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Example of power including reactive power
With DC the instantanesous values of voltage and current are
constant with respect to time, hence the power is constant.
In contrast to this the instantaneous value of power of AC or
AC + DC signals will fluctuate, its amplitude and polarity will
periodically change. If the phase angle is zero this is the special
case of pure active power which remains positive (exclusively
directed from source to load) at all times.
If there is a reactive component in the circuit there will be a
phase difference between voltage and current. The inductive
or capacitive element will store and release energy periodically
which creates an additional current component, the reactive
part. The product of voltage and current will therefore become
negative for portions of a period which means that energy will
flow back to the source.
Apparent power (unit VA)
The apparent power is equal to the product of voltage and cur-
rent. The apparent power is further equal to the geometric
sum of active and reactive power as shown in this diagram:
With the designations:
S
= apparent power
P
= active power
Q
= reactive power
V
= rms voltage
rms
I
= rms current
rms
the apparent power is derived:
2
2
S =
P
+ Q
= V
rms
Power factor
In general the power factor PF is derived:
P
PF = – – – –
S
PF
= power factor
S
= apparent power
P
= active power
In the very special case of sinusoidal voltage and
current the power factor equals
HINT
PF = cosϕ
x J
rms
B a s i c s o f P o w e r M e a s u r e m e n t
If e.g. the current is rectangular while the voltage is sinusoidal
the power factor will be P/S. Also in such case the reactive
power can be determined as demonstrated in the following
example:
û = 325,00 V
î = 12,25 A
How to calculate the power factor (example):
rms voltage is:
û
= — — = 229,8 V ≈ 230 V
V
rms
√
2
The rms current is given by:
1
2π
∫
2
I
=
î
· dϕ
––
rms
2π
0
π
2
î
π –
[(
)
(
J
=
––
·
––
+
rms
2π
3
2
2
J
=
î
î
· –– =
·
rms
3
2
I
= 12,25 A
·
–– = 10,00 A
rms
3
The apparent power S:
S = V
· I
= 230 V · 10,0 A = 2300 VA
rms
rms
The active power is derived from:
π
∫
1
û · î sin ϕ · dϕ = ––––
P = ––
π
π
3
û · î
[(
)
P
= ––––
– (-1)
– (-0,5)
π
1,5
P
= –––– · 325 V · 12,25 A = 1900 W
π
The power factor thus becomes:
P
1900 W
PF = ––– = ––––––––––– = 0,826
S
2300 VA
Obviously there is a reactive power component as the
apparent power exceeds the active power:
2
2
Q =
S
– P
= (2300 VA)
4π
)]
2π –
–––
3
2
––
3
π
û · î
– cos ϕ
[
]
π
π
3
1,5
]
=
–––– · û · î
π
2
2
– (1900 W)
= 1296 var
Subject to change without notice
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