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

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

N2 = Sqrt (3) x N1

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.

answer:

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