174 Higher Engineering Mathematics
2
(^0) t (seconds)
v
p
i
p
i
v
1
2
Figure 17.9
The waveforms ofv,iandpare shown in Fig. 17.9.
The frequency of power is twice that of voltage and
current. For the power curve shown inFig. 17.9, the area
above the horizontal axis is equal to the area below, thus
over a complete cycle the average powerPis zero. It
is noted that whenvandiare both positive, powerpis
positive and energy is delivered from the source to the
inductance; whenvandihave opposite signs, powerp
is negative and energy is returned from the inductance
to the source.
In general, when the current through an inductance
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 powerP in a purely
inductive a.c. circuit is zero.
(c) Purely capacitive a.c. circuits
Let a voltagev=Vmsinωt be applied to a circuit
containing pure capacitance. The resulting current is
i=Imsin
(
ωt+π 2
)
, since current leads voltage by 90◦
in a purely capacitive 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
)
=cosωt
thus p=VmImsinωtcosωt
Rearranging givesp=^12 VmIm(2sinωtcosωt).
Thuspower,p=^12 VmImsin2ωt.
The waveforms ofv,iandpare shown in Fig. 17.10.
Over a complete cycle the average powerPis 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 powerP in 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–Ccircuit and negative for anR–Lcircuit.
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 differ-
ences of cosine functions as shown in Section 17.4,
i.e. sinAsinB=−^12 [cos(A+B)−cos(A−B)].