Question

The gross mechanical power developed by the motor is maximum when the back EMF is equal to _______ of the supply voltage.
1. twice
2. \(\frac 2 3\)
3. \(\frac 1 3\)
4. \(\frac 1 2\)

Answer 1

Correct Answer - Option 4 : \(\frac 1 2\)

Power Equation of DC Motor:

For a DC Motor,

V = E_b + IR     ---(1)

Where V is the terminal voltage

E_b 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 E_b acts in opposition to the applied voltage V, the net voltage across the armature circuit is (V - E_b).

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

VI = IE_b + I²R

Where,

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

IE_b (= P_m) power developed by armature (armature output) 

I²R is copper loss

P_m = 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 = E_b + 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.