ω t (seconds)

p

v

i

Figure 18.9

below, thus over a complete cycle the average power

Pis zero. It is noted that whenvandiare both posi-

tive, powerpis positive and energy is delivered from

the source to the inductance; whenvandihave oppo-

site signs, powerpis negative and energy is returned

from the inductance to the source.

In general, when the current through an induc-

tance is increasing, energy is transferred from the

circuit to the magnetic field, but this energy is

returned when the current is decreasing.

Summarizing,the average powerPin a purely

inductive a.c. circuit is zero.

(c) Purely capacitive a.c. circuits

Let a voltagev=Vmsinωtbe applied to a circuit

containing pure capacitance. The resulting current

isi=Imsin

(

ωt+π 2

)

, since current leads voltage by

90 ◦in a purely capacitive circuit, and the correspond-

ing 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

)

=cosωt

thus p=VmImsinωtcosωt