Reducing the voltage supplied to a squirrel cage motor forces it to draw more current to maintain the same load. This is because power (P) is the product of voltage (V) and current (I): P = V x I. If V decreases, I must increase to keep P constant. The heat generated in the motor windings is proportional to the square of the current (I²R, where R is the resistance of the windings). Therefore, a significant increase in current due to reduced voltage leads to excessive heat generation. This overheating can damage the winding insulation, potentially causing a motor burnout.

The starting current of a squirrel cage induction motor is directly proportional to the supply voltage. Reducing the voltage proportionately reduces the starting current. This reduced starting current is beneficial because it minimizes stress on the motor windings and reduces voltage dips on the electrical distribution system. A lower power surge also prevents excessive power factor reduction. This gentler "cushion start" stabilizes line voltage. For example, a 50% voltage reduction results in approximately a 50% reduction in starting current.

The starting torque (Ta) of a squirrel cage induction motor is proportional to the square of the voltage (Ta ∝ V²). Therefore, a 50% voltage reduction results in only 25% of the normal starting torque. This can make it difficult or impossible to start motors driving high inertia loads. If the starting torque is insufficient to overcome the load torque, the motor will stall, leading to excessive current flow and potential damage to the windings.

When the voltage is reduced, the motor draws more current to try to maintain its speed under load. However, with a significant voltage reduction, the motor's speed will decrease. If the speed drops below a critical point (typically near the maximum torque point on the motor's torque-speed curve), the motor will lose synchronization and stall, resulting in a very low speed or complete stop.