Why does an induction motor draw 6 to 8 times rated current at start, and how much does star-delta help?

At standstill the motor is electrically a transformer with a short-circuited secondary. The rotating stator field sweeps past the stationary rotor bars at full synchronous speed (120 × f / poles = 1500 rpm here), induces a large voltage in them, and the rotor current is limited only by the winding resistances and leakage reactances. In the equivalent circuit the rotor branch resistance is R2 / s. At start the slip s is 1; at full load it is about 0.03, which makes that term some thirty times larger and brings the current down to its rated value.

The starting current is also at a poor power factor, typically 0.2 to 0.4, so seven times rated current produces only about one and a half to two and a half times rated torque.

Star-delta cuts the line current to one third. In star each winding gets 400 / √3 = 230 V instead of 400 V, so the winding current falls to 58 percent. On top of that, in star the line current equals the winding current, whereas in delta it is √3 times the winding current. The two effects multiply: 1 / 3. Your 7 × 29 = 203 A becomes about 68 A.

Torque goes with voltage squared, so it also falls to one third. If the motor gives 2.2 times rated torque direct-on-line, it gives about 0.73 times rated torque in star. That is fine for a fan or centrifugal pump, but a loaded conveyor or piston compressor may not accelerate at all.

Conditions and pitfalls for star-delta: the motor must be wound for delta at your line voltage (a nameplate reading 400/690 V, Δ/Y) and have all six winding ends brought out. A motor marked 230/400 V must never be run in delta on a 400 V supply.

The changeover timer should be set so the motor is close to full speed in star before switching, usually above 80 to 90 percent. If it switches early, the motor takes nearly the full direct-on-line current anyway at the transition, and the brief disconnection between star and delta contactors can cause a current spike and a torque jolt. If the load cannot reach that speed on one third torque, star-delta is the wrong method.

A soft starter reduces the voltage smoothly with thyristors. Current is proportional to voltage and torque to voltage squared, so limiting the current to 3 times rated instead of 7 gives (3 / 7)² = 18 percent of direct-on-line torque, about 0.4 times rated torque with the figures above. It avoids the transition jolt but obeys the same physics: less current always means much less torque.

A variable-frequency drive is different in kind. It starts at low frequency with voltage in proportion, so the motor always runs at small slip, and it can deliver around rated torque at around rated current from zero speed. It is the only one of the three that reduces current without sacrificing torque.

For the dipping lights, estimate the voltage drop from the starting kVA against the fault level of the supply. Direct-on-line, the motor takes √3 × 400 V × 203 A = 141 kVA. If it is fed from a 250 kVA transformer with 4 percent impedance, the fault level at the transformer terminals is 250 / 0.04 = 6250 kVA, and the dip there is roughly 141 / 6250 = 2.3 percent. The cable between transformer and motor adds its own drop, often more than the transformer does.

With star-delta the same sum gives under 1 percent. Also check the permitted starts per hour in the motor data: every start heats the rotor, and frequent starting is a thermal limit of its own.