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856 ⏐⏐⏐ POWER (ac)
i
+ v –
p C C
FIG. 19.8
Defining the power level for a purely
capacitive load.
Power
delivered to
element by
source
θ θ = –90°
Power
returned to
source by
element
i
Energy
returned
VI
T 2
19.5 CAPACITIVE CIRCUIT
P q
s
For a purely capacitive circuit (such as that in Fig. 19.8), i leads v by
90°, as shown in Fig. 19.9. Therefore, in Eq. (19.1), v ��90°. Substituting
v ��90° into Eq. (19.1), we obtain
pC � VI cos(�90°)(1 � cos 2qt) � VI sin(�90°)(sin 2qt)
� 0 � VI sin 2qt
or pC ��VI sin 2qt
(19.19)
where �VI sin 2qt is a negative sine wave with twice the frequency of
either input (v or i) and a peak value of VI. Again, note the absence of
an average or constant term.
Energy
absorbed
–VI
T 1
p C
v
Energy
returned
Energy
absorbed
FIG. 19.9
The power curve for a purely capacitive load.
Plotting the waveform for pC (Fig. 19.9) gives us
T1 � period of either input quantity
T2 � period of pC curve
Note that the same situation exists here for the pC curve as existed for the
pL curve. The power delivered by the source to the capacitor is exactly
equal to that returned to the source by the capacitor over one full cycle.
The net flow of power to the pure (ideal) capacitor is zero over a full
cycle,
and no energy is lost in the transaction. The power absorbed or returned
by the capacitor at any instant of time t1 can be found by substituting t1 into Eq. (19.19).
The reactive power associated with the capacitor is equal to the peak
value of the pC curve, as follows:
Q C � VI
�t �
(VAR) (19.20)
But, since V � IX C and I � V/X C, the reactive power to the capacitor
can also be written
Q C � I 2 X C
(VAR) (19.21)