Start the insurance to start the input, and the operation exits the circuit

The starting current of a three-phase motor is very large, usually about 3 times the rated current of the motor. If the selected fuse (also called fuse) has a large rated current, it will be detrimental to protecting the motor. If the size is appropriate, the fuse will often burn out. This contradiction can be solved by using the circuit as shown in the figure. In the figure, FU1 is the motor main fuse of appropriate size; FU2 is the starting fuse. The rated current of FU1 should be equal to the rated current of the motor, and FU2 is generally equal to FU1. When the start button ST is pressed, the AC contactor 2KM is energized, and FU1 and FU2 are connected in parallel. At the same time, the time relay 1KT is energized. After a short period of time (less than 1 second), the normally open contact of 1KT is closed, 1KM is closed, and the motor M is started. When 2KM is energized, time relay 2KT is also energized at the same time. After 1 to 30 seconds (adjusted to motor M to start, during normal operation, check the stopwatch to confirm), the normally closed contact of 2KT is disconnected and 2KM is released. FU2 exits immediately. At the same time, time relays 1KT and 2KT lose power at the same time. The AC contactor is self-protected for 1KM, and the motor M runs normally until the stop button SSTP is pressed.