How many kW of motor can I drive with a 10 A relay output?
It would not be correct to give a single kW figure, because the ampere rating given for a relay contact generally applies to a resistive load condition. On inductive loads such as motors the starting current rises to roughly 10 times the running current, and the contact rating is calculated with a substantial reduction. With large motors the correct method is to drive a contactor coil with the relay.
The 2x10A relay output of the LF100R is, as stated in the product description, capable of providing forward/reverse direction control of small DC motors without a contactor. However, converting this figure directly into a motor power would be misleading.
The reasons:
- Load category: AC-1 is for resistive or lightly inductive loads with a power factor above 0.95; inductive control loads such as contactor coils fall into category AC-15.
- Starting current: on inductive AC loads a current of up to roughly 10 times the running current flows at the first instant.
- Derating: when using a contact for which only a resistive rating is given on an inductive load, common practice is to apply a substantial derating; some sources recommend using only 20-40 percent of the resistive rating.
A practical road map: read the motor's nameplate current, take the starting current into account, and then use the relay contact not directly but to drive a contactor coil. The contactor carries the motor current, a thermal overload relay protects against overload, and the relay contact switches only the few hundred milliamperes of coil current. This arrangement lasts longer and is cheaper to service.
The full topic: Wireless remote control installation: pairing, relays, safety