A motor control circuit has two sides: the power circuit, which carries the motor current through the contactor and protection, and the control circuit, which energises the contactor coil. The control circuit usually operates at a lower voltage and includes start and stop buttons, auxiliary contacts and protective interlocks.
Direct-on-line starting
In a direct-on-line starter the contactor applies full line voltage to the motor. Starting current is substantially higher than running current, so the supply, protection and mechanical system must tolerate the start. For larger motors, reduced-voltage starting or a variable-frequency drive limits the starting current and the mechanical shock.
Motor line current
Problem. A 7.5 kW three-phase motor has an efficiency of 0.90 and a power factor of 0.85 on a 400 V supply. Estimate its line current.
Solution. P_input = 7500 ÷ 0.90 = 8333 W. I = 8333 ÷ (√3 × 400 × 0.85) = 8333 ÷ 588.9 = 14.15 A, approximately 14.2 A.
Overload protection
An overload relay protects the motor against sustained overcurrent, not against a short circuit — the short-circuit protection is provided by a fuse or circuit breaker. The relay’s setting must match the motor’s rated current and the manufacturer’s instructions, and it must be coordinated with the motor’s starting profile.
- The contactor switches the motor circuit under the command of the control circuit.
- The control circuit is typically at a lower voltage and includes interlocks.
- An overload relay protects against sustained overcurrent, not short circuits.
- A variable-frequency drive controls speed and limits starting current.
- Local standards govern control-circuit voltages, colour codes and isolation requirements.
- A contactor switches the power circuit; the control circuit drives its coil.
- P_input = P_output/η and I = P_input/(√3 V cos φ).
- Overload relays protect against sustained overcurrent.
- VFDs and soft starters reduce starting stress.
Worked example
A 15 kW motor with efficiency 0.92 and power factor 0.88 at 400 V draws P_input = 15 000 ÷ 0.92 = 16 304 W and I = 16 304 ÷ (√3 × 400 × 0.88) = 26.7 A.