Question

Consider a lossless DC series fed motor. (i.e. one where there is no resistance in either the field or the armature windings.) If this motor is operating at a constant speed under load, and the load suddenly changes so that the load torque drops by a factor of 3, by what factor will the speed of the motor change ? Support your conclusion with analysis based on the theory of this kind of motor

Answer 1

The speed of DC series motor will increase by Sqrt (3) times.

N2 = Sqrt (3) x N1

Step-by-step explanation:

Since, Torque of a DC series motor is given as:

T = K ∅ Ia²

Where:

K = Machine Constant

∅ = Flux in the machine

Ia = Armature Current

For Initial Condition:

T1 = K ∅ Ia1²--------------(1)

For Final Condition:

T2 = K ∅ Ia2² --------------(2)

Dividing Equations (1) and (2),

T1 / T2 = K ∅ Ia1² / K ∅ Ia2²

T1 / T2 = Ia1² / Ia2² -----------------(3)

Since, T2 = T1 / 3 (as given in question that torque drops by the factor of 3),

Then, equation 3 implies:

⇒ T1 / (T1/3) = Ia1² / Ia2²

⇒ 3 = Ia1² / Ia2²

⇒ Ia1 = Sqrt (3) x Ia2 ---------------- (4)

Since,

N2 / N1 = Ia1 / Ia2

Therefore,

N2 / N1 = [Sqrt (3) x Ia2] / Ia2

⇒ N2 / N1 = Sqrt (3)

⇒ N2 = Sqrt (3) x N1

Hence, the with the drop in load torque by the factor of 3, the speed of DC series motor will increase by Sqrt (3) times.