For a DC motor, gross mechanical power developed by the motor is maximum and the motor is supplied through a 220 V source. What will be back emf?

Power Equation of DC Motor:

For a DC Motor,

V = Eb + IR     ---(1)

Where V is the terminal voltage

Eb is back-EMF

I is current through armature in Shunt motor and current through both armature and field in series motor

R is armature resistance in Shunt motor and both armature and field resistance in series motor

Since back EMF Eb acts in opposition to the applied voltage V, the net voltage across the armature circuit is (V - Eb).

If Equation (1) above is multiplied by 'l' throughout, we get,

VI = IEb + I2R

Where,

VI is electric power supplied to the armature (armature input)

IEb (= Pm) power developed by armature (armature output) 

I2R is copper loss

Pm = VI - I2R

Since V and R are fixed, the power developed by the motor depends upon armature current.

For maximum power, \(\frac{dP_m}{dI}=0\)

or, \(\frac{dP_m}{dI}=V-2IR=0\)

or, V = 2IR

or, IR = V/2     ---(2)

From equation (1) & (2),

V = Eb + V/2

or, \(E_b =\frac{V}{2}\)

Hence, The gross mechanical power developed by the motor is maximum when the back EMF is equal to 50% (1/2) of the supply voltage.

Solution:

Given V = 220 V

E_b = V / 2 = 220 / 2 = 110 V