Why is the starting current high in a DC motor?

Question Posted / ramesh

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Why is the starting current high in a DC motor?..

Answer / bharat

In DC motors, Voltage equation is V=Eb-IaRa
V = Terminal voltage
Eb = Back emf in Motor
Ia = Armature current
Ra = Aramture resistance

At starting, Eb is zero. Therefore, V=IaRa
Ia = V/Ra where Ra is very less like 0.01ohm.
i.e, Ia will become enoromously increased.

Ref. Electrical Technology Auth : BL Theraja, Vol-I & II.

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Why is the starting current high in a DC motor?..

Answer / rama krishna

The simplest way to start a three-phase induction motor is
to connect its terminals to the line. In an induction motor,
the magnitude of the induced EMF in the rotor circuit is
proportional to the stator field and the slip speed (the
difference between synchronous and rotor speeds) of the
motor, and the rotor current depends on this EMF. When the
motor is started, the slip speed is equal to the synchronous
speed, as the rotor speed is zero (slip equal to 1), so the
induced EMF in the rotor is large. As a result, a very high
current flows through the rotor. This is similar to a
transformer with the secondary coil short circuited, which
causes the primary coil to draw a high current from the
mains. When an induction motor starts DOL, a very high
current is drawn by the stator, on the order of 5 to 9 times
the full load current. This high current can, in some
motors, damage the windings; in addition, because it causes
heavy line voltage drop, other appliances connected to the
same line may be affected by the voltage fluctuation. To
avoid such effects, several other strategies are employed
for starting motors.