# DC motor at open load

Why at no load condition armature current is very small in DC series motor. Please explain me, what exactly no load mean.

Thank you.

• What have you learned in your studies so far? Did you learn about back EMF? Commented Sep 28, 2018 at 16:26
• I think a load is just a power consumer from output with high resistance.If it is no load then there must flow high current in the circuit,because there is no resistance Commented Sep 28, 2018 at 16:26
• @saisridatta Your thought is incorrect. Commented Sep 28, 2018 at 16:27
• @Transistor ...sir, what is link between back emf and current supplied to armature during initial stage.i dont understand ur idea Commented Sep 28, 2018 at 16:30
• For a motor, "no load" means that the motor is not driving anything - there is nothing connected to the motor output shaft. Commented Sep 28, 2018 at 17:42

Remember that a DC motor acts as a generator as well. When you spin the motor the generated voltage is in the opposite direction as the driving voltage.

• If the motor-generator is perfect - they never are - the generated voltage will be equal to the driving voltage. Due to losses it will always be a little less.

Let's say you have a 12 V, 1 Ω motor.

• If you connect it to a 12 V battery you can expect an initial current of $$\ \frac {V}{R} = \frac {12}{1} = 12 \ \text A \$$. The motor speed is zero so the generated voltage - also called the back EMF (electro-motive force) is zero.
• Let's say the motor speeds up to 1/4 of full speed. The back EMF will rise to 3 V and the current will fall to $$\ \frac {12 -3}{1} = 9 \ \text A \$$.
• If the motor speeds up to 1/2 of full speed. The back EMF will rise to 6 V and the current will fall to $$\ \frac {12 -6}{1} = 6 \ \text A \$$.
• If the motor speeds up to 3/4 of full speed. The back EMF will rise to 9 V and the current will fall to $$\ \frac {12 - 9}{1} = 3 \ \text A \$$.
• If the motor was perfect and could speed up to full speed the back EMF will rise to 12 V and the current would fall to $$\ \frac {12 - 12}{1} = 0 \ \text A \$$.

Real motors aren't perfect so the speed will settle down at the point where enough current flows to balance the losses and friction.

Power

The power drawn for each of the above situations is important. $$\ P = VI \$$.

• At stall we have $$\ P = 12 \times 12 = 144\ \text W \$$.
• At 1/4 speed the motor will consume $$\ 12 \times 9 = 108\ \text W \$$.
• At 1/2 speed the motor will consume $$\ 12 \times 6 = 72\ \text W \$$.
• At 3/4 speed the motor will consume $$\ 12 \times 3 = 36\ \text W \$$.
• At full speed (if possible or if driven) $$\ 12 \times 0 = 0 \ \text W\$$.

These figures are very important. The motor might be designed for 3 A continuous although it will take 12 A at start. This relies on the fact that it will take time for the temperature to rise enough to burn out the motor. If the rotor is locked (jammed, stalled, stuck or overloaded) our motor would draw 12 A, overheat and burn out. Some protection such as a slow-blow fuse is required.