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P q
s
The reason for rating some electrical equipment in kilovolt-amperes
rather than in kilowatts can be described using the configuration of Fig.
19.5. The load has an apparent power rating of 10 kVA and a current
rating of 50 A at the applied voltage, 200 V. As indicated, the current
demand of 70 A is above the rated value and could damage the load element,
yet the reading on the wattmeter is relatively low since the load
is highly reactive. In other words, the wattmeter reading is an indication
of the watts dissipated and may not reflect the magnitude of the current
drawn. Theoretically, if the load were purely reactive, the wattmeter
reading would be zero even if the load was being damaged by a high
current level.
I = 70 A > 50 A
+
200 V
–
P = VI cos θθ
0 Wattmeter 10
(kW)
±
I
±
V
19.4 INDUCTIVE CIRCUIT AND
REACTIVE POWER
[10 kVA = (200 V)(50 A)]
S = VI
Load
R
X L
(X L >> R)
FIG. 19.5
Demonstrating the reason for rating a load in kVA rather than kW.
For a purely inductive circuit (such as that in Fig. 19.6), v leads i by
90°, as shown in Fig. 19.7. Therefore, in Eq. (19.1), v � 90°. Substituting
v � 90° into Eq. (19.1) yields
pL � VI cos(90°)(1 � cos 2qt) � VI sin(90°)(sin 2qt)
� 0 � VI sin 2qt
Power
delivered to
element by
source
v
θ θ = 90°
Power
returned to
source by
element
Energy
absorbed
–VI
T 2
p L
i
Energy
returned
T 1
INDUCTIVE CIRCUIT AND REACTIVE POWER ⏐⏐⏐ 853
Energy
absorbed
FIG. 19.7
The power curve for a purely inductive load.
i
p L
+ v –
FIG. 19.6
Defining the power level for a purely inductive
load.
VI
Energy
returned
�t �