Specifications

Transducers
118
“Inherent active power” of active power transducers
* The primary-side active power (kW) mentioned here is not the full-load active power based on the VT-CT rating. It refers to the active power value kW (i.e.,
primary-side active power value corresponding to the rated output value) to be controlled according to the load state (e.g., light load). (equivalent to the scale of
the indicator)
Input/Output relationships
Inherent active power value calculation example
0
0
2.5
5
V
0
0.5
1
mA
500 1000W
0
0
3
5
V
0
14
12
20
mA
500 1000W
-1000
-5
0
5
V
-1
0
1
mA
0 1000W
-1000
0
3
5
V
0
14
12
20
mA
0 1000W
DC output
AC input
DC output
AC input
DC output
AC input
DC output
AC input
Connection diagrams (Refer to p.156 for outer dimensions.)
Please specify in the case of S Series.
In the case of a 3-phase, 4-wire unit,
please specify the phase
voltage and line voltage.
Although manufacturing will be performed even when specified in kW units,
the specifications will be converted to W units on the rating nameplate.
Model name
T-101HW
Frequency Number of units
3
Auxiliary power supply
110VAC
Phase-wire
Input
Voltage Current
3P3W 110V 5A
Output
Voltage or current
0–5V
Transducer inherent
active power value
1000W
VT
OUTPUT
Output
Auxiliary power
supply
MA MB
PP CC
Load Load
OUTPUT
Output
Auxiliary power
supply
P1
MA MB
P0 C1P2
2
0
1
CT
C2C1 C2
Load
VT
CT
OUTPUT
Output
Auxiliary power
supply
MA MB
P2P1 C1P3
3
2
1
CT
C3C1 C3
Load
VT
CT
OUTPUT
Output
Auxiliary power
supply
MA MB
P0P1 C1P3
3
2
0
1
C3
CT
CT
C2C1 C3C2
Fig. 1 T-101HW 1P2W
T-101SW 1P2W
Fig. 4 T-101HW 3P4W
T-101SW 3P4W
Fig. 3 T-101HW 3P3W
T-101SW 3P3W
T-101HW 1P3W
T-101SW 1P3W
Fig. 2
Note 1 Note 1 Note 1
An active power transducer can be manufactured if the transducer’s inherent active power
is within the range of the table on the left.
In the case of positive/negative bidirectional input, calculate using the larger of the positive or negative active powers.
In the case of a 3-phase, 3-wire, VT 6600/110V, CT 200/5A arrangement with the primary side power being 2000kW:
Transducer inherent active power P
0
= = = 0.833 (kW)
primary-side active power (kW)
VT ratio
✕ CT ratio
2000kW
6600/110
✕ 200/5
Note 1.
For low-voltage circuits, grounding of the secondary side of the instrument voltage transfomer and current transformer is unnecessary.
Ordering method
P
0
=
primary-side active power (kW)
VT ratio
✕ CT ratio