App-23
IM WT310E-01EN
Appendix
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App
Index
Power Factor Influence (Power Factor Error)
The previous example was for when the phase difference between the voltage and current signals was
0°, or in other words, when the power factor was 1. Next, we will calculate the error for an example in
which the power factor is not 1.
When the Power Factor is 0
This is an example for when the phase difference is 90°, or in other words, when the power factor is
0. Theoretically, the active power is 0 W, apparent power is 80 VA, and the reactive power is 80 var.
This assumes an ideal capacitive (C) load or an ideal inductive (L) load. For details, see appendix 2.
When the power factor (λ) = 0, the power error on this instrument is defined as follows:
±0.1% of S (S: apparent power) in the 45 Hz ≤ f ≤ 66 Hz range
When the measured apparent power is 80.00 [VA], the error in the measured power (0.00 W) is as
follows:
80.00 × ±0.1% = ±0.08 [W]
When the Power Factor Is Greater Than 0 but Less Than 1
As an example, let us calculate the error for when the power factor is 0.5, or in other words, when
the phase difference between the voltage and current (Φ) is 60°.
• Voltage measurement range: 150 V, measured voltage: 100.00 V
• Current measurement range: 1A, measured current: 0.800 A
• Power measurement range: 150 W, measured power: 40.00 W, measured apparent power:
80.00 VA, measured reactive power: 69.28 var
When 0 < λ < 1, the power error on this instrument is defined as follows:
(Power reading) × [(power reading error %) + (power range error %) × (power range/indicated
apparent power value) + {tan Φ × (influence when λ = 0)%}],
If we substitute the above value into this equation, the power error becomes as follows:
40.00 [W] × [ 0.1% + 0.05% × (150/80.00) + {tan60° × (influence (%) when λ = 0)}]
= 40.00 [W] × {0.1 + 0.05 × (150/80.00) + √3 × 0.1}%
= 0.1468 [W]
The error in the measured power (40.00 W) is ±0.1468 [W].
Appendix 5 Measurement Accuracy and Measurement Error
