Let in a d.c. motor (See Fig. 4.3),

V = applied voltage
E_b = back e.m.f.
R_a = armature resistance
I_a = armature current

Since back e.m.f. Eb acts in opposition to the applied voltage V, the net voltage across the armature circuit is V- E_b. The
armature current I_a is given by;

I_a = (V – E_b)/ R_a

or V = E_{b +} I_aR_{a ……………………………..}(i)

This is known as voltage equation of the d.c. motor.

Power Equation

If Eq.(i) above is multiplied by I_a throughout, we get,

VI_a = E_bI_{a +}I²_aR_a

VI_a= electric power supplied to armature (armature input)
E_bI_a = power developed by armature (armature output)
I²_aR_a = electric power wasted in armature (armature Cu loss)

Thus out of the armature input, a small portion (about 5%) is wasted as a I²_aRa and the remaining portion E_bI_a is converted into mechanical power within the armature.

Condition For Maximum Power

The mechanical power developed by the motor is P_m= E_bI_a

Now P_m=VI_{a -}I²_aR_a

Since, V and R_a are fixed, power developed by the motor depends upon armature current. For maximum power, dP_m/dI_a should be zero.

dP_m/dI_{a =} V – 2I_aR_a

or I_aR_{a =} V/2

Now, V = E_{b +} I_aRa =E_{b +} V/2

therefore Eb=  V/2

Hence mechanical power developed by the motor is maximum when back e.m.f. is equal to half the applied voltage.

Limitations
In practice, we never aim at achieving maximum power due to the following reasons:
(i) The armature current under this condition is very large—much excess of rated current of the machine.
(ii) Half of the input power is wasted in the armature circuit. In fact, if we take into account other losses (iron and mechanical), the efficiency will be well below 50%.

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