App-22
IM WT310E-01EN
Example 2: Measuring 1 Arms Using the 5 A Measurement Range
In this example, we measure the same input using the 5 A range. When the measured value is 1.0000
[A], the reading error and range error are as follows:
• Reading error: 1.0000 [A]×0.1% = 0.001 [A]
• Range error: 5 [A]×0.05% = 0.0025 [A]
The error included in 1.0000 [A] is the sum of the reading and range errors, which is ±0.0035 [A]. This
corresponds to 0.35% of the measured value.
The error has increased even though the same current signal as example 1 was measured. As this
example illustrates, using a measurement range that is unnecessarily large for an input signal results
in larger measurement errors. It is import to measure using a measurement range that is appropriate
for the input signal.
Note
If the input signal is not a sine wave and includes distortions and spikes, select a somewhat large
measurement range that would not cause peak over-ranges to occur.
Example 3: Measuring 0.5 Arms Using the 1 A Measurement
Range
Next, we measure 0.5 A using the 1 A measurement range (the same as example 1). When the
measured value is 0.5000 [A], the reading error and range error are as follows:
• Reading error: 0.5000 [A]×0.1% = 0.0005 [A]
• Range error: 1 [A]×0.05% = 0.0005 [A]
The error included in 0.5000 [A] is the sum of the reading and range errors, which is ±0.001 [A]. This
corresponds to 0.2% of the measured value.
When we compare this result with that of example 1, we notice the following:
• The reading error has been reduced in accordance with the input amplitude.
• The range error has not changed.
As a result, the error is 0.2%, which is slightly larger than 0.15% of example 1. This is also because
the measurement range is large relative to the input signal. In this case, we should use the 0.5 A
measurement range.
Measurement Error of Active Power
On this instrument, the power accuracy in the range of 45 Hz to 66 Hz is ±(0.1% of r 0.05 %
of range).
Let us calculate the error for the following example.
• Voltage measurement range: 150 V, measured voltage: 100.00 V
• Current measurement range: 1A, measured current: 0.800 A
• Measured power: 80.00 W
• 60 Hz sine wave for both voltage and current
• Phase difference between the voltage and current signals = 0°
Power Range
The power measurement range is defined as voltage measurement range × current measurement
range. In this example, the power measurement range is 150 V×1 A = 150 W. We use this power
measurement range to calculate the range error.
The reading error and range error included in the measured power (80.00 W) are as follows:
• Reading error: 80.00 [W]×0.1% = 0.08 [W]
• Range error: 150 [W]×0.05% = 0.075 [W]
The error included in 80.00 [W] is the sum of the reading and range errors, which is ±0.155 [W]. This
corresponds to 0.19375% of the displayed value.
Appendix 5 Measurement Accuracy and Measurement Error
