**7.6. ****Mechanical Power Developed ****by ****Motor Armature **

Refer Fig. 56. The voltage V applied across the motor armature has to (i) overcome back e.m.f E_{b} and (ii) supply the armature ohmic drop I_{a}R_{a}

V = E_{b} + I_{a}R_{a}

This is known as voltage equation of a motor.

Multiplying both sides by I_{a}, we get

VI_{a} = E_{b}I_{a} + I_{a}^{2}R_{a}

Here VI_{a} = electrical input to the armature,

E_{b}I_{a} = electrical equivalent of mechanical power P_{m} developed in the armature, and

I_{a}^{2}R_{a} = copper loss in the armature.

The power available at the pulley for doing useful work is somewhat less than the mechanical power developed by the armature.

This is evident, since there are certain mechanical/asses (such as bearing and windage field friction and iron losses) that must be supplied by the driving power of the motor.

**Condition for maximum power**. We know that, the mechanical power developed by the motor,

P_{m} = VI_{a} – I_{a}^{2} R_{a}

Hence, mechanical power developed by a motor is maximum when back e.m.f. is equal to half the applied voltage. In practice, however, this condition is not realized because in that case current would be much beyond the normal current of the motor. Moreover, half the input would be wasted in the form of heat and taking other losses, such as mechanical and magnetic, into consideration, the efficiency of the motor will be below 50 per cent.

## Recent Comments