Answer Posted / guest

When voltage is first applied to the motor field winding, only the conductor resistance opposes the flow of current to the motor winding. For example a motor winding rated at 240V with a winding resistance of 0.6 ohms. Using Ohms law we find I (current)= E/R or 240V/0.6 ohms = 400Amps.
DURING MOTOR STARTING = LOW RESISTANCE, HIGH CURRENT
Because the conductor resistance is so low, the motor has a very large inrush current.
Once the rotor begins turning, the rotor bars (winding) are increasingly cut by the stationary magnetic field, resulting in increasing counter electromotive force (CEMF).
CEMF opposes the applied voltage, resulting in an increased opposition to current flow within the conductor, which is also called inductive reactance.
The increase in inductive reactance, because of self induction, causes the impedance of the conductor winding to increase and this results to reduced current flow.
Same motor, R=0.6 ohm, XL=6 ohms, V=240 volts

Z = √(R^2+〖XL〗^2 ) = √(〖0.6〗^2+6^2 ) = 6 ohms
I = E/Z = 240V/6Ω = 40Amps

DURING MOTOR RUNNING = LOW RESISTANCE, HIGH INDUCTANCE, LOW CURRENT

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