# Starting of direct current Motor

what happens if starting arm of starter does not return to "OFF" position after a temporary supply failure?

• It starts when the supply comes back on ... – Solar Mike May 23 '19 at 7:38
• @SolarMike are you serious – user19926 May 28 '19 at 15:11
• The dc starter adds resistance to armature winding to limit starting current. As Mike says, motor starts with current depending on position of starting arm. – StainlessSteelRat May 28 '19 at 20:25
• @StainlessSteelRat that is the point if position of starting arm is stuck to its position after a temporary power supply failure what would happen. Resistance offered by starter would be zero, ... Right? – user19926 May 29 '19 at 4:06
• Yes. The purpose of the dc starter is to limit starting current by inserting a resistance. If starter was locked in on position, motor would start as a shunt (compound, etc) motor. – StainlessSteelRat May 29 '19 at 11:12

## 1 Answer

Schematic Diagram of a Manual Face Plate Starter is shown for a DC Shunt Motor.

From Quora: If the field from a DC motor is removed, what would happen? Original Source: Electrical Machines And Power Systems by Wildi. As arm moves from M to N, full voltage is supplied to field winding and electromagnet. Field current $$I_x$$ flows. Drawing is a little misleading, since as long as any contact is engaged, full source voltage is applied to field winding.

But starting current $$I$$ ($$I_a$$ - armature current), is limited by series connection of $$R_1$$, $$R_2$$, $$R_3$$ and $$R_4$$ into the armature winding. Contact arm conducts. As the arm moves through the contacts, each resistor is removed from the armature circuit until the full voltage is applied to the armature. Armature starting current is governed by: $$I_a = \frac {E_S - E_o}{R_a + R_1 + R_2 + R_3 + R_4}$$

At locked-rotor (LR), the Counter EMF $$E_o = 0$$. No rotation, means no induced EMF (Electromotive Force).

$$I_{a_{LR}} = \frac {E_S}{R_a + R_1 + R_2 + R_3 + R_4}$$

The resistors limit the LR current to 1.5 to 2 times rated current.

Factor in Counter EMF $$E_o \ne 0$$ as rotor starts to move and the removal of resistors as contact arm moves clockwise means armature current builds up in a controlled manner to rated full-load current (if driving full-load).

$$I_a = \frac {E_S - E_o}{R_a}$$

When the steel contact arm gets to the electromagnet, it is held in place until field loses power, for example when the power is disconnected. Then spring 3 should bring it back to contact M.

Answer time: If the arm does not return to starting position, then the armature current that flows is governed by where the arm is. The Starter is mechanical and things fail.

If it stays on the last position, then LR current is:

$$I_{a_{LR}} = \frac {E_S}{R_a}$$

This could be 5 to 8 times normal full-load current. The resistance of the armature winding $$R_a$$ is small, which could (from Wildi):

a. Burning out the armature;

b. Damaging the commutator and brushes, due to heavy sparking;

c. Overloading the feeder;

d. Snapping off the shaft due to mechanical shock;

e. Damaging the driven equipment because of the sudden mechanical hammerblow.

Now, if it stops at any other contact, say between $$R_3$$ and $$R_4$$, then resistor $$R_4$$ will limit some of the LR current.

$$I_{a_{LR}} = \frac {E_S}{R_a + R_4}$$

There still may be damage, but less than with no resistance.